ThreeDimensionalGeometry
Question1
Iftheshortestdistancebetweenthelines x−4
1=y+1
2=z
−3and
x−λ
2=y+1
4=z−2
−5is 6
√5,thenthesumofallpossiblevaluesofλis:
[27-Jan-2024Shift1]
Options:
A.5
B.8
C.7
D.10
Answer:B
Solution:
Question2
Thedistance,ofthepoint(7, −2,11)fromtheline x−6
1=y−4
0=z−8
3
alongtheline x−5
2=y−1
−3=z−5
6,is:
[27-Jan-2024Shift1]
Options:
A.12
B.14
C.18
D.21
Answer:B
Solution:
Question3
ThepositionvectorsoftheverticesA,BandCofatriangleare
2^
i−3^
j+3^
k,2^
i+2^
j+3^
kand−^
i+^
j+3^
krespectively.Letl denotes
thelengthoftheanglebisectorADof∠BACwhereDisontheline
segmentBC,then2l 2equals:
[27-Jan-2024Shift2]
Options:
A.49
B.42
C.50
D.45
Answer:D
Solution:
Question4
LetthepositionvectorsoftheverticesA,BandCofatrianglebe
2^
i+2^
j+^
k,^
i+2^
j+2^
kand2^
i+^
j+2^
krespectively.Letl 1,l2andl 3be
thelengthsofperpendicularsdrawnfromtheorthocenterofthe
triangleonthesidesAB,BCandCArespectively,thenl 1
2+l2
2+l3
2
equals:
[27-Jan-2024Shift2]
Options:
A. 1
5
B. 1
2
C. 1
4
D. 1
3
Answer:B
Solution:
Question5
Lettheimageofthepoint(1,0,7)intheline x
1=y−1
2=z−2
3bethepoint
(α,β,γ).Thenwhichoneofthefollowingpointsliesonthelinepassing
through(α,β,γ)andmakingangles 2π
3and 3π
4withy-axisandz-axis
respectivelyandanacuteanglewithx-axis?
[27-Jan-2024Shift2]
Options:
A.(1, −2,1+ √2)
B.(1,2,1− √2)
C.(3,4,3−2√2)
D.(3, −4,3+2√2)
Answer:C
Solution:
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Question6
Thelines x−2
2=y
−2=z−7
16 and x+3
4=y+2
3=z+2
1intersectatthepointP.If
thedistanceofPfromtheline x+1
2=y−1
3=z−1
1isl ,then14l 2isequal
to.................
[27-Jan-2024Shift2]
Answer:108
Solution:
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Question7
LetObetheoriginandthepositionvectorofAandBbe2^
i+2^
j+^
kand
2^
i+4^
j+4^
krespectively.Iftheinternalbisectorof∠AOBmeetstheline
ABatC,thenthelengthofOCis
[29-Jan-2024Shift1]
Options:
A. 2
3√31
B. 2
3√34
C. 3
4√34
D. 3
2√31
Answer:B
Solution:
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Question8
LetPQRbeatrianglewithR(−1,4,2).SupposeM(2,1,2)isthemid
pointofPQ.Thedistanceofthecentroidof△PQRfromthepointof
intersectionoftheline x−2
0=y
2=z+3
−1and x−1
1=y+3
−3=z+1
1is
[29-Jan-2024Shift1]
Options:
A.69
B.9
C.√69
D.√99
Answer:C
Solution:
Solution:
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Question9
Alinewithdirectionratios2,1,2meetsthelinesx =y+2=zand
x+2=2y =2zrespectivelyatthepointPandQ.ifthelengthofthe
perpendicularfromthepoint(1,2,12)tothelinePQisl ,thenl 2is
[29-Jan-2024Shift1]
Answer:65
Solution:
Solution:
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Question10
LetP(3,2,3), Q(4,6,2)andR(7,3,2)betheverticesof△PQR.Then,
theangle∠QPRis
[29-Jan-2024Shift2]
Options:
A. π
6
B.cos−17
18
C.cos−11
18
D. π
3
Answer:D
( )
( )
Solution:
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Question11
LetObetheorigin,andMandNbethepointsonthelines
x−5
4=y−4
1=z−5
3and x+8
12 =y+2
5=z+11
9respectivelysuchthatMNisthe
shortestdistancebetweenthegivenlines.Then →
OM ⋅→
ONisequalto
[29-Jan-2024Shift2]
Answer:9
Solution:
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Question12
Let(α,β,γ)bethefootofperpendicularfromthepoint(1,2,3)onthe
line x+3
5=y−1
2=z+4
3.then19(α+β+γ)isequalto:
[30-Jan-2024Shift1]
Options:
A.102
B.101
C.99
D.100
Answer:B
Solution:
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Question13
Ifd 1istheshortestdistancebetweenthelines
x+1=2y = −12z,x=y+2=6z −6andd 2istheshortestdistance
betweenthelines x−1
2=y+8
−7=z−4
5,x−1
2=y−2
1=z−6
−3,thenthevalueof
32√3d1
d2is:__
[30-Jan-2024Shift1]
Answer:16
Solution:
Question14
LetL1:→
r=^
i−^
j+2^
k+λ^
i−^
j+2^
k,λ∈R
L2:→
r=^
j−^
k+µ 3^
i+^
j+p^
k,µ∈Rand
L3:→
r=δℓ^
i+m hat j +n^
kδ∈R
BethreelinessuchthatL1isperpendiculartoL2andL3is
perpendiculartobothL1andL2.ThenthepointwhichliesonL3is
[30-Jan-2024Shift2]
Options:
A.(−1,7,4)
B.(−1, −7,4)
C.(1,7, −4)
D.(1, −7,4)
Answer:A
Solution:
( ) ( )
( ) ( )
( )
Question15
Letalinepassingthroughthepoint(−1,2,3)intersectthelines
L1:x−1
3=y−2
2=z+1
−2atM(α,β,γ)andL2:x+2
−3=y−2
−2=z−1
4atN(a,b,c).
Thenthevalueof (α+β+γ)2
(a+b+c)2equals
[30-Jan-2024Shift2]
Answer:196
Solution:
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Question16
ThedistanceofthepointQ(0,2, −2)formthelinepassingthroughthe
pointP(5, −4,3)andperpendiculartothelines→
r= −3^
i+2^
k+
λ 2^
i+3^
j+5^
k,λ∈ ℝ and →
r=^
i−2^
j+^
k+µ −^
i+3^
j+2^
k,µ∈ ℝis
[31-Jan-2024Shift1]
Options:
A.√86
B.√20
C.√54
D.√74
Answer:D
Solution:
( )
( ) ( ) ( )
Solution:
Avectorinthedirectionoftherequiredlinecanbeobtainedbycrossproductof
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Question17
LetQandRbethefeetofperpendicularsfromthepointP(a,a,a)on
thelinesx =y,z=1andx = −y,z = −1respectively.If∠QPRisaright
angle,then12a2isequalto____
[31-Jan-2024Shift1]
Answer:12
Solution:
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Question18
Let(α,β,γ)bemirrorimageofthepoint(2,3,5)intheline
x−1
2−y−2
3−z−3
4.
Then2α +3β +4γisequalto
[31-Jan-2024Shift2]
Options:
A.32
B.33
C.31
D.34
Answer:B
Solution:
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Question19
TheshortestdistancebetweenlinesL1andL2,where
L1:x−1
2=y+1
−3=z+4
2andL2isthelinepassingthroughthepoints
A(−4,4,3) ⋅ B(−1,6,3)andperpendiculartotheline x−3
−2=y
3=z−1
1,is
[31-Jan-2024Shift2]
Options:
A. 121
√221
B. 24
√117
C. 141
√221
D. 42
√117
Answer:C
Solution:
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Question20
AlinepassesthroughA(4, −6, −2)andB(16, −2,4).ThepointP(a,b,c)
wherea,b,carenon-negativeintegers,onthelineABliesatadistance
of21units,fromthepointA.ThedistancebetweenthepointsP(a,b,c)
andQ(4, −12,3)isequalto
[31-Jan-2024Shift2]
Answer:22
Solution:
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Question21
Iftheshortestdistancebetweenthelines x−λ
−2=y−2
1=z−1
1and
x− √3
1=y−1
−2=z−2
1is1,thenthesumofallpossiblevaluesofλis:____
[1-Feb-2024Shift1]
Options:
A.0
B.2√3
C.3√3
D.−2√3
Answer:B
Solution:
Solution:
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Question22
Letthelineoftheshortestdistancebetweenthelines
L1:→
r=^
i+2^
j+3^
k+λ^
i−^
j+^
kand
L2:→
r=4^
i+5^
j+6^
k+µ^
i+^
j−^
k
intersectL1andL2atPandQrespectively.If(α,β,γ)isthemidpointof
thelinesegmentPQ,then2(α+β+γ)isequalto____
[1-Feb-2024Shift1]
( ) ( )
( ) ( )
Answer:21
Solution:
Solution:
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Question23
LetPandQbethepointsontheline x+3
8=y−4
2=z+1
2whichareata
distanceof6unitsfromthepointR(1,2,3).Ifthecentroidofthe
trianglePQRis(α,β,γ),thenα2+β2+γ2is:
[1-Feb-2024Shift2]
Options:
A.26
B.36
C.18
D.24
Answer:C
Solution:
Solution:
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Question24
IfthemirrorimageofthepointP(3,4,9)intheline x−1
3=y+1
2=z−2
1is
(α,β,γ),then14(α+β+γ)is:
[1-Feb-2024Shift2]
Options:
A.102
B.138
C.108
D.132
Answer:C
Solution:
Solution:
P(8λ −3,2λ +4,2λ −1)
PR =6
(8λ −4)2+ (2λ +2)2+ (2λ −4)2=36
λ=0,1
Hence P(−3,4, −1)&Q(5,6,1)
Centroidof ∆PQR = (1,4,1)≡(α,β,γ)
α2+β2+γ2=18
→
PN ⋅→
b=0?
3(3λ −2) + 2(2λ −5) + (λ−7) = 0
14λ =23 ⇒λ=23
14
N83
14,32
14,51
14
∴α+3
2=83
14 ⇒α=62
7
β+4
2=32
14 ⇒β=4
7
( )
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Question25
Thedistanceofthepoint(−1,9, −16)fromtheplane2x +3y −z=5
measuredparalleltotheline x+4
3=2−y
4=z−3
12 is
[24-Jan-2023Shift1]
Options:
A.13√2
B.31
C.26
D.20√2
Answer:C
Solution:
Solution:
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Question26
Thedistanceofthepoint(7, −3, −4)fromtheplanepassingthrough
thepoints(2, −3,1), (−1,1, −2)and(3, −4,2)is:
[24-Jan-2023Shift1]
Options:
A.4
B.5
C.5√2
D.4√2
Answer:C
Solution:
γ+9
2=51
14 ⇒γ=−12
7
Ans.14(α+β+r) = 108
Equationofline
x+1
3=y−9
−4=z+16
12
G.Ponline(3λ −1, −4λ +9,12λ −16)
pointofintersectionofline&plane
6λ −2−12λ +27 −12λ +16 =5
λ=2
Point(5,1,8)
Distance = √36 +64 +576 =26
Solution:
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Question27
Theshortestdistancebetweenthelines x−2
3=y+1
2=z−6
2and
x−6
3=1−y
2=z+8
0isequalto
[24-Jan-2023Shift1]
Answer:14
Solution:
Solution:
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Question28
Ifthefootoftheperpendiculardrawnfrom(1,9,7)tothelinepassing
throughthepoint(3,2,1)andparalleltotheplanesx +2y +z=0and
3y −z=3is(α,β,γ),thenα +β+γisequalto
[24-Jan-2023Shift2]
Options:
A.−1
B.3
C.1
EquationofPlaneis
=
x−2 y +3 z −1
−3 4 −3
4−5 4
=0
x−z−1=0
DistanceofP(7, −3, −4)fromPlaneis
d=7+4−1
√2=5√2
| |
| |
=
4 2 −14
3 2 2
3−20
^
i^
jk
3 2 2
3−20
= 16 +12 +168
−4^
i+6^
j−12k
=196
14 =14
| |
| |
| |
D.5
Answer:D
Solution:
Solution:
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Question29
Lettheplanecontainingthelineofintersectionoftheplanes
P1 :x+ (λ+4)y+z=1and
P2 :2x +y+z=2passthroughthepoints(0,1,0)and(1,0,1).
Thenthedistanceofthepoint(2λ,λ, −λ)fromtheplaneP2is
[24-Jan-2023Shift2]
Lettheplanecontainingthelineofintersectionoftheplanes
P1 :x+ (λ+4)y+z=1and
P2 :2x +y+z=2passthroughthepoints(0,1,0)and(1,0,1).
Thenthedistanceofthepoint(2λ,λ, −λ)fromtheplaneP2is
[24-Jan-2023Shift2]
Options:
A.5√6
B.4√6
C.2√6
D.3√6
Answer:D
Directionratioofline =
^
i^
j^
k
1 2 1
0 3 −1
=^
i(−5) − ^
j(−1) + ^
k(3)
= −5^
i+^
j+3^
k
M(−5λ +3,λ+2,3λ +1)
→
PM⟂−5^
i+^
j+3^
k
−5(−5λ +2) + (λ−7) + 3(3λ −6) = 0
⇒25λ +λ+9λ −10 −7−18 =0
⇒λ=1
Point M = (−2,3,4) = (α,β,γ)
α+β+γ=5
| |
( )
Solution:
Solution:
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Question30
Iftheshortestbetweenthelines x+ √6
2=y− √6
3=z− √6
4and
x−λ
3=y−2√6
4=z+2√6
5is6,thenthesquareofsumofallpossiblevaluesof
λis
[24-Jan-2023Shift2]
Answer:384
Solution:
-------------------------------------------------------------------------------------------------
Question31
EquationofplanepassingthroughpointofintersectionofP1andP2
P=P1 +kP 2
(x+ (λ+4)y+z−1) + k(2x +y+z−2) = 0
Passingthrough(0,1,0)and(1,0,1)
(λ+4−1) + k(1−2) = 0
(λ+3) − k=0...(1)
Alsopassing(1,0,1)
(1+1−1) + k(2+1−2) = 0
1+k=0
k= −1
putin(1)
λ+3+1=0
λ= −4
Thenpoint(2λ,λ, −λ)
d=−16 −4, −4,4)
√6
d=18
√6×√6
√6=3√6
| |
Shortestdistancebetweenthelines
x+ √6
2=y− √6
3=z− √6
4x−λ
3=y−2√6
4=2+2√6
5is6
Vectoralonglineofshortestdistance
=
i j k
234
345
, ⇒− ^
i+2^
j−k(itsmagnitudeis√6)
Now 1
√6
√6+λ√6−3√6
2 3 4
3 4 5
= ±6
⇒λ= −2√6,10√6
So,squareofsumofthesevaluesis384.
| |
| |
ConsiderthelinesL1andL2givenby
L1:x−1
2=y−3
1=z−2
2
L2:x−2
1=y−2
2=z−3
3
AlineL3havingdirectionratios1, −1, −2,intersectsL1andL2atthe
pointsPandQrespectively.ThenthelengthoflinesegmentPQis
[25-Jan-2023Shift1]
Options:
A.2√6
B.3√2
C.4√3
D.4
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question32
ThedistanceofthepointP(4,6, −2)fromthelinepassingthroughthe
point(−3,2,3)andparalleltoalinewithdirectionratios3,3, −1is
equalto:
[25-Jan-2023Shift1]
Options:
A.3
B.√6
C.2√3
D.√14
Answer:D
Solution:
LetP= (2λ +1,λ+3,2λ +2)
LetQ= (µ+2,2µ +2,3µ +3)
⇒2λ −µ−1
1=λ−2µ +1
−1=2λ −3µ −1
−2
⇒λ=µ=3⇒P(7,6,8)and Q(5,8,12)
PQ =2√6
Question33
Lettheequationoftheplanepassingthroughtheline
x−2y −z−5=0=x+y+3z −5andparalleltotheline
x+y+2z −7=0=2x +3y +z−2beax +by +cz =65.Thenthedistance
ofthepoint(a,b,c)fromtheplane2x +2y −z+16 =0is_______.
[25-Jan-2023Shift1]
Answer:9
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question34
Thefootofperpendicularofthepoint(2,0,5)ontheline
x+1
2=y−1
5=z+1
−1is(α,β,γ).Then.WhichofthefollowingisNOT
correct?
[25-Jan-2023Shift2]
Options:
A. αβ
γ=4
15
Equationoflineis x+3
3=y−2
3=z−3
−1=λ
M(3λ −3,3λ +2,3−λ)
D.RofPM(3λ −7,3λ −4,5−λ)
SincePMisperpendiculartoline
⇒3(3λ −7) + 3(3λ −4) − 1(5−λ) = 0
⇒λ=2
⇒M(3,8,1) ⇒ PM = √14
Equationofplaneis
(x−2y −z−5) + b(x+y+3z −5) = 0
1+b−2+b−1+3b
1 1 2
2 3 1
=0
⇒b=12
∴planeis 13x +10y +35z =65
Distancefromgivenpointtoplane = 9
| |
B. α
β= −8
C. β
γ= −5
D. γ
α=5
8
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question35
Theshortestdistancebetweenthelinesx +1=2y = −12zand
x=y+2=6z −6is
[25-Jan-2023Shift2]
Options:
A.2
B.3
C. 5
2
D. 3
2
Answer:A
Solution:
L:x+1
2=y−1
5=z+1
−1=λ(let)
Letfootofperpendicularis
P(2λ −1,5λ +1, −λ−1)
→
PA = (3−2λ)^
i− (5λ +1)^
j+ (6+λ)^
k
Directionratioofline⇒→
b=2^
i+5^
j−^
k
Now, ⇒→
PA ⋅→
b=0
⇒2(3−2λ) − 5(5λ +1) − (6+λ) = 0
⇒λ=−1
6
P(2λ −1,5λ +1, −λ−1) ≡ P(α,β,γ)
⇒α=2−1
6−1= − 4
3⇒α= − 4
3
⇒β=5−1
6+1=1
6⇒β=1
6
⇒γ= −λ−1=1
6−1⇒γ= − 5
6
∴Checkoptions
( )
( )
Solution:
-------------------------------------------------------------------------------------------------
Question36
Iftheshortestdistancebetweenthelinejoiningthepoints(1,2,3)and
(2,3,4),andtheline x−1
2=y+1
−1=z−2
0isα,then28α2isequalto_______.
[25-Jan-2023Shift2]
Answer:18
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question37
Lettheco-ordinatesofonevertexof△ABCbeA(0,2,α)andtheother
twoverticeslieontheline x+α
5=y−1
2=z+4
3.Forα ∈ ℤ,iftheareaof
x+1
1=y
1
2
=z
−1
12
and x
1=y+2
1=z−1
1
6
⇒Shortestdistance =
→
b−→
a⋅→
p×→
q
→
p×→
q
S.D. = − ^
i+2^
j−^
k⋅
→
p×→
q
→
p×→
q
→
p×→
q≡^
i,^
j,^
k;1,1
2,−1
12 ;1,1,1
6=1
6
^
i−1
4
^
j+1
2
^
k or 2^
i−3^
j+6^
k
S.D. =−^
i+2^
j−^
k⋅2^
i−3^
j+6^
k
22+32+62=−14
7=2
( ) ( )
| |
( ) ( )
| |
{ | |}
( ) ( )
√| |
→
r=^
i+2^
j+3^
k+λ^
i+^
j+^
k→
r=→
a+λ→
p
→
r= +^
i−^
j+2^
k+µ 2^
i−^
j→
r=→
b+µ→
q
→
p×→
q=
^
i^
j^
k
111
2−10
= ^
i+2^
j−3^
k
d=
→
b−→
a⋅→
p×→
q
→
p×→
q
d=−3^
j−^
k⋅^
i+2^
j−3^
k
√14
=−6+3
√14 =3
√14
α=3
√14
Now,28α2=2∕2×9
14 =18
( ) ( )
( ) ( )
| |
|( ) ( )
| | |
|( ) ( ) |
| |
△ABCis21sq.unitsandthelinesegmentBChaslength2√21units,
thenα2isequalto_______.
[29-Jan-2023Shift1]
Answer:9
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question38
LettheequationoftheplanePcontainingthelinex +10 =8−y
2=zbe
ax +by +3z =2(a+b)andthedistanceoftheplanePfromthepoint
(1,27,7)bec.Thena2+b2+c2isequalto________.
[29-Jan-2023Shift1]
Answer:355
Solution:
-------------------------------------------------------------------------------------------------
Question39
Theplane2x −y+z=4intersectsthelinesegmentjoiningthepoints
A.(O12,α)
(2α +5)2+ (2α +20)2+ (2α −5)2= √21√38
⇒12α2+80α +450 =798
⇒12α2+80α −348 =0
⇒α=3⇒α2=9
√
Theline x+10
1=y−8
−2=z
1haveapoint(−10,8,0)withd.r.(1, −2,1)
∵theplaneax +by +3z =2(a+b)
⇒b=2a
&dotproductofd.r.'siszero
∴a−2b +3=0
∴a=1&b=2
Distancefrom(1,27,7)is
c=1+54 +21 −6
√14 =70
√14 =5√14
∴a2+b2+c2=1+4+350
=355
A(a, −2,4)andB(2,b, −3)atthepointCintheratio2 :1andthe
distanceofthepointCfromtheoriginis√5.Ifab <0andPisthepoint
(a−b,b,2b −a)thenCP2isequalto:
[29-Jan-2023Shift2]
Options:
A. 17
3
B. 16
3
C. 73
3
D. 97
3
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question40
Shortestdistancebetweenthelines x−1
2=y+8
−7=z−4
5and
x−1
2=y−2
1=z−6
−3is
[29-Jan-2023Shift2]
Options:
A.2√3
B.4√3
A(a, −2,4), B(2,b, −3)
AC :CB =2:1
⇒C≡a+4
3,2b −2
3,−2
3
Clieson2x −y+2=4
⇒2a +8
3−2b −2
3−2
3=4
⇒a−b=2... (1)
AlsoOC = √5
⇒a+4
3
2+2b −2
3
2+4
9=5...(2)
Solving,(1)and(2)
(b+6)2+ (2b −2)2=41
⇒5b2+4b −1=0
⇒b= −1 or 1
5
⇒a=1 or 11
5
Butab <0⇒ (a,b) = (1, −1)
C≡5
3,−4
3,−2
3,P≡ (2, −1, −3)
CP2=1
9+1
9+49
9=51
9=17
3
( )
( ) ( )
( )
C.3√3
D.5√3
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question41
Ifthelines x−1
1=y−2
2=z+3
1and x−a
2=y+2
3=z−3
1intersectsatthepoint
P,thenthedistanceofthepointPfromtheplanez =ais:
[29-Jan-2023Shift2]
Options:
A.16
B.28
C.10
D.22
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
x−1
2=y+8
−7=z−4
5
→
a=i−8
^
^
j+4^
k
x−1
2=y−2
1=z−6
−3
→
b=^
i+2^
j+6^
k
→
p=2^
i−7^
j+5^
k,→
q=2^
i+^
j−3^
k
→
p×→
q=
^
i^
j^
k
2−75
2 1 −3
=^
i(16) − ^
j(−16) + ^
k(16)
=16 ^
i+^
j+^
k
=−12
√3=4√3∣
| |
( )
| |
PointonL1≡ (λ+1,2λ +2,λ−3)
PointonL2≡ (2µ +a,3µ −2,µ+3)
λ−3=µ+3⇒λ=µ+6...(1)
2λ +2=3µ −2⇒2λ =3µ −4...(2)
Solving,(1)and(2)
⇒λ=22&µ=16
⇒P≡ (23,46,19)
⇒a= −9
DistanceofPfromz= −9is28
Question42
Letaunitvector
◠
OPmakeangleα,β,γwiththepositivedirectionsofthe
co-ordinateaxesOX,OY,OZrespectively,whereβ ∈0,π
2
◠
OPis
perpendiculartotheplanethroughpoints(1,2,3),(2,3,4)and
(1,5,7),thenwhichoneofthefollowingistrue?
[30-Jan-2023Shift1]
Options:
A.α ∈π
2,π andγ ∈π
2,π
B.α ∈0,π
2andγ ∈0,π
2
C.α ∈π
2,π andγ ∈0,π
2
D.α ∈0,π
2andγ ∈π
2,π
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question43
Thelinel 1passesthroughthepoint(2,6,2)andisperpendiculartothe
plane2x +y−2z =10.Thentheshortestdistancebetweenthelinel 1
andtheline x+1
2=y+4
−3=z
2is:
[30-Jan-2023Shift1]
Options:
A.7
( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
Equationofplane:-
x−1 y −2 z −3
1 1 1
0 3 4
=0
⇒ [x−1] − 4[y−2] + 3[z−3] = 0
⇒x−4y +3z =2
D.R'sofnormalofplane<1, −4,3>
D.C'sof ± 1
√26 , ∓ 4
√26 , ± 3
√26
cos β =4
√26
cos α =−1
√26
π
2<α<π
cos γ =−3
√26
π
2<γ<π
| |
⟨ ⟩
B. 19
3
C. 19
3
D.9
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question44
Iftheequationoftheplanepassingthroughthepoint(1,1,2)and
perpendiculartothelinex −3y +2z −1=0 4x −y+zis
Ax +By +Cz =1,then140(C−B+A)isequalto_______.
[30-Jan-2023Shift1]
Answer:15
Solution:
Solution:
Lineℓ,isgivenby
L1:x−2
2=y−6
1=z−2
−2
Given,
L2:x+1
2=y+4
−3=z
2
Shortestdistance =→
AB ⋅
→
MN
MN
AB =3^
i+10 ^
j+2^
k
MN =
^
i^
j^
k
2 1 −2
2−32
= −4^
i−8^
j−8^
k
MN = √16 +64 +64 =12
∴Shortestdistance =−12 −80 −16
12 =9
∴ Option(4)iscorrect.
| |
| |
| |
x−3y +2z −1=0
4x −y+z=0
-------------------------------------------------------------------------------------------------
Question45
Ifλ1<λ2aretwovaluesofλsuchthattheanglebetweentheplanes
P1:r 3 ^
i−5^
j+^
k=7andP2:→
r⋅λ^
i+^
j−3^
k=9issin −12√6
5,then
thesquareofthelengthofperpendicularfromthepoint(38λ1,10λ2,2)
totheplaneP1is_______.
[30-Jan-2023Shift1]
Answer:315
Solution:
( ) ( ) ( )
n1×n2=
^
i^
j^
k
1−3 2
4−1 1
= − ^
i+7^
j+11 ^
k
Dreofnormaltotheplaneis−1,7,11
Equationofplane:
−1(x−1) + 7(y−1) + 11(z−2) = 0
−x+7y +11z =28
−1
28 x+7y
28 +11z
28 =1
Ax +By +Cz =1
140(C−B+A) = 140 11
28 −7
28 −1
28
=140×3
28 =15
| |
( )
-------------------------------------------------------------------------------------------------
Question46
Avector→
vinthefirstoctantisinclinedtothexaxisat60∘,tothey-axis
at45∘andtothez-axisatanacuteangle.Ifaplanepassingthroughthe
points(√2, −1,1)and(a,b,c),isnormalto→
v,then
[30-Jan-2023Shift2]
Options:
A.√2a+b+c=1
B.a +b+ √2c=1
C.a + √2b+c=1
D.√2a−b+c=1
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question47
Ifaplanepassesthroughthepoints(−1,k,0), (2,k, −1),(1,1,2)andis
paralleltotheline x−1
1=2y +1
2=z+1
−1,thenthevalueof k2+1
(k−1)(k−2)is
Perpendiculardistanceofpoint
(38λ1,10λ2,2) ≡ (50,50,2)fromplaneP1
=|3×50 −5×50 +2−7|
√35 =105
√35
Square =105 ×105
35 =315
^
v=cos 60∘^
i+cos 45∘^
j+cos γ ^
k
⇒1
4+1
2+cos2γ=1(γ→Acute )
⇒cos γ =1
2
⇒γ=60∘
Equationofplaneis
1
2(x− √2) + 1
√2(y+1) + 1
2(z−1) = 0
⇒x+ √2y +z=1
(a,b,c)liesonit.
⇒a+ √2b +c=1
[30-Jan-2023Shift2]
Options:
A. 17
5
B. 5
17
C. 6
13
D. 13
6
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question48
LetalineLpassthroughthepointP(2,3,1)andbeparalleltotheline
x+3y −2z −2=0=x−y+2z.IfthedistanceofLfromthepoint
(5,3,8)isα,then3α2isequalto_______.
[30-Jan-2023Shift2]
Answer:158
Solution:
x−1
1=2y +1
2=z+1
−1
x−1
1=
y+1
2
1=z+1
−1
Points:A (−1,k,0), B(2,k, −1), C(1,1,2)
→
CA = −2^
i+ (k−1)^
j−2^
k
→
CB =^
i+ (k−1)^
j−3^
k
→
CA ×→
CB =
^
i^
j^
k
−2 k −1−2
1k−1−3
=^
i(−3k +3+2k −2) − ^
j(6+2) + ^
k(−2k +2−k+1)
= (1−k)^
i−8^
j+ (3−3k)^
k
Theline x−1
1=
y+1
2
1=z+1
−1isperpendiculartonormalvector.
∴1⋅ (1−k) + 1(−8) + (−1)(3−3k) = 0
⇒1−k−8−3+3k =0
⇒2k =10 ⇒k=5
∴k2+1
(k−1)(k−2)=26
4⋅3=13
6
| |
Solution:
-------------------------------------------------------------------------------------------------
Question49
LettheshortestdistancebetweenthelinesL :x−5
−2=y−λ
0=z+λ
1,λ≥0
andL1:x+1=y−1 =4−zbe2√6.If(α,β,γ)liesonL,thenwhichof
thefollowingisNOTpossible?
[31-Jan-2023Shift1]
Options:
A.α +2γ =24
B.2α +γ=7
C.2α −γ=9
D.α −2γ =19
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question50
LetθbetheanglebetweentheplanesP1=→
r⋅
⋀
i+
⋀
i+2
⋀
k=9and
( )
^
i^
j^
k
1 3 −2
1−1 2
=4^
i−4^
j−4^
k
∴Equationoflineis x−2
1=y−3
−1=z−1
−1
LetQbe(5,3,8)andfootof⟂fromQonthislinebeR.
Now,R≡ (k+2, −k+3, −k+1)
DRofQRare(k−3, −k, −k−7)
∴ (1)(k−3) + (−1)(−k) + (−1)(−k−7) = 0
⇒k= − 4
3
∴α2=13
3
2+4
3
2+17
3
2=474
9
∴3α2=158
| |
( ) ( ) ( )
→
b1×→
b2=
^
i^
j^
k
−20 1
1 1 −1
= − ^
i−^
j−2^
k
a2−a1=6^
i+ (λ−1)^
j+ (−λ−4)^
k
2√6=−6−λ+1+2λ +8
√1+1+4
|λ+3| = 12 ⇒λ=9, −15
α= −2k +5,γ=k−λ where k ∈R
⇒α+2γ =5−2λ = −13,35
| |
| |
P2=→
r⋅2
⋀
i−
⋀
i+
⋀
k=15.
LetLbethelinethatmeetsP2atthepoint(4, −2,5)andmakesan
angleθwiththenormalofP2.IfαistheanglebetweenLandP2then
(tan2θ)(cot2α)isequalto_______.
[31-Jan-2023Shift1]
Answer:9
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question51
LetthelineL :x−1
2=y+1
−1=z−3
1intersecttheplane2x +y+3z =16at
thepointP.LetthepointQbethefootofperpendicularfromthepoint
R(1, −1, −3)onthelineL.IfαistheareaoftrianglePQR.thenα2is
equalto________.
[31-Jan-2023Shift1]
Answer:180
Solution:
( )
cos θ =
^
i+^
j+2^
k⋅2^
i−^
j+^
k
6=2−1+2
6=1
2
θ=π∕3 α =π∕6
(tan2θ)(cot2α)
(3)(3)=9
( ) ( )
AnypointonL((2λ +1), (−λ−1), (λ+3))
2(2λ +1) + (−λ−1) + 3(λ+3) = 16
-------------------------------------------------------------------------------------------------
Question52
IfapointP(α,β,γ)satisfying(αβγ)
2 10 8
9 3 8
8 4 8
= ( 000)liesonthe
plane2x +4y +3z =5,then6α +9β +7γisequalto:
[31-Jan-2023Shift2]
Options:
A.−1
B. 11
5
C. 5
4
D.11
Answer:D
Solution:
Solution:
( )
6λ +10 =16 ⇒λ=1
∴P= (3, −2,4)
DRofQR = ⟨2λ, −λ,λ+6⟩
DRofL= ⟨2, −1,1⟩
4λ +λ+λ+6=0 6λ +6=0⇒λ= −1
Q= (−1,0,2)
→
QR =2^
i−^
j−5^
k→
QP =4^
i−2^
j+2^
k
→
QR ×→
QP =
^
i^
j^
k
2−1−5
4−2 2
= −12 ^
i−24 ^
j
α=1
2× √144 +576 ⇒α2=720
4=180
| |
2α +4β +3γ =5...(1)
-------------------------------------------------------------------------------------------------
Question53
LettheplaneP :8x +α1y+α2z+12 =0beparalleltotheline
L:x+2
2=y−3
3=z+4
5.IftheinterceptofPonthey-axisis1,thenthe
distancebetweenPandLis:
[31-Jan-2023Shift2]
Options:
A.√14
B. 6
√14
C. 2
7
D. 7
2
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
√
√
2α +9β +8γ =0...(2)
10α +3β +4γ =0...(3)
8α +8β +8γ =0...(4)
Subtract(4)from(2)
−6α +β=0
β=6α...(5)
Fromequation(4)
8α +48α +8γ =0
γ= −7α...(6)
Fromequation(1)
2α +24α −21α =5
5α =5
α=1
β= +6,γ= −7
∴6α +9β +7γ
=6+54 −49
=11
P:8x +α1y+α2z+12 =0
L: x+2
2=y−3
3=z+4
5
∵PisparalleltoL
⇒8(2) + α1(3) + 5(α2) = 0
⇒3α1+5(α2) = −16
Alsoy-interceptofplanePis1
⇒α1= −12
Andα2=4
⇒EquationofplanePis2x −3y +z+3=0
⇒DistanceoflineLfromPlanePis
=0−3(6) + 1+3
√4+9+1
= √14
| |
Question54
LetPbetheplane,passingthroughthepoint(1, −1, −5)and
perpendiculartothelinejoiningthepoints(4,1, −3)and(2,4,3).Then
thedistanceofPfromthepoint(3, −2,2)is
[31-Jan-2023Shift2]
Options:
A.6
B.4
C.5
D.7
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question55
Theshortestdistancebetweenthelines x−5
1=y−2
2=z−4
−3and
x+3
1=y+5
4=z−1
−5is
[1-Feb-2023Shift1]
Options:
A.7√3
B.5√3
C.6√3
D.4√3
Answer:C
Solution:
Solution:
EquationofPlane:
2(x−1) − 3(y+1) − 6(z+5) = 0
Or 2x −3y −6z =35
⇒Requireddistance =
|2(3) − 3(−2) − 6(2) − 35|
√4+9+36
=5
Shortestdistancebetweentwolines
x−x1
a1
=y−y1
a2
=z−z1
a3&
-------------------------------------------------------------------------------------------------
Question56
LettheimageofthepointP(2, −1,3)intheplanex +2y −z=0beQ.
Thenthedistanceoftheplane3x +2y +z+29 =0fromthepointQis
[1-Feb-2023Shift1]
Options:
A. 22√2
7
B. 24√2
7
C.2√14
D.3√14
Answer:D
Solution:
Solution:
x−x2
b1
=y−y2
b2
=z−z2
b3
isgivenas
x1−x2y1−y2z1−z2
a1a2a3
b1b2b3
(a1b3−a3b2)2+ (a1b3−a3b1)2+ (a1b2−a2b1)2
5− (3)2− (−5)4−1
1 2 −3
1 4 −5
(−10 +12)2+ (−5+3)2+ (4−2)2
8 7 3
12−3
14−5
(2)2+ (2)2+ (2)2
=|8(−10 +12) − 7(−5+3) + 3(4−2)|
√4+4+4
=16 +14 +6
√12 =36
√12 =36
2√3
=18
√3=6√3
| |
√
| |
√
| |
√
-------------------------------------------------------------------------------------------------
Question57
LettheplanePpassthroughtheintersectionoftheplanes
2x +3y −z=2andx +2y +3z =6,andbeperpendiculartotheplane
2x +y−z+1=0.Ifd isthedistanceofPfromthepoint(−7,1,1),then
d2isequalto:
[1-Feb-2023Shift2]
Options:
A. 250
83
B. 15
53
C. 25
83
D. 250
82
Answer:A
Solution:
Solution:
eq.oflinePM x−2
1=y+1
2=z−3
−1=λ
anypointonline = (λ+2,2λ −1, −λ+3)
forpoint'm'(λ+2) + 2(2λ −1) − (3−λ) = 0
λ=1
2
Pointm1
2+2,2×1
2−1,−1
2+3
=5
2,0,5
2
ForImageQ(α,β,γ)
α+2
2=5
2,β−1
2=0,
γ+3
2=5
2
Q: (3,1,2)
d=3(3) + 2(1) + 2+29
32+22+12
d=42
√14 =3√14
( )
( )
|√|
P≡P1+λP2=0
(2+λ)x+ (3+2λ)y+ (3λ −1)z−2−6λ =0
Plane P isperpendicularto P3∴→
n⋅→
n3=0
2(λ+2) + (2λ +3) − (3λ −1) = 0
λ= −8
P≡ −6x −13y −25z +46 =0
6x +13y +25z −46 =0
Distfrom (−7,1,1)
d= ∣ −42 +13 +25 −46
√36 +169 +625∣ = 50
√830
d2=50 ×50
830 =250
83
Question58
ThepointofintersectionCoftheplane8x +y+2z =0andtheline
joiningthepointsA(−3, −6,1)andB(2,4, −3)dividesthelinesegment
ABinternallyintheratiok :1.Ifa,b,c( |a|, | b|,|c|arecoprime)are
thedirectionratiosoftheperpendicularfromthepointContheline
1−x
1=y+4
2=z+2
3,then|a+b+c|isequalto_______.
[1-Feb-2023Shift2]
Answer:10
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question59
Letαx +βy +yz =1betheequationofaplanepassingthroughthepoint
(3, −2,5)andperpendiculartothelinejoiningthepoints(1,2,3)and
(−2,3,5).Thenthevalueofαβyisequalto______.
[1-Feb-2023Shift2]
Plane:8x +y+2z =0
GivenlineAB :x−2
5=y−4
10 =z+3
−4=λ
Anypointonline(5λ +2,10λ +4, −4λ −3)
Pointofintersectionoflineandplane
8(5λ +2) + 10λ +4−8λ −6=0
λ= − 1
3
C1
3,2
3, − 5
3
L:x−1
−1=y+4
2=z+2
3=µ
( )
→
CD = −µ+2
3
^
i+2µ −14
3
^
j+3µ −1
3
^
k
−µ+2
3(−1) + 2µ −14
32+3µ −1
33=0
µ=11
14
→
CD =−5
42 ,−130
42 ,85
42
Directionratios → (−1, −26,17)
|a+b+c| = 10
( ) ( ) ( )
( ) ( ) ( )
Answer:6
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question60
OnevertexofarectangularparallelepipedisattheoriginOandthe
lengthsofitsedgesalongx,yandzaxesare3,4and5units
respectively.LetPbethevertex(3,4,5).Thentheshortestdistance
betweenthediagonalOPandanedgeparalleltozaxis,notpassing
throughOorPis:
[6-Apr-2023shift1]
Options:
A. 12
5√5
B.12√5
C. 12
5
D. 12
√5
Answer:C
Solution:
GivenEquationisnotequationofplaneasyzispresent.Ifweconsideryisγthenanswerwouldbe6.
Normalvectorofplane = 3
⋀
i−
⋀
j−2
⋀
k
Plane:3x −y−2z +λ=0
Point(3, −2,5)satisfiestheplane
λ= −1
3x −y−2z =1
αβy =6
EquationofOPis x
3=y
4=z
5
a1= (0,0,0)a2= (3,0,5)
b1= (3,4,5)b2= (0,0,1)
Equationofedgeparalleltozaxis
x−3
0=y−0
0=z−5
1
S.D=
→
a2⋅→
a1⋅→
b1×→
b2
→
b1×→
b2
( ) ( )
| |
-------------------------------------------------------------------------------------------------
Question61
Iftheequationoftheplanepassingthroughthelineofintersectionof
theplanes2x −y+z=3,4x −3y +5z+9 =0andparalleltotheline
x+1
−2=y+3
4=z−2
5isax +by +cz +6=0,thena +b+cisequalto:
[6-Apr-2023shift1]
Options:
A.15
B.14
C.13
D.12
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question62
LettheimageofthepointP(1,2,3)intheplane2x −y+z=9beQ.If
thecoordinatesofthepointRare(6,10,7).thenthesquareofthearea
ofthetrianglePQRis_______.
[6-Apr-2023shift1]
305
345
001
^
i^
j^
k
345
001
= 3(4)
4^
i−3^
j
=12
5
| |
| | | |
Usingfamilyofplaner
P:P1+λP2=0⇒P(2+4λ)x− (1+3λ)y+ (1+5λ)z= (3−9λ)
Pis|to x+1
−2=y+3
4=z−2
5
Thenforλ:→
np⋅→
vL=0
−2(2+4λ) − 4(1+3λ) + 5(1+5λ) = 0
−3+5λ =0⇒λ=3
5
Hence:P:22x −14y +20z = −12
P:11x −7y +10z +6=0
⇒a=11
b= −7
c=10
⇒a+b+c=14
Answer:594
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question63
LetthelineLpassthroughthepoint(0,1,2),intersecttheline
x−1
2=y−2
3=z−3
4andbeparalleltotheplane2x +y−3z =4.Thenthe
distanceofthepointP(1, −9,2)fromthelineLis:
[6-Apr-2023shift2]
Options:
A.9
B.√54
C.√69
D.√74
Answer:D
Solution:
Solution:
LetQ(α,β,γ)betheimageofP,abouttheplane
2x −y+z=9
α−1
2=β−2
−1=γ−3
1=2
⇒α=5,β=0,γ=5
ThenareaoftrianglePQRis = 1
2
→
PQ ×→
PR
= −12^
i−3^
j+21^
k= √144 +9+441 = √594
Squareofarea = 594
| |
| |
→
PQ = (2λ +1,3λ +1,4λ +1)
→
PQ ⋅→
n=0⇒ (2λ +1) ⋅ (2) + (3λ +1)(1) + (4λ +1)(−3) = 0
⇒ −5λ =0
⇒λ=0
Q = (1,2,3)
x−0
1=y−1
1=z−2
1=µ
-------------------------------------------------------------------------------------------------
Question64
AplanePcontainsthelineofintersectionoftheplane→
r⋅^
i+^
j+^
k=6
and→
r⋅2^
i+3^
j+4^
k= −5.IfPpassesthroughthepoint(0,2, −2),then
thesquareofdistanceofthepoint(12,12,18)fromtheplanePis:
[6-Apr-2023shift2]
Options:
A.620
B.1240
C.310
D.155
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question65
( )
( )
distanceoflinefrom(1, −9,2)
(P′Q) ⋅ (1,1,1) = 0
⇒ [µ−1,µ+10,µ] ⋅ [1,1,1] = 0
⇒µ−1+µ+10 +µ=0
µ= −3
Q′= (−3, −2,1)
P′Q′= √16 +49 +9= √74
eqnofplane P1+λP2=0
(x+y+z−6) + λ(2x +3y +4z +5) = 0
passth.(0,2, −2)
(−6) + λ(6−8+5) = 0
(−6) + λ[3] = 0⇒λ=2
eqnofplane
5x +7y +9z +4=0
distancefrom(12,12,18)
d=60 +84 +162 +4
√25 +49 +81
d=310
√155
d2=310 ×310
155
d2=620
Ans.Option1
| |
Ifthelines x−1
2=2−y
−3=z−3
αand x−4
5=y−1
2=z
βintersect,thenthe
magnitudeoftheminimumvalueof8αβis_______.
[6-Apr-2023shift2]
Answer:18
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question66
Theshortestdistancebetweenthelines x−4
4=y+2
5=z+3
3and
x−1
3=y−3
4=z−4
2is
[8-Apr-2023shift1]
Options:
A.2√6
B.3√6
C.6√3
D.6√2
Answer:B
Solution:
Ifthelines x−1
2=2−y
−3=z−3
αand x−4
5=y−1
2=z
βintersectPointoffirstline(1,2,3)andpointonsecondline
(4,1,0).
Vectorjoiningbothpointsis−3^
i+^
j+3k
Nowvectoralongsecondlineis2^
i+3^
j+αk
Alsovectoralongsecondlineis5^
i+2^
j+βk
Nowthesethreevectorsmustbecoplanar
⇒
2 3 α
5 2 β
−31 3
⇒2(6−β) − 3(15 +3β) + α(11) = 0
⇒α−β=3
Now α =3+β
Givenexpression 8(3+β) ⋅ β=8(β2+3β)
=8 β2+3β +9
4−9
4=8 β +3
2
2−18
Somagnitudeofminimumvalue = 18
| |
( ) ( )
-------------------------------------------------------------------------------------------------
Question67
Iftheeqationoftheplanecontainingtheline
x+2y +3z −4=02x +y−z+5andperpendiculartotheplane
→
r=^
i−^
j+λ^
i+^
j+^
k+µ^
i−2^
j+3^
kisax+by+cz =4,then
(a−b+c)isequalto
[8-Apr-2023shift1]
Options:
A.22
B.24
C.20
D.18
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question68
Letλ1,λ2bethevaluesofλforwhichthepoints 5
2,1,λ and
(−2,0,1)areatequaldistancefromtheplane2x+3y −6z +7=0.If
( ) ( ) ( )
( )
Sd=
→
a−→
b×→
n1×→
n2
→
n1×→
n2
a= (4, −2, −3)
b= (1,3,4)
n1= (4,5,3)
n2= (3,4,2)
n1×n2=
i j k
453
342
=^
i(−2) − ^
j(−1) + ^
k(1) = (−2,1,1)
Sd=(3, −5, −7) ⋅ (−2,1,1)
√6=−6−5−7
√6=3√6
|( ) ( )
| | |
| |
| |
D.R'sofline→
n1= −5^
i+7^
j−3^
k
D.R'sofnormalofsecondplane
→
n2=5^
i−2^
j−3^
k
→
n1×→
n2= −27^
i−30^
j−25^
k
Apointontherequiredplaneis 0, − 11
5,14
5
Theequationofrequiredplaneis
27x +30y +25z =4
∴a−b+c=22
( )
λ1>λ2,thenthedistanceofthepoint(λ1−λ2,λ2,λ1)fromtheline
x−5
1=y−1
2=z+7
2is________.
[8-Apr-2023shift1]
Answer:9
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question69
Fora,b∈Zand|a−b| ≤ 10,lettheanglebetweentheplane
P:ax +y−z=bandtheline1 :x−1=a−y=z+1becos−11
3.Ifthe
distanceofthepoint(6, −6,4)fromtheplanePis3√6,thena4+b2is
equalto
[8-Apr-2023shift2]
Options:
A.85
B.48
C.25
D.32
Answer:D
( )
2x +3y −6z +7=05
2,1,λ, (−2,0,1)
d1=5+3−6λ +7
7=d2=−4−6+7
7
⇒ | 15 −6λ |=|3|
15 −6λ =3 or 15 −6λ = −3
6λ =12 6λ =18
λ=2 λ =3
λ1=3,λ2=2
∴P(1,2,3)x−5
1=y−1
2=z+7
2
( )
| | | |
d=(4, −1, −10) × (1,2,2)
3=18^
i−18^
j+9^
k
3=9
| | | |
Solution:
-------------------------------------------------------------------------------------------------
Question70
LetP1betheplane3x −y−7z =11andP2betheplanepassingthrough
thepoints(2, −1,0), (2,0, −1),and(5,1,1).Ifthefootofthe
perpendiculardrawnfromthepoint(7,4, −1)onthelineofintersection
oftheplanesP1andP2is(α,β,γ),thenα +β+γisequalto_______.
[8-Apr-2023shift2]
Answer:11
Solution:
Solution:
θ=cos−11
3∴sin θ =1−1
9=2√2
3
sin θ =a⋅1+1(−1) + (−1) ⋅ 1
a2+1+1⋅ √3
=2√2
3
⇒ {3(a−2)}2=24(a2+2)
⇒3(a2−4a +4) = 8a2+16
⇒5a2+12a +4=0
⇒5a2+10a +2a +4=0
∴a= −2,−2
5∵a∈z
∴a= −2
Distanceof(6, −6,4)from
−2x +y−z−b=0is3√6
∴−12 −6−4−b
√4+1+1=3√6
⇒ | b+22 | = 18 ∴b= −40, −4
∵ | a−b| ≤10
∴b= −4
∴a4+b2
=32 Ans.
√
√
| |
P2isgivenby
x−5 y −1 z −1
3 2 1
3 1 2
=0
x−y−z=3
DRoflineintersectionofP1&P2
i j k
1−11
3−1−7
+6^
i+4^
j+2^
k
Let
z=0,x−y=3
3x −y=112x =8
| |
| |
-------------------------------------------------------------------------------------------------
Question71
LetPbetheplanepassingthroughtheline x−1
1=y−2
−3=z+5
7andthe
point(2,4, −3).Iftheimageofthepoint(−1,3,4)intheplanePis
(α,β,γ)thenα +β+γisequalto
[8-Apr-2023shift2]
Options:
A.12
B.9
C.10
D.11
Answer:C
Solution:
Solution:
x=4
y=1
SoLineis
x−4
6=y−1
4=z−0
2=r
(α,β,γ) = (6r +4,4r +1,2r)
6(α−7) + 4(β−4) + 2(γ+1) = 0
6α −42 +4β −16 +2γ +2=0
36r +24 +16r +4+4r −56 =0
56r =28
r=1
2α+β+γ=12r +5
=6+5=11
Equationofplaneisgivenby
x−1 y −2 z +5
1 2 2
1−37
=0
4x −y−z=7
α+1
4=β−3
−1=γ−4
−1=−2(−4−3−4−7)
16 +1+1=2
α=7,β=1,γ=2
α+β+γ=10(Option 3)
| |
-------------------------------------------------------------------------------------------------
Question72
LetObetheoriginandthepositionvectorofthepointPbe
−^
i−2^
j+3^
k.IfthepositionvectorsoftheA,BandCare
−2^
i+^
j−3^
k,2^
i+4^
j−2^
kand−4^
i+2^
j−^
krespectively,thenthe
projectionofthevector →
OPonavectorperpendiculartothevectors →
AB
and →
ACis:
[10-Apr-2023shift1]
Options:
A. 10
3
B. 8
3
C. 7
3
D.3
Answer:D
Solution:
Solution:
PositionvectorofthepointP(−1, −2,3), A(−2,1, −3)B(2,4, −2),andC(−4,2, −1)
Then →
OP⋅
→
AB ×→
AC
∣→
AB ×→
AC ∣
→
AB ×→
AC =
^
i^
j^
k
4 3 1
−21 2
=^
i(5) − ^
j(8+2) + ^
k(4+6)
=5^
i−10^
j+10^
k
Now
)
| |
-------------------------------------------------------------------------------------------------
Question73
LettwoverticesofatriangleABCbe(2,4,6)and(0, −2, −5),andits
centroidbe(2,1, −1).Iftheimageofthethirdvertexintheplane
x+2y +4z =11is(α,β,γ),thenαβ +βγ +γαisequalto:
[10-Apr-2023shift1]
Options:
A.76
B.74
C.70
D.72
Answer:B
Solution:
Solution:
→
OP⋅
→
AB ×→
AC
∣→
AB ×→
AC ∣
= −^
i−2^
j+3^
k⋅
^
i−10^
j+10^
k
(5)2+ (−10)2+ (10)2
=−5+20 +30
√25 +100 +100
=45
√225 =45
15 =3
)( ) ( )
√
GivenTwoverticesofTriangleA(2,4,6)andB(0, −2, −5)andifcentroidG(2,1, −1)
LetThirdverticesbe(x,y,z)
Now 2+0+x
3=2,4−2+y
3=1,6−5+z
3= −1
x=4,y=1,z= −1
ThirdverticesC(4,1, −4)
Now,ImageofverticesC(4,1, −4)inthegivenplaneisD(α,β,γ)
Now
α−4
1=β−1
2=γ+4
4= −2(4+2−16 −11)
1+4+16
-------------------------------------------------------------------------------------------------
Question74
Theshortestdistancebetweenthelines x+2
1=y
−2=z−5
2and
x−4
1=y−1
2=z+3
0is:
[10-Apr-2023shift1]
Options:
A.8
B.7
C.6
D.9
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question75
LetPbethepointofintersectionoftheline x+3
3=y+2
1=1−z
2andthe
planex +y+z=2.IfthedistanceofthepointPfromtheplane
3x −4y +12z =32isq,thenqand2qaretherootsoftheequation:
[10-Apr-2023shift1]
Options:
A.x2+18x −72 =0
B.x2+18x +72 =0
C.x2−18x −72 =0
α−4
1=β−1
2=γ+4
4=42
21 ⇒2
α=6,β=5,γ=4
Then αβ +βγ +γα
= (6×5) + (5×4) + (4×6)
=30 +20 +24
=74
x+2
1=y
−2=z−5
2and x−4
1=y−1
2=z+3
0
=|−54|
−4^
i+2^
j+4k
=54
√16 +4+16
=54
6
=9
| |
D.x2−18x +72 =0
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question76
LettimeimageofthepointP(1,2,6)intheplanepassingthroughthe
pointsA(1,2,0), B(1,4,1)andC(0,5,1)beQ(α,β,γ).Then
(α2+β2+γ2)isequalto
[10-Apr-2023shift2]
Options:
A.70
B.76
C.62
D.65
Answer:D
Solution:
x+3
3=y+2
1=1−z
2=λ
x=3λ −3,y=λ−2,z=1−2λ
P(3λ −3,λ−2,1−2λ)willsatisfytheequationofplanex+y+z=2.
3λ −3+λ−2+1−2λ =2
2λ −4=2
λ=3
P(6,1, −5)
PerpendiculardistanceofPfromplane3x −4y +12z −32 =0is
q=3(6) − 4(1) + 12(−5) − 32
√9+16 +144
q=6
2q =12
Sumofroots = 6+12 =18
Productofroots = 6(12) = 72
∴Quadraticequationhavingqand2qasrootsisx2−18x +72.
| |
EquationofplaneA(x−1) + B(y−2) + C(z−0) = 0
Put(1,4,1) ⇒ 2B +C=0
Put(0,5,1) ⇒ −A+3B +C=0
Sub:B−A=0⇒A=B,C= −2B
1(x−1) + 1(y−2) − 2(z−0) = 0
x+y−2z −3=0
Imageis (α,β,γ)pt ≡ (1,2,6)
α−1
1=β−2
1=γ−6
−2=−2(1+2−12 −3)
6
α−1
1=β−2
1=γ−6
−2=4
α=5,β=6,γ= −2⇒α2+β2+γ2
=25 +36 +4=65
Question77
Lettheline x
1=6−y
2=z+8
5intersectthelines x−5
4=y−7
3=z+2
1and
x+3
6=3−y
3=z−6
1atthepointsAandBrespectively.Thenthedistanceof
themid-pointofthelinesegmentABfromtheplane2x −2y +z=14is
[10-Apr-2023shift2]
Options:
A.3
B. 10
3
C.4
D. 11
3
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question78
LetthefootofperpendicularfromthepointA(4,3,1)ontheplane
P:x−y+2z +3=0beN.IfB(5,α,β),α,β∈ZisapointonplaneP
suchthattheareaofthetriangleABN is3√2,thenα2+β2+αβisequal
to_______.
[10-Apr-2023shift2]
x
1=y−6
−2=z+8
5=λ...(1)
x−5
4=y−7
3=z+2
1=µ...(2)
x+3
6=y−3
−3=z−6
1=γ...(3)
Intersectionof (1)&(2)“A”
(λ, −2λ +6,5λ −8)&(4µ +5,3µ +7,µ−2)
λ=1,µ= −1
A(1,4, −3)
Intersection(1)&(3)B ”"
(λ, −2λ +6,5λ −8)&(6γ −3, −3γ +3,γ+6)
λ=3
γ=1
B(3,0,7)
ModpointofA&B ⇒ (2,2,2)
Perpendiculardistancefromtheplane
2x −2y +z=14
2(2) − 2(2) + 2−14
√4+4+1∣ = 4
Answer:7
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question79
Let(α,β,γ)betheimageofthepointP(2,3,5)intheplane
2x +y−3z =6.Thenα +β+γisequalto:
[11-Apr-2023shift1]
Options:
A.5
B.9
C.10
D.12
Answer:C
Solution:
AN = √6
5−α+2β +3=0
⇒α=8+2β
Nisgivenby
x−4
1=y−3
−1=z−1
2=−(4−3+2+3)
1+1+4
x=3,y=4,z= −1
N
(3,4, −1)
BN =4+ (α−4)2+ (β+1)2
=4+ (2β +4)2+ (β+1)2
Areaof △ABN =1
2AN ×BN =3√2
1
2× √6×BN =3√2
BN =2√3
4+ (2β +4)2+ (β+1)2=12
(2β +4)2+ (β+1)2−8=0
5β2+18β +9=0
(5β +3)(β+3) = 0
β= −3
α=2
α2+β2+αβ =9+4−6=7
√
√
-------------------------------------------------------------------------------------------------
Question80
Iftheequationoftheplanethatcontainsthepoint(−2,3,5)andis
perpendiculartoeachoftheplanes2x +4y +5z =8and3x −2y +3z =5
isαx +βy +γz +97 =0thenα +β+γ=:
[11-Apr-2023shift1]
Options:
A.15
B.18
C.17
D.16
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
α−2
2=β−3
1=γ−5
−3= −22×2+3−3×5−6
22+12+1−32=2
α−2
2=2 β −3=2 γ −5= −6
α=6 β =5 γ = −1
α+β+γ=10
( )
Theequationofplanethrough(−2,3,5)is
a(x+2) + b(y−3) + c(z−5) = 0
itisperpendicularto2x +4y +5z =8&3x −2y +3z =5
∴2a +4b +5c =0
3a −2b +3c =0
∴a
4 5
−23
=−b
2 5
3 3
=c
2 4
3−2
⇒a
22 =b
9=c
−16
∴Equationofplaneis
22(x+2) + 9(y−3) − 16(z−5) = 0
⇒22x +9y −16z +97 =0
Comparingwith αx +βy +γx +97 =0
Weget α +β+γ=22 +9−16 =15
| | | | | |
Question81
Letalinel passthroughtheoriginandbeperpendiculartothelines
11:→
r=^
i−11^
j−7^
k+λ^
i+2^
j+3^
k,λ∈ ℝand
12:→
r= −^
i+^
k+µ2^
i+2^
j+^
k,µ∈ ℝ.
IfPisthepointofintersectionofl andl 1,andQ(α,β,γ)isthefootof
perpendicularfromPonl 2,then9(α+β+γ)isequalto______.
[11-Apr-2023shift1]
Answer:5
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question82
LetPbetheplanepassingthroughthepoints(5,3,0), (13,3, −2)and
(1,6,2).Forα ∈N,ifthedistancesofthepointsA(3,4,α)and
B(2,α,a)fromtheplanePare2and3respectively,thenthepositive
valueofais
[11-Apr-2023shift2]
Options:
A.5
Let ℓ = 0^
i+0^
j+0^
k+γ a^
i+b^
j+c^
k
=γ a^
i+b^
j+c^
k
a^
i+b^
j+c^
k=
^
i^
j^
k
123
221
=^
i(2−6) − ^
j(1−6) + ^
k(2−4)
= −4^
i−5^
j−2^
k
ℓ = γ−4^
i+5^
j−2^
k
Pisintersectionofℓandℓ1
−4γ =1+λ,5γ = −11 +2λ, −2γ = −7+3λ
Bysolvingtheseequationγ= −1,P(4, −5,2)
LetQ(−1+2µ,2µ,1+µ)
→
PQ ⋅2^
i+2^
j+^
k=0
−2+4µ +4µ +1+µ=0
9µ =1
µ=1
9
Q−7
9,2
9,10
9
9(α+β+γ) = 9−7
9+2
9+10
9
=5
( ) ( )
( )
| |
( )
( )
( )
( )
B.6
C.4
D.3
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question83
LetthelinepassingthroughthepointP(2, −1,2)andQ(5,3,4)meet
theplanex −y+z=4atthepointT .ThenthedistanceofthepointR
fromtheplanex +2y +3z +2=0measuredparalleltotheline
x−7
2=y+3
2=z−2
1isequalto
[11-Apr-2023shift2]
Options:
A.3
B.√61
C.√31
D.√189
Answer:A
Solution:
Solution:
^
i^
j^
k
8 0 −2
4−3−2
=^
i(−6) + 8^
j−24^
k
Normaloftheplane = 3^
i−4^
j+12^
k
Plane:3x −4y +12z =3
DistancefromA(3,4,α)
9−16 +12α −3
13 =2
α=3
α= −8 (rejected)
DistancefromB(2,3,a)
6−12 +12a −3
13 =3
a=4
| |
| |
| |
Line: x−5
3=y−3
4=z−4
2=λ
R(3λ +5,4λ +3,2λ +4)
∴3λ +5−4λ −3+2λ +4=4
λ+6=4∴λ= −2
∴R≡ (−1, −5,0)
-------------------------------------------------------------------------------------------------
Question84
Letthelineℓ : x=1−y
−2=z−3
λ,λ∈RmeettheplaneP :x+2y +3z =4at
thepoint(α,β,γ).IftheanglebetweenthelineℓandtheplanePis
cos−15
14 ,thenα +2β +6γisequalto_______.
[11-Apr-2023shift2]
Answer:11
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question85
Letthelines11:x+5
3=y+4
1=z−α
−2and
l2:3x +2y +z−2=0=x−3y +2z −13becoplanar.Ifthepoint
P(a,b,c)onl 1isnearesttothepointQ(−4, −3,2),then|a|+ | b|+|c|
isequalto
[12-Apr-2023shift1]
Options:
A.10
(√)
Line: x+1
2=y+5
2=z−0
1=µ
PointT= (2µ −1,2µ −5,µ)
Itliesonplane
2µ −1+2(2µ −5) + 3µ +2=0
µ=1
∴T= (1, −3,1)
∴RT =3
ℓ : x=y−1
2=z−3
λ,λ∈ℝ
Dr'soflineℓ(1,2,λ)
Dr'sofnormalvectorofplaneP:x+2y +3z =4are(1,2,3)
Now,anglebetweenlineℓandplanePisgivenby
sin θ =1+4+3λ
5+λ2⋅ √14 =3
√14 given cos θ =5
14
⇒λ=2
3
Letvariablepointonlineℓis t,2t +1,2
3t+3
lineofplaneP.
⇒t= −1
⇒ −1, −1,7
3≡ (α,β,γ)
⇒α+2β +6γ =11
|√|(√)
( )
( )
B.8
C.12
D.14
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question86
LettheplaneP:4x −y+z=10berotatedbyanangle π
2aboutitslineof
intersectionwiththeplanex +y−z =4.Ifαisthedistanceofthepoint
(2,3, −4)fromthenewpositionoftheplaneP,then35αisequalto
[12-Apr-2023shift1]
Options:
A.90
B.105
C.85
D.126
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
(3x +2y +z−2) + µ(x−3y +2z −13) = 0
3(3+µ) + 1⋅ (2−3µ) − 2(1+2µ) = 0
9−4µ =0
µ=9
4
4(−15 −8+α−2) + 9(−5+12 +2α −13) = 0
−100 +4α −54 +18α =0
⇒α=7
Let P ≡ (3λ −5,λ−4, −2λ +7)
DirectionratioofPQ(3λ −1,λ−1, −2λ +5)
ButPQ ⟂ℓ1
⇒3(3λ −1) + 1⋅ (λ−1) − 2(−2λ +5) = 0
⇒λ=1
P(−2, −3,5)⇒ | a|+|b|+|c| = 10
Letequationinnewpositionis
(4x −y+z−10) + λ(x+y−z−4) = 0
4(4+λ) − 1⋅ (−1+λ) + 1⋅ (1−λ) = 0
⇒λ= −9
Soequationinnewpositionis
−5x −10y +10z +26 =0
⇒α=54
15
Question87
Lettheequationofplanepassingthroughthelineofintersectionofthe
planesx +2y +az =2andx −y+z=3be5x−11y +bz =6a −1.For
c∈Z ,ifthedistanceofthisplanefromthepoint(a, −c,c)is 2
√a,then
a+b
cisequalto
[13-Apr-2023shift1]
Options:
A.-4
B.2
C.-2
D.4
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question88
Thedistanceofthepoint(−1,2,3)fromtheplane→
r⋅^
i−2^
j+3^
k=10
paralleltothelineoftheshortestdistancebetweenthelines
→
r=^
i−^
j+λ 2^
i+^
kand→
r=2^
i−^
j+µ^
i−^
j+^
kis
[13-Apr-2023shift1]
Options:
A.2√5
B.3√5
C.3√6
D.2√6
Answer:D
Solution:
( )
( ) ( ) ( ) ( )
(x+2y +az −2) + λ(x−y+z−3) = 0
1+λ
5=2−λ
−11 =a+λ
b=2+3λ
6a −1
λ= − 7
2,a=3,b=1
2
√a=5a +11c +bc −6a +1
√25 +121 +1
c= −1
∴a+b
c=3+1
−1= −4
| |
Solution:
-------------------------------------------------------------------------------------------------
Question89
Lettheimageofthepoint 5
3,5
3,8
3intheplanex −2y +z−2=0beP.
IfthedistanceofthepointQ(6, −2,α ), α>0,fromPis13,thenαis
equalto________.
[13-Apr-2023shift1]
Answer:15
Solution:
-------------------------------------------------------------------------------------------------
Question90
( )
LetL1:→
r=^
i−^
j+λ 2^
i+^
k
L2:→
r=2^
i−^
j+µ^
i−^
j+^
k
→
n=
^
i^
j^
k
201
1−1 1
→
n=^
i−^
j−2^
k
EquationoflinealongshortestdistanceofL1andL2
x+1
1=y−2
−1=z−3
−2=r
⇒ (x,y,z) ≡ (r−1,2−r,3−2r)
⇒ (r−1) − 2(2−r) + 3(3−2r) = 10
⇒r= −2
⇒Q(x,y,z) ≡ (−3,4,7)
⇒PQ = √4+4+16 =2√6
( ) ( )
( ) ( )
| |
Imageofpoint 5
3,5
3,8
3
x−5
3
1=
y−5
3
−2=
z−8
3
1=
−2 1 ×5
3+ (−2) × 8
3+1×8
3−2
12+22+12
=1
3
∴x=2,y=1,z=3
132= (6−2)2+ (−2−1)2+ (α−3)2
⇒(α−3)2=144 ⇒α=15(∵α>0)
( )
( )
Theplane,passingthroughthepoints(0, −1,2)and(−1,2,1)and
paralleltothelinepassingthrough(5,1, −7)and(1, −1, −1),also
passesthroughthepoint
[13-Apr-2023shift2]
Options:
A.(0,5, −2)
B.(−2,5,0)
C.(2,0,1)
D.(1, −2,1)
Answer:B
Solution:
-------------------------------------------------------------------------------------------------
Question91
Theline,thatiscoplanartotheline x+3
−3=y−1
1=z−5
5,is
[13-Apr-2023shift2]
Options:
A. x+1
−1=y−2
2=z−5
5
B. x+1
1=y−2
2=z−5
5
C. x−1
−1=y−2
2=z−5
4
D. x+1
−1=y−2
2=z−5
4
Answer:A
Solution:
Planepassingthrough(0, −1,0)and(−1,2,1)
Thenvectorinplane⟨−1,3, −1⟩vectorparalleltoplaneis⟨4,2, −6⟩
Normalvectortoplane →
n=
^
i^
j^
k
−13−1
4 2 −6
=^
i(16) − ^
j(10) + ^
k(−14)
→
n= ⟨8,5,7⟩
Equationofplane
8(x−0) + 5(y+1) + 7(z−2) = 0
⇒8x +5y +7z =9
Fromgivenoptionspoint(−2,5,0)liesonplane.
( ) | |
-------------------------------------------------------------------------------------------------
Question92
LetNbethefootofperpendicularfromthepointP(1, −2,3)ontheline
passingthroughthepoints(4,5,8)and(1, −7.5).Thenthedistanceof
Nfromtheplane2x −2y +z+5=0is
[13-Apr-2023shift2]
Options:
A.6
B.7
C.9
D.8
Answer:B
Solution:
Conditionofco-planarity
x2−x1a1a2
y2−y1b1b2
z2−z1c1c2
=0
Wherea1,b1,c1aredirectioncosineof1st lineanda2,b2,c2aredirectioncosineof2nd line.Now.Solvingoptions
Point(−3,1,5)&point(−1,2,5)
(1)
−3 1 5
1 2 5
−2−1 0
= −3(5) − (10) + 5(−1+4)
= −15 −10 +15 = −10
(2)point(−1,2,5)
−3 1 5
−1 2 5
−2−1 0
=3(5) − (10) + 5(1+4)
−25 +5=0
(3)point(−1,2,5)
−3 1 5
−1 2 4
−2−1 0
−3(4) − (8) + 5(1+4)
−12 −8+25 =5
(4)point(−1,2,5)
−315
−125
4 1 0
−3(−5) − (−20) + 5(−1−8)
15 +20 −45 = −10
| |
| |
| |
| |
| |
-------------------------------------------------------------------------------------------------
Question93
LetthefootofperpendicularofthepointP(3, −2, −9)ontheplane
passingthroughthepoints(−1, −2, −3), (9,3,4), (9, −2,1)be
Q(α,β,γ).ThenthedistanceofQfromtheoriginis
[15-Apr-2023shift1]
Options:
A.√29
B.√38
C.√42
D.√35
Answer:C
Solution:
→
PN = ⟨λ,4λ −5,λ+2⟩
→
PN. < 1,4,1> = 0
⇒λ+16λ −20 +λ+2=0
⇒λ=1
N(2, −3,6)
DistanceofNfrom2x −2y +z+5=0is
d=2(2) − 2(−3) + 6+5
22+ (−2)2+ (1)2
=21
3=7
|√|
| |
P(3, −2, −9)
EquationofplanethroughA,B,C.
x+1 y +2 z +3
10 5 7
10 0 4
=0
2x +3y −5z −7=0
FootofIrofP(3, −2, −9)is
x−3
2=y+2
3=z+9
−5= − (6−6+45 −7)
4+9+25
x−3
2=y+2
3=z+9
−5= −1
| |
-------------------------------------------------------------------------------------------------
Question94
LetSbethesetofallvaluesofλ,forwhichtheshortestdistance
betweenthelines x−λ
0=y−3
4=z+6
1and x+λ
3=y
−4=z−6
0is13.Then
8∑
λ∈Sλ isequalto
[15-Apr-2023shift1]
Options:
A.302
B.306
C.304
D.308
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question95
Ifthelinex =y=zintersectstheline
xsin A +ysin B +zsin C −18 =0=xsin 2A +ysin 2B +zsin 2C −9,where
A,B,CaretheanglesofatriangleABC,then80 sin A
2sin B
2sin C
2is
equalto________
[15-Apr-2023shift1]
| |
( )
Q(1, −5, −4) ≡ (α,β,γ)
OQ =α2+β2+γ2= √42
√
Shorttestdistance =
0 4 1
3−40
2λ 3 −12
∣
^
i^
j^
k
041
3−40
∣
13 =|153 +8λ|
4^
i+3^
j−12^
k
=|153 +8λ|
13
|153 +8λ | = 169
153 +8λ =169, −169
λ=16
8,−322
8
8∑
λ∈S
λ=306
| |
| |
| |
| |
Answer:5
Solution:
-------------------------------------------------------------------------------------------------
Question96
LettheplanePcontaintheline2x +y−z−3=0=5x −3y +4z +9and
beparalleltotheline x+2
2=3−y
−4=z−7
5Thenthedistanceofthepoint
A(8, −1, −19)fromtheplanePmeasuredparalleltotheline
x
−3=y−5
4=2−z
−12 isequalto_______
[15-Apr-2023shift1]
Answer:26
Solution:
sin A +sin B +sin C =18
x
sin 2A +sin 2B +sin 2C =9
x
∴sin A +sin B +sin C =2(sin 2A +sin 2B +sin 2C)
4 cos A ∕2 cos B ∕2 cos C ∕2=2(4sin Asin Bsin C)
16sin A ∕2sin B ∕2sin C ∕2=1
HenceAns. =5.
Plane ≡P1=λP2=0
(2x +y−z−3) + λ(5x −3y) + 4z +9) = 0
(5λ +2)x+ (1−3λ)y+ (4λ −1)z+9λ −3=0
→
n⋅→
b=0 where →
b(2,4,5)
2(5λ +2) + 4(1−3λ) + 5(4λ −1) = 0
λ= − 1
6
Plane 7x +9y −10z −27 =0
EquationoflineABis
x−8
−3=y+1
4=z+19
12 =λ
Let B = (8−3λ, −1+4λ, −19 +12λ)liesonplane P
∴7(8−3λ) + 9(4λ −1) − 10(12λ −19) = 27
λ=2
∴Point B = (2,7,5)
-------------------------------------------------------------------------------------------------
Question97
Iftheshortestdistancebetweenthelines x−1
2=y−2
3=z−3
λand
x−2
1=y−4
4=z−5
5is 1
√3,thenthesumofallpossiblevalueofλis:
[24-Jun-2022-Shift-2]
Options:
A.16
B.6
C.12
D.15
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question98
LetthepointsontheplanePbeequidistantfromthepoints(−4,2,1)
and(2, −2,3).ThentheacuteanglebetweentheplanePandtheplane
2x +y+3z =1is
[24-Jun-2022-Shift-2]
Options:
A. π
6
AB =62+82+242=26
√
B. π
4
C. π
3
D. 5π
12
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question99
LetQbethemirrorimageofthepointP(1,0,1)withrespecttothe
planeS :x+y+z=5.IfalineLpassingthrough(1, −1, −1),parallelto
thelinePQmeetstheplaneSatR,thenQR2isequalto:
[25-Jun-2022-Shift-1]
Options:
A.2
B.5
C.7
D.11
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question100
Letthelines
L1:→
r=λ∧
i+2∧
j+3 widehat k ,λ∈R
L2:→
r=∧
i+3∧
j+widehat k +µ∧
i+∧
j+5 widehat k ;µ∈R
intersectatthepointS.Ifaplaneax +by −z+d=0passesthroughS
andisparalleltoboththelinesL1andL2,thenthevalueofa +b+d is
equalto
[25-Jun-2022-Shift-1]
Answer:5
Solution:
-------------------------------------------------------------------------------------------------
Question101
( )
( ) ( )
Letpbetheplanepassingthroughtheintersectionoftheplanes
→
r⋅∧
i+3∧
j−∧
k=5and→
r⋅2∧
i−∧
j+∧
k=3,andthepoint(2,1, −2).Letthe
positionvectorsofthepointsX andY be∧
i−2∧
j+4∧
kand5∧
i−∧
j+2∧
k
respectively.Thenthepoints
[25-Jun-2022-Shift-2]
Options:
A.X andX +Y areonthesamesideofP
B.Y andY −X areontheoppositesidesofP
C.X andY areontheoppositesidesofP
D.X +Y andX −Y areonthesamesideofP
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question102
LetI1bethelineinxy-planewithxandyintercepts 1
8and 1
4√2
respectively,andI2bethelineinzx-planewithxandzintercepts−1
8
and−1
6√3respectively.Ifd istheshortestdistancebetweenthelineI1
andI2,thend−2isequalto____
[25-Jun-2022-Shift-2]
Answer:51
Solution:
Solution:
( ) ( )
Lettheequationofrequiredplane
π: (x+3y −z−5) + λ(2x −y+z−3) = 0
∵(2,1, −2)liesonitso,2+λ(−2) = 0
⇒λ=1
Hence,π:3x +2y −8=0
∵πx= −9,πy=5,πx+y=4
πx−y= −22andπy−x=6
ClearlyXandYareonoppositesidesofplaneπ
x−1
8
1
8
=y
−1
4√2
=z
0L1
-------------------------------------------------------------------------------------------------
Question103
Ifthetwolinesl 1:x−2
3=y+1
−2,z=2andl 2:x−1
1=2y +3
α=z+5
2are
perpendicular,thenananglebetweenthelinesI2and
l3:1−x
3=2y −1
−4=z
4is:
[26-Jun-2022-Shift-1]
Options:
A.cos−129
4
B.sec−129
4
C.cos−12
29
D.cos−12
√29
Answer:B
Solution:
( )
( )
( )
( )
or
x−1
8
1=y
−√2=z
0......(i)
EquationofL2
x+1
8
−6√3=y
0=z
8......(ii)
d=
→
c−→
a⋅→
(b×→
d
→
b×→
d
=
1
4
∧
i⋅4√2
∧
i+4
∧
j+3 6
∧
k
(4√2)2+42+ (3√6)2
=√2
√32 +16 +54 =1
√51
d−2=51
|( ) )
| | |
( ) (√)
√
Question104
Lettheplane2x +3y +z+20 =0berotatedthrougharightangleabout
itslineofintersectionwiththeplanex −3y +5z =8.Ifthemirrorimage
ofthepoint 2, − 1
2,2 intherotatedplaneisB(a,b,c),then:
[26-Jun-2022-Shift-1]
Options:
A. a
8=b
5=c
−4
B. a
4=b
5=c
−2
C. a
8=b
−5=c
4
D. a
4=b
5=c
2
Answer:A
Solution:
( )
Question105
Iftheplane2x +y−5z =0isrotatedaboutitslineofintersectionwith
theplane3x −y+4z −7=0byanangleof π
2,thentheplaneafterthe
rotationpassesthroughthepoint:
[26-Jun-2022-Shift-2]
Options:
A.(2, −2,0)
B.(−2,2,0)
C.(1,0,2)
D.(−1,0, −2)
Answer:C
Solution:
Question106
Ifthelines→
r=∧
i−∧
j+∧
k+λ 3∧
j−∧
kand→
r=α∧
i−∧
j+µ 2∧
i−3∧
kareco-
planar,thenthedistanceoftheplanecontainingthesetwolinesfrom
thepoint(α,0,0)is:
[26-Jun-2022-Shift-2]
Options:
A. 2
9
B. 2
11
C. 4
11
D.2
Answer:B
Solution:
Solution:
( ) ( ) ( ) ( )
Question107
Iftwostraightlineswhosedirectioncosinesaregivenbytherelations
l+m−n=0,3l 2+m2+cnl =0areparallel,thenthepositivevalueofc
is:
[27-Jun-2022-Shift-1]
Options:
A.6
B.4
C.3
D.2
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question108
Letthemirrorimageofthepoint(a,b,c)withrespecttotheplane
3x −4y +12z +19 =0be(a−6,β,γ).Ifa +b+c=5,then7β−9γis
equalto
[27-Jun-2022-Shift-1]
Answer:137
Solution:
l+m−n=0⇒n=l+m
3l 2+m2+cnl =0
3l 2+m2+cl (l+m) = 0
= (3+c)l2+cl m +m2=0
= (3+c)l
m
2+cl
m+1=0
∴Linesareparallel
D=0
c2−4(3+c) = 0
c2−4c −12 =0
(c−4)(c+3) = 0
c=4(asc>0)
( ) ( )
-------------------------------------------------------------------------------------------------
Question109
Letthefootoftheperpendicularfromthepoint(1,2,4)ontheline
x+2
4=y−1
2=z+1
3beP.ThenthedistanceofPfromtheplane
3x +4y +12z +23 =0is
[27-Jun-2022-Shift-2]
Options:
A.5
B. 50
13
C.4
D. 63
13
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
x−a
3=y−b
−4=z−c
12 =−2(3a −4b +12c +19)
32+ (−4)2+122
x−a
3=y−b
−4=z−c
12 =−6a +8b −24c −38
169
(x,y,z) ≡ (a−6,β,γ)
(a−6) − a
3=β−b
−4=γ−c
12 =−6a +8b −24c −38
169
β−b
−4= −2⇒β=8+b
γ−c
12 = −2⇒γ= −24 +c
−6a +8b −24c −38
169 = −2
⇒3a −4b +12c =150
a+b+c=5
3a +3b +3c =15
Applying(1)-(2)
−7b +9c =135
7b −9c = −135
7β −9γ =7(8+b) − 9(−24 +c)
=56 +216 +7b −9c
=56 +216 −135 =137
L:x+2
4=y−1
2=z+1
3=t
LetP= (4t −2,2t +1,3t −1)
∵Pisthefootofperpendicularof(1,2,4)
∴4(4t −3) + 2(2t −1) + 3(3t −5) = 0
⇒29t =29 ⇒t=1
∴P= (2,3,2)
Now,distanceofPfromtheplane
3x +4y +12z +23 =0,is
6+12 +24 +23
√9+16 +144 =65
13 =5
||
Question110
Theshortestdistancebetweenthelines x−3
2=y−2
3=z−1
−1and
x+3
2=y−6
1=z−5
3,is
[27-Jun-2022-Shift-2]
Options:
A. 18
√5
B. 22
3√5
C. 46
3√5
D.6√3
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question111
IftwodistinctpointQ,Rlieonthelineofintersectionoftheplanes
−x+2y −z=0and3x −5y +2z =0andPQ =PR = √18wherethepoint
Pis(1, −2,3),thentheareaofthetrianglePQRisequalto
[28-Jun-2022-Shift-1]
Options:
A. 2
3√38
B. 4
3√38
C. 8
3√38
D. 152
3
√
L1:x−3
2=y−2
3=z−1
−1
L2:x+3
2=y−6
1=z−5
3
Now, →
p×→
q=begin array ccc
∧
i
∧
j
∧
k
2 3 −1
2 1 3 end array =10
∧
i−8
∧
j−4
∧
k
and →
a2−→
a1=6
∧
i−4
∧
j−4
∧
k
∴S.D. = 60 +32 +16
√100 +64 +16 =108
√180 =18
√5
|
|
||
Answer:B
Solution:
-------------------------------------------------------------------------------------------------
Question112
TheacuteanglebetweentheplanesP1andP2,whenP1andP2arethe
planespassingthroughtheintersectionoftheplanes
5x +8y +13z −29 =0and8x −7y +z−20 =0andthepoints(2,1,3)
and(0,1,2),respectively,is
[28-Jun-2022-Shift-1]
Options:
A. π
3
B. π
4
C. π
6
D. π
12
Answer:A
Solution:
LineLisx=y=z
→
PQ ⋅∧
i+∧
j+∧
k=0
⇒(α−3) + α+2+α−1=0
⇒α=2
3so, T =2
3,2
3,2
3
PT =38
3
⇒QT =4
√3
So,Area =1
2×4
√3×√38
√3⋅2
=4√38
3sq.units
( )
( )
√
( )
FamilyofPlane'sequationcanbegivenby
(5+8λ)x+ (8−7λ)y+ (13 +λ)z− (29 +20λ) = 0
P1passesthrough (2,1,3)
-------------------------------------------------------------------------------------------------
Question113
LettheplaneP :→
r⋅→
a=d containthelineofintersectionoftwoplanes
→
r⋅∧
i+3∧
j−∧
k=6and→
r⋅ −6∧
i+5∧
j−∧
k=7IftheplanePpassesthrough
thepoint 2,3,1
2,thenthevalueof 13→
a2
d2isequalto
[28-Jun-2022-Shift-1]
Options:
A.90
B.93
C.95
D.97
Answer:B
Solution:
( ) ( )
( ) | |
⇒(10 +16λ) + (8−7λ) + (39 +3λ) − (29 +20λ) = 0
⇒−8λ +28 =0⇒λ=7
2
d.r,sofnormalto P1
33,−33
2,33
2or 1, − 1
2,1
2
P2passesthrough (0,1,2)
⇒8−7λ +26 +2λ − (29 +20λ) = 0
⇒5−25λ =0
⇒λ=1
5
d.r,sofnormalto P2
33
5,33
5,66
5or ⟨1,1,2⟩
Anglebetweennormals
=
∧
i−1
2
∧
j+1
2
∧
k⋅∧
i+∧
j+2
∧
k
√3
2
cos θ =
1−1
2+1
3=1
2
θ=π
3
⟨ ⟩ ⟨ ⟩
⟨ ⟩
( ) ( )
P1:x+3y −z=6
P2: −6x +5y −z=7
FamilyofplanespassingthroughlineofintersectionofP1andP2isgivenby
x(1−6λ) + y(3+5λ) + z(−1−λ) − (6+7λ) = 0
Itpassesthrough 2,3,1
2
So, 2(1−6λ) + 3(3+5λ) + 1
2(−1−λ) − (6+7λ) = 0
⇒2−12λ +9+15λ −1
2−λ
2−6−7λ =0
⇒9
2−9λ
2=0⇒λ=1
( )
-------------------------------------------------------------------------------------------------
Question114
Lettheplaneax +by +cz =d passthrough(2,3, −5)andis
perpendiculartotheplanes
2x +y−5z =10and3x +5y −7z =12.Ifa,b,c,d areintegersd >0
andgcd (|a|, | b|,|c| ,d) = 1,thenthevalueofa +7b +c+20d is
equalto:
[28-Jun-2022-Shift-2]
Options:
A.18
B.20
C.24
D.22
Answer:D
Solution:
-------------------------------------------------------------------------------------------------
Question115
LettheimageofthepointP(1,2,3)inthelineL :x−6
3=y−1
2=z−2
3beQ.
LetR(α,β,γ)beapointthatdividesinternallythelinesegmentPQin
theratio1 :3.Thenthevalueof22(α+β+γ)isequalto___
[28-Jun-2022-Shift-2]
Options:
Requiredplaneis
−5x +8y −2z −13 =0
Or →
r⋅ −5
∧
i+8
∧
j−2
∧
k=13
13→
a2
|d|2=132
(13)2⋅→
a|2=93
( )
| | |
Equationofpanethroughpoint(2,3, −5)andperpendiculartoplanes2x +y−5z =10and3x +5y−7z =12is
x−2 y −3 z +5
2 1 −5
3 5 −7
=0
∴Equationofplaneis(x−2)(−7+25) − (y−3)
(−14 +15) + (z+5) ⋅ 7=0
∴18x −y+7z +2=0
⇒18x −y+7z = −2
∴−18x +y−7z =2
Oncomparingwithax +by +cz =dwhered>0isa= −18,b=1,c= −7,d=2
∴a+7b +c+20d =22
| |
Answer:125
Solution:
-------------------------------------------------------------------------------------------------
Question116
Ifthemirrorimageofthepoint(2,4,7)intheplane3x −y+4z =2is
(a,b,c),then2a +b+2cisequalto:
[29-Jun-2022-Shift-1]
Options:
A.54
B.50
C.−6
D.−42
Answer:C
Solution:
Solution:
ThepointdividingPQintheratio1:3willbemid-pointofP&footofperpendicularfromPontheline.
∴Letapointonlinebeλ
⇒x−6
3=y−1
2=z−2
3=λ
⇒P′(3λ +6,2λ +1,3λ +2)
asP′isfootofperpendicular
(3λ +5)3+ (2λ −1)2+ (3λ −1)3=0
⇒22λ +15 −2−3=0
⇒λ=−5
11
∴P′51
11,1
11,7
11
Mid-pointof PP′≡
51
11 +1
2,
1
11 +2
2,
7
11 +3
2
≡62
22,23
22,40
22 ≡ (α,β,γ)
⇒22(α,β,γ) = 62 +23 +40 =125
( )
( )
( )
Weknowmirrorimageofpoint(x1,y1,z1)intheplaneax +by +cz =d
x−x1
a=y−y1
b=z−z1
c=−2(ax1+by1+cz1−d)
a2+b2+c2
Heregivenpoint(2,4,7)andplane3x −y+4z =2thenmirrorimageis
x−2
3=y−4
−1=z−7
4=−2(6−4+28 −2)
9+1+16
⇒x−2
3=y−4
−1=z−7
4= − 28
13
∴x= − 58
13 =a
y=80
13 =b
z= − 21
13 =c
∴2a +b+2c
-------------------------------------------------------------------------------------------------
Question117
Letd bethedistancebetweenthefootofperpendicularsofthepoints
P(1,2, −1)andQ(2, −1,3)ontheplane−x+y+z=1.Thend 2isequal
to___
[29-Jun-2022-Shift-1]
Answer:26
Solution:
-------------------------------------------------------------------------------------------------
Question118
LetP1:→
r⋅2∧
i+∧
j−3∧
k=4beaplane.LetP2beanotherplanewhich
passesthroughthepoints(2, −3,2), (2, −2, −3)and(1, −4,2).Ifthe
directionratiosofthelineofintersectionofP1andP2be16,α,β,then
thevalueofα +βisequalto____
[29-Jun-2022-Shift-1]
Answer:28
Solution:
( )
=2−58
13 +80
13 +2−21
13
=−116 +80 −42
13 =−78
13 = −6
( ) ( )
FootofperpendicularfromP
x−1
−1=y−2
1=z+1
1=−(−1+2−1−1)
3
⇒p′≡2
3,7
3,−2
3
andfootofperpendicularfromQ
x−2
−1=y+1
1=z−3
1=−(−2−1+3−1)
3
⇒Q′≡5
3,−2
3,10
3
P′Q′=(1)2+ (3)2+ (4)2=d= √26
⇒d2=26
( )
( )
√
DirectionratioofnormaltoP1≡<2,1, −3>
-------------------------------------------------------------------------------------------------
Question119
Let x−2
3=y+1
−2=z+3
−1lieontheplanepx −qy +z=5,forsomep,q∈R.
Theshortestdistanceoftheplanefromtheoriginis:
[29-Jun-2022-Shift-2]
Options:
A. 3
109
B. 5
142
C. 5
√71
D. 1
√142
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question120
LetQbethemirrorimageofthepointP(1,2,1)withrespecttothe
planex +2y +2z =16.LetT beaplanepassingthroughthepointQand
containstheline→
r= −∧
k+λ∧
i+∧
j+2∧
k,λ∈R.Then,whichofthe
followingpointsliesonT?
[29-Jun-2022-Shift-2]
√
√
( )
andthatofP2≡
∧
i
∧
j
∧
k
0 1 −5
−1−2 5
= −5
∧
i−∧
j(−5) + ∧
k(1)
i.e.⟨ −5,5,1>
d.r'soflineofintersectionarealongvector
∧
i
∧
j
∧
k
2 1 −3
−5 5 1
=∧
i(16) − ∧
j(−13) + ∧
k(15)
i.e.<16,13,15>
∴α+β=13 +15 =28
| |
| |
(2, −1, −3)satisfythegivenplane.
So2p +q=8
Alsogivenlineisperpendiculartonormalplaneso3p +2q −1=0
⇒p=15,q= −22
Eq.ofplane15x −22y +z−5=0
itsdistancefromorigin = 6
√710 =5
142
√
Options:
A.(2,1,0)
B.(1,2,1)
C.(1,2,2)
D.(1,3,2)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question121
Letalinehavingdirectionratios1, −4,2intersectthelines
x−7
3=y−1
−1=z+2
1and x
2=y−7
3=z
1atthepointsAandB.Then(AB)2is
equalto__
[24-Jun-2022-Shift-1]
Answer:84
Solution:
-------------------------------------------------------------------------------------------------
ImageofP(1,2,1)inx+2y +2z −16 =0
isgivenbyQ(4,8,7)
Eq.ofplaneT=
x y z+1
4 8 6
1 1 2
=0
⇒2x −z=1soB(1,2,1)liesonit.
| |
Question122
Letαbetheanglebetweenthelineswhosedirectioncosinessatisfythe
equationsI +m−n=0andI 2+m2−n2=0.Then,thevalueof
sin4α+cos4αis
[25Feb2021Shift1]
Options:
A. 3
4
B. 3
8
C. 5
8
D. 1
2
Answer:C
Solution:
:
Given,l+m−n=0...(i)
andI2+m2−n2=0...(ii)
OnsquaringEq.(i),weget
(l+m)2=n2
⇒l2+m2+2l m =n2...(iii)
FromEqs.(ii)and(iii),
I2+m2−n2=0
I2+m2+2l m =n2
− − − −
−n2−2l m = −n2
⇒2l m =0⇒I m =0
⇒I=0 or m =0
CaseIWhenI=0
⇒0+m−n=0
⇒m=n
and I 2+m2+n2=1
⇒m2+m2=1[ ∵n=mandl=0]
⇒m2=1
2
m= ± 1
√2=n
∴ (I,m,n) = 0,1
√2,1
√2or 0,−1
√2,−1
√2
CaseIIWhenm=0
then,l+m−n=0
⇒I=nandl2+m2+n2=1
[ ∵n=Iandm=0]
⇒I2+0+I2=1
I= ± 1
√2[ ∵n=landm=0]
⇒∴ (I,m,n) = 1
√2,0,1
√2or −1
√2,0,−1
√2
⇒a=0,1
√2,1
√2andb=1
√2,0,1
√2
Thencos α =a⋅b
|a||b|=1
2
∴sin α = ± √3
2
( ) ( )
( ) ( )
( ) ( )
-------------------------------------------------------------------------------------------------
Question123
Aline'l 'passingthroughoriginisperpendiculartothelines
l1:r= (3+t)∧
i+ (−1+2t)∧
j+ (4+2t)∧
k
l2:r(3+2s)∧
i+ (3+2s)∧
j+ (2+s)∧
k
Ifthecoordinatesofthepointinthefirstoctanton'I 2'atadistanceof
√17fromthepointofintersectionof'l 'and'l 1'are(a,b,c),then
18(a+b+c)isequalto............
[25Feb2021Shift2]
Answer:44
Solution:
-------------------------------------------------------------------------------------------------
Question124
Theequationofthelinethroughthepoint(0,1,2)andperpendicularto
Now,cos4α+sin4α=1
16 +9
16 =10
16 =5
8
LetL1⇒x−3
1=y−3
2=z−4
2=u(say)
⇒Directionratiosof L1=1,2,2
L2⇒x−3
2=y−3
2=z−2
1=v (say)
DirectionratiosofL2=2,2,1
LineLpassingthroughoriginisperpendiculartoL1andL2.
Hence,directionratiosofLisparallelto(L1×L2).
⇒ (−2,3, −2)
EquationofL⇒x
2=y
−3=z
2=λ(say)
SolveLandL1,weget
(2λ, −3λ,2λ) = (µ+3,2µ −1,2µ +4)
Gives,λ=1,µ= −1
So,intersectionpointP(2, −3,2).
LetQ(2v +3,2v +3,v+2)berequiredpointonL2.
Now,PQ = √17(given)
Now, PQ = √17 (given)
PQ =(2v +1)2+ (2v +6)2+ (v)2
= √17
⇒(2v +1)2+ (2v +6)2+v2=17 (squaringonbothsides)
⇒9v2+28v +20 =0
Onsolving,wegetv= −2(rejected), −10
9(accepted)
∴Qis 7
9,7
9,8
9
∴18(a+b+c) = 18 7
9+7
9+8
9=44
√
( )
( )
theline x−1
2=y+1
3=z−1
−2is
[25Feb2021Shift1]
Options:
A. x
3=y−1
4=z−2
3
B. x
3=y−1
−4=z−2
3
C. x
3=y−1
4=z−2
−3
D. x
−3=y−1
4=z−2
3
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question125
Letλbeaninteger.Iftheshortestdistancebetweenthelines
x−λ=2y −1= −2zandx =y+2λ =z−λis √7
2√2,thenthevalueof|λ|is
..........
[24Feb2021Shift2]
Answer:1
Solution:
Given,line⇒ x−1
2=y+1
3=z−1
−2=λ(let)
AnypointonthislineisB(2λ +1,3λ −1, −2λ +1)anddirectionratiosofthisline = (2,3, −2) = d1
LetgivenpointbeA(0,1,2).
Thendirectionratioof
AB = (2λ +1,3λ −2, −2λ −1) = d2
∵Bothlinesareperpendiculartoeachother.
∴d1⋅d2=0
2(2λ +1) + 3(3λ −2) − 2(−2λ −1) = 0
⇒4λ +2+9λ −6+4λ +2=0
⇒17λ =2
λ λ =2∕17
∴Directionratioofrequiredline d 2= (21, −28, −21)
= (3, −4, −3) = (−3,4,3)
ThislinepassesthroughA(0,1,2).
∴Requiredequationofline⇒ x−0
−3=y−1
4=z−2
3
Given,equationofline⇒x−λ=2y −1=
-------------------------------------------------------------------------------------------------
Question126
Leta,b∈R.IfthemirrorimageofthepointP(a,6,9)withrespectto
theline x−3
7=y−2
5=z−1
−9is(20,b, −a−9),then|a+b|isequalto
[24Feb2021Shift2]
Options:
A.88
B.86
C.84
D.90
Answer:A
Solution:
Solution:
⇒x−λ
1=y−1∕2
1
2
=z
−1
2
or x−λ
2=y−1∕2
1=z
−1
Pointonthislinethroughwhichitpassesis λ,1
2,0.
Equationofanotherline⇒x=y+2λ =z−λ
⇒x
1=y− (−2λ)
1=z−λ
1...(ii)
Apointthroughwhichthislinepassesis(0, −2λ,λ).
Now,distancebetweentwoskewlines
=|(a2−a1) ⋅ (b1×b2)|
|b1×b2|
Accordingtothequestion,
−5λ −3
2
√14 =√7
2√2
⇒ | 10λ +3| = 7
10λ +3= ±7
⇒10λ =4, −10
⇒λ=2
5and λ = −1
∴λ= −1
( λ =2
5isnotpossibleas λ isaninteger)
∴
Hence, |λ| = | −1| = 1
( )
| |
Given,P(a,6,9)
Equationofline x−3
7=y−2
5=z−1
−9
ImageofpointPwithrespecttolineispointQ(20,b, −a−9)
Mid-pointofPandQ=a+20
2,6+b
2,−a
2
Thispointliesonline
∴
a+20
2−3
7=
6+b
2−2
5=
−a
2−1
−9
⇒a+14
14 =b+2
10 =a+2
18
( )
-------------------------------------------------------------------------------------------------
Question127
ConsiderthethreeplanesP1:3x +15y +21z =9,P2:x−3y −z=5and
P3:2x +10y +14z =5Then,whichoneofthefollowingistrue?
[26Feb2021Shift1]
Options:
A.P1andP2areparallel
B.P1andP3areparallel
C.P2andP3areparallel
D.P1,P2andP3allareparallel
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question128
Ifthemirrorimageofthepoint(1,3,5)withrespecttotheplane
4x −5y +2z =8is(α,β,γ),then5(α+β+γ)equals
[26Feb2021Shift2]
Options:
A.47
B.43
C.39
D.41
Answer:A
⇒a+14
14 =a+2
18 and b+2
10 =a+2
18
Solving,wegeta= −56,b= −32
∴ | a+b| = | −56 −32 | = 88
Given,P1⇒3x +15y +21z =9
P2⇒x−3y −z=5
P3⇒2x +10y +14z =5
ConsiderplaneP1,itcanbewrittenas
3x +15y +21z =9 or x +5y +7z =3
Again,considerplaneP3,itcanbewrittenas,
2x +10y +14z =5orx+5y +7z =5∕2
Hence,P1andP3areparallel.
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question129
LetLbealineobtainedfromtheintersectionoftwoplanes
x+2y +z=6andy +2z =4.IfpointP(α,β,γ)isthefootof
perpendicularfrom(3,2,1)onL,thenthevalueof21(α+β+γ)equals
[26Feb2021Shift2]
Options:
A.142
B.68
C.136
D.102
Answer:D
Solution:
Given,pointP(1,3,5)hasthemirrorimageQ(α,β,γ)withrespecttoplane4x −5y +2z =8
Then,Mbethemid-pointoflinejoiningpointPandQ.andMliesongivenplane.
CoordinatesofM(a,b,c)areasfollows
a=α+1
2,b=β+3
2,c=γ+5
2
∵Mliesonplane,then
4a −5b +2c =8
4α+1
2−5β+3
2+2γ+5
2=8...(i)
Also,PQisperpendiculartoplane
⇒α−1
4=β−3
−5=γ−5
2=λ (say)
⇒α=4λ +1,β=3−5λ,γ=2λ +5...(ii)
UseEq.(ii)inEq.(i),weget
2(4λ +2) − 56−5λ
2+2λ +10 =8
⇒8λ +4−15 +25λ
2+2λ +10 =8⇒λ=2
5
∴α=4λ +1=42
5+1=13
5
β=3−52
5=5
5=1
and γ=5+22
5=29
5
∴5(α+β+γ) = 513
5+1+29
5=47
( ) ( ) ( )
( )
( )
( )
( )
( )
-------------------------------------------------------------------------------------------------
Question130
Let(λ,2,1)beapointontheplanewhichpassesthroughthepoint
(4, −2,2).Iftheplaneisperpendiculartothelinejoiningthepoints
(−2, −21,29)and(−1, −16,23),then λ
11
2−4λ
11 −4is..........
[26Feb2021Shift1]
Answer:8
Solution:
Solution:
( )
Given,x+2y +z=6...(i)
and y+2z =4...(ii)
Puty=4−2zfromEq.(ii)Eq.in(i),weget
x+8−4z +z=6
⇒x= −2+3z
⇒x+2
3=z...(iii)
y=4−2z
⇒y−4
−2=z...(iv)
FromEqs.(iii)and(iv),lineofintersectionoftwoplanesis
x+2
3=y−4
−2=z
1=λ
Then,DirectionratiosofPQis
x−3λ −4−2λ,z=λ
(3λ −5, −2λ +2,λ−1)
(3λ −5, −2λ +2,λ−1)
Since,PQisperpendiculartotheline,then
3(3λ −5) − 2(−2λ +2) + 1(λ−1) = 0
∴λ=10
7
∴P16
7,8
7,10
7
Then, 21(α+β+γ) = 21 16
7+8
7+10
7
=21 34
7=3×34 =102
( )
( )
( )
Given(λ,2,1)bepointontheplanewhichpassesthrough(4, −2,2)andplaneisperpendiculartolinejoiningPandQ.
Given,ABisperpendiculartoPQi.e.,AB ⋅PQ =0
(
-------------------------------------------------------------------------------------------------
Question131
AplanepassesthroughthepointsA(1,2,3), B(2,3,1)andC(2,4,2).If
0istheoriginandPis(2, −1,1),thentheprojectionofOPonthis
planeisoflength
[202125FebShift2]
Options:
A. 2
3
B. 2
11
C. 2
7
D. 2
5
Answer:B
Solution:
Solution:
√
√
√
√
Now,AB = (4−λ)∧
i+ (−2−2)∧
j+2−∧
k
= (4−λ)∧
i−4
∧
j+∧
k
PQ = (−1+2)∧
i+ (−16 +21)∧
j+ (23 −29)∧
k
=∧
i+5
∧
j−6
∧
k
= (4−λ)∧
i−4
∧
j+∧
k
PQ = (−1+2)∧
i+ (−16 +21)∧
j+ (23 −29)∧
k
=∧
i+5
∧
j−6
∧
k
Hence,AB ⋅PQ =0
⇒ (4−λ)(1) + (−4)(5) + (1)(−6) = 0
⇒4−λ−20 −6=0
⇒λ= −22
Then, λ
11
2−4λ
11 −4=−22
11
2−4× (−22)
11 −4
=4− (−8) − 4=8
(
( ) ( ) ( ) ( )
Referdiagram,thenormalvectorbenanditisperpendiculartobothABandAC
AB ×AC =n
Now,A(1,2,3), B(2,3,1)andC(2,4,2)
Then,AB = (2−1)∧
i+ (3−2)∧
j+ (1−3)∧
k
=∧
i+∧
j−2
∧
k
−1)∧
i+ (4−2)∧
j+ (2−3)∧
k
AC = (2−1)∧
i+ ( 4
-------------------------------------------------------------------------------------------------
Question132
Thevectorequationoftheplanepassingthroughtheintersectionofthe
planesr ⋅∧
i+j+∧
k=1andr ⋅∧
i−2∧
j= −2andthepoint(1,0,2)is
[24Feb2021Shift2]
Options:
A.r ⋅∧
i+7∧
j+3∧
k=7
3
B.r⋅3∧
i+7∧
j+3∧
k=7
C.r ⋅∧
i+7∧
j+3∧
k=7
D.r ⋅∧
i−7∧
j+3∧
k=7
3
Answer:C
Solution:
( ) ( )
()
( )
( )
()
=∧
i+2
∧
j−∧
k
Now,AB ×AC =
∧
i
∧
j
∧
k
11−2
12−1
=∧
i(−1+4) − ∧
j(−1+2) + ∧
k(2−1)
=3
∧
i−∧
j+∧
k
n=3
∧
i−∧
j+∧
k
LetPbeanypointonnormalvectorandObeorigin.Thenreferthediagram,projectionofOPonplanehavelengthOM .
OP =2
∧
i−∧
j+∧
kand n =3
∧
i−∧
j+∧
k
OP. n = | OP||n|cos θ
(6+1+1) = √4+1+1(√9+1+1)cos θ
8= √6√11 cos θ ⇒cos θ =8
√66
Again, sin θ =|OM |
|OP|,gives OM =sin θ⋅OP
⇒ | OM =1−cos2θ|OP|
=1−64
66 √4+1+1(use cos θ =8√66
=2
66 ⋅ √6
∴ | OM =2
11
| |
| | | |
|√
√)
√
|√
Solution:
-------------------------------------------------------------------------------------------------
Question133
Thedistanceofthepoint(1,1,9)fromthepointofintersectionofthe
line x−3
1=y−4
2=z−5
2andtheplanex +y+z=17is:
24Feb2021Shift1
Options:
A.2√19
B.19√2
C.38
D.√38
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question134
Theequationoftheplanepassingthroughthepoint(1,2, −3)and
perpendiculartotheplanes3x +y−2z =5and2x −5y −z=7,is
Given,point(1,0,2)
Equationofplane=
r⋅∧
i+∧
j+∧
k=1 and r ⋅∧
i−2
∧
j= −2
Equationofplanepassingthroughtheintersectionofgivenplanesis
r⋅∧
i+∧
j+∧
k−1+λ r ⋅∧
i−2
∧
j+2=0
∵Thisplanepassesthroughpoint(1,0,2)i.e.,
vector
∧
i+2
∧
k
∴∧
i+2
∧
k⋅∧
i+∧
j+∧
k−1+λ
∧
i+2
∧
k⋅∧
i−2
∧
j+2=0
⇒ (3−1) + λ(1+2) = 0
⇒2+λ×3=0
⇒λ= −2∕3
Hence,equationofrequiredplaneis
r⋅∧
i+∧
j+∧
k−1+−2
3r⋅∧
i−2
∧
j+2=0
or 3 r ⋅∧
i+∧
j+∧
k−1−2 r ⋅∧
i−2
∧
j+2=0
or r ⋅∧
i+7
∧
j+3
∧
k=7
( ) ( )
[ ( ) ] [ ( ) ]
( )
[ ( ) ( ) ] [ ( ) ( ) ]
[ ( ) ] ( ) [ ( ) ]
[ ( ) ] [ ( ) ]
( )
Let x−3
1=y−4
2=z−5
2=t
⇒x=3+t,y=2t +4,z=2t +5
3+t+2t +4+2t +5=17
⇒5t =5⇒t=1
⇒Pointofintersectionis(4,6,7)
Distancebetween(1,1,9)and(4,6,7)is
(4−1)2+ (6−1)2+ (7−9)2
= √9+25 +4= √38.
√
24Feb2021Shift1
Options:
A.3x −10y −2z +11 −0
B.6x −5y −2z −2=0
C.11x +y+17z +38 =0
D.6x −5y +2z +10 =0
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question135
LetthepositionvectorsoftwopointsPandQbe3∧
i−∧
j+2∧
kand
∧
i+2∧
j−4∧
k,respectively.LetRandSbetwopointssuchthatthedirection
ratiosoflinesPRandQSare(4, −1,2)and(−2,1, −2),respectively.Let
linesPRandQSintersectatT .IfthevectorTAisperpendiculartoboth
PRandQSandthelengthofvectorTAis√5units,thenthemodulusof
apositionvectorofAis
[16Mar2021Shift1]
Options:
A.√482
B.√171
C.√5
D.√227
Answer:B
Solution:
Normalvector:
∧
i j
∧
k
3 1 −2
2−5−1
= −11
∧
i−∧
j−17
∧
k
So,directionratiosofnormaltotherequiredplaneare<11,1,17>
Planepassesthrough(1,2, −3)
So,equationofplane:
11(x−1) + 1(y−2) + 17(z+3) = 0
⇒11x +y+17z +38 =0
| |
-------------------------------------------------------------------------------------------------
Question136
Ifthefootoftheperpendicularfrompoint(4,3,8)ontheline
L1:x−a
1=y−2
3=z−b
4,I ≠0is(3,5,7),thentheshortestdistance
betweenthelineL1andlineL2:x−2
3=y−4
4=z−5
5isequalto
[16Mar2021Shift2]
Options:
A.1/2
B.1/√6
C.√2/3
D. 1
√3
P=3
∧
i−∧
j+2
∧
kand Q =∧
i+2
∧
j−4
∧
k
VPR = (4, −1,2)and V QS(−2,1, −2)
EquationoflinePR =3
∧
i−∧
j+2
∧
k+λ 4
∧
i−∧
j+2
∧
k
EquationoflineQS = ∧
i+2
∧
j−4
∧
k+µ−2
∧
i+∧
j−2
∧
k
LetTbethepointofintersection.
T= (3+4λ, −1−λ,2+2λ)
T= (1−2µ,2+µ, −4−2µ)
3+4λ =1−2µ
⇒2λ +µ= −1.......(i)
−1−λ=2+µ
⇒λ+µ= −3.....(ii)
FromEqs.(i)and(ii),
λ=2andµ= −5
T= [3+4(2)], −1− (2), 2+2(2) = (11, −3,6)
Now,DCofTAwillbeVPR ×VQS
∧
i,∧
j,∧
k; −2,1, −2;4, −1,2=0
∧
i−4
∧
j−2
∧
k
LT A ⇒11
∧
i−3
∧
j+6
∧
k+x−4
∧
j−2
∧
k
LetA= (11, −3−4x,6−2x)
T A = √5
⇒ (11 −11)2+ (−3−4x +3)2+ (6−2x −6)2= √5
⇒(4x)2+ (2x)2=5⇒20x2=5
⇒x2=1
4⇒x= ± 1
2
A= [11, −3−4(1∕2), 6−2(1∕2)]
A= (11, −5,5)
Or
A= [11, −3+4(1∕2), 6+2(1∕2)]
A= (11, −1,7)
∴ | A=112+52+52or
⇒ | A=112+12+72
⇒ | A| = √171or√171
∴ | A| = √171
( ) ( )
( ) ( )
( ) ( )
||
( ) ( )
√
|√
|√
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question137
LetPbeanarbitrarypointhavingsumofthesquaresofthedistance
fromtheplanesx +y+z=0,I x −nz =0andx −2y +z=0,equalto9.
IfthelocusofthepointPisx2+y2+z2=9,thenthevalueofI −nis
equalto
[17Mar2021Shift2]
Answer:0
Solution:
L1⇒x−a
I=y−2
3=z−b
4
FootofperpendicularfromA(4,3,8)toL1isB(3,5,7).
AB =OB −OA
=3
∧
i+5
∧
j+7
∧
k−4
∧
i+3
∧
j+8
∧
k
= −∧
i+2
∧
j−∧
k
Now,ABisperpendiculartodirectioncosineofL1,
So, −∧
i+2
∧
j−∧
k⋅∧
i+3
∧
j+4
∧
k=0
⇒ − I+6−4=0⇒I=2
As,(3,5,7)liesonL1,
3−a
2=5−2
3=7−b
4
3−a=2
So,a=1,7−b=4
So,b=3
L1⇒x−1
2=y−2
3=z−3
4
andL2=x−2
3=y−4
4=z−5
5
∴Shortestdistancewillbealongcommonnormal.
So,commonnormal= ∧
i,∧
j,∧
k;2,3,4;3,4,5
⇒n= −∧
i+2
∧
j−∧
k⇒∧
n=1
√6−∧
i+2
∧
j−∧
k
Shortestdistancewillbetheprojectionof(2−1)∧
i+ (4−2)∧
j+ (5−3)∧
kor∧
i+2
∧
j+2
∧
kalong∧
n
⇒∧
i+2
∧
j+2
∧
k−∧
i+2
∧
j−∧
k
√6=−1+4−2
√6=1
√6
( ) ( )
( ) ( )
||
( )
( ) ( )
LetP= (α,β,γ)
DistanceofpointPfromtheplanex+y+z=0is
=α+β+γ
√3
DistanceofpointPfromtheplanel x −nz =0is
-------------------------------------------------------------------------------------------------
Question138
Lettheplane,ax +by +cz +d=0bisectthelinejoiningthepoints
(4, −3,1)and(2,3, −5)attherightangles.Ifa,b,c,d areintegers,
thentheminimumvalueof(a2+b2+c2+d2)is.........
[18Mar2021Shift1]
Answer:28
Solution:
-------------------------------------------------------------------------------------------------
=l α −nγ
12+n2
anddistanceofpointPfromtheplanex−2y +z=0is
=α−2β +γ
√6
Accordingtothequestion,
α+β+γ
√3
2+l x −nz
I2+n2
2+α−2β +γ
√6
2=9
∴Locusis
(x+y+z)2
3+(l x −nz)2
I2+n2+(x−2y +z)2
6=9
⇒x21
2+l2
l2+n2+y2+z21
2+n2
l2+n2+xz 1 −2l n
l2+n2=0
Comparingitwiththegivenequationoflocus,weget
2l n =I2+n2
⇒ (I−n)2=0
⇒I−n=0
√
( ) (√)( )
( ) ( ) ( )
LetP≡ (4, −3,1)
andQ≡ (2,3, −5)
∴M=P+Q
2
⇒M≡4+2
2,−3+3
2,1−5
2
⇒M≡ (3,0, −2)
Also,directionratiosofPQ = {4−2, −3−3,1+5}
= {2, −6,6}
⇒DirectionratiosofPQ = {1, −3,3} = directionratiosofnormaltotheplane.
∴Equationoftheplaneis
1(x−3) − 3(y−0) + 3(z+2) = 0
⇒x−3y +3z +3=0
Comparingthistoax +by +cz +d=0,weget
a=1,b−3,c=3,d=3
∴Minimumvalueof(a2+b2+c2+d2) = 28
( )
Question139
Theequationoftheplanesparalleltotheplanex −2y +2z −3=0which
areatunitdistancefromthepoint(1,2,3)isax +by +cz +d=0.If
(b−d) = K(c−a),thenthepositivevalueofK is..........
[18Mar2021Shift1]
Answer:4
Solution:
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Question140
TheequationoftheplanewhichcontainstheY -axisandpassesthrough
thepoint(1,2,3)is
[17Mar2021Shift1]
Options:
A.x +3z =10
B.x +3z =0
C.3x +z=6
D.3x −z=0
Answer:D
Solution:
Equationofanyplaneparalleltotheplanex−2y +2z −3=0is
x−2y +2z +λ=0....(i)
Given,distancefrom(1,2,3)is1.
∴1−2×2+2×3+λ
(1)2+ (−2)2+ (2)2=1
⇒ | λ+3| = 3
⇒λ+3= ±3
⇒λ=0, −6
Consider,λ= −6
∴Equationofrequiredplaneis
x−2y +2z −6=0
Oncomparingthisequationto
ax +by +cz +d=0,weget
a=1,b= −2,c=2andd= −6
∵ (b−d) = k(c−a)
∴4=k× (1)
⇒k=4
|√|
Equationofplanepassingthroughapoint(x1,y1,z1)is
a(x−x1) + b(y−y1) + c(z−z1) = 0
-------------------------------------------------------------------------------------------------
Question141
Iffora >0,thefeetofperpendicularsfromthepointsA(a, −2a,3)and
B(0,4,5)ontheplaneI x +my +nz =0arepointsC(0, −a, −1)andD
respectively,then,thelengthoflinesegmentCDisequalto
[16Mar2021Shift1]
Options:
A.√31
B.√41
C.√55
D.√66
Answer:D
Solution:
Solution:
Here,(x1,y1,z1) = (1,2,3)
So,a(x−1) + b(y−2) + c(z−3) = 0
Now,Y-axisliesonit.
DirectionratioofY-axisis(0,1,0).
Normalvectortotheplane = a
∧
i+b
∧
j+c
∧
k
So,thenormalvectoroftheplanewillbeperpendiculartodirectionratioofY-axis.
a⋅0+b⋅1+c⋅0=0⇒b=0
Equationofplanebecomes
a(x−1) + c(z−3) = 0
Now,x=0,z=0alsosatisfiestheequation.
a(0−1) + c(0−3) = 0
⇒t−a−3c =0⇒a= −3c
So,−3c(x−1) + c(z−3) = 0
−3x +3+z−3=0
[as,C≠0]
⇒3x −z=0
Given,A⇒ (a, −2a,3)
B⇒ (0,4,5)
C⇒ (0, −a, −1)
EquationofplaneP⇒I x +my +nz =0
As,CisfootofperpendicularfromAtoplaneP.So,CA|N,whereNisthenormalvectortotheplane.
CA = (a−0)∧
i+ (−2a +a)∧
j+ (3+1)∧
k
=a
∧
i−a
∧
j+4
∧
k
Now,CA|N
So, a
l=−a
m=4
n=λ
whereλisanyrealnumber.
P⇒a
λx−a
λy+4
λz=0
P⇒ax −ay +4z =0
( ) ( ) ( )
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Question142
If(x,y,z)beanarbitrarypointlyingonaplaneP,whichpassesthrough
thepoints(42,0,0), (0,42,0)and(0,0,42),thenthevalueof
expression
3+x−11
(y−19)2(z−12)2+y−19
(x−11)2(z−12)2
+z−12
(x−11)2(y−19)2−x+y+z
14(x−11)(y−19)(z−12)
isequalto
[16Mar2021Shift2]
Options:
A.0
B.3
C.39
D.−45
Answer:B
Solution:
Solution:
Cliesonplane.
So,a⋅0−a(−a) + 4(−1) = 0
a2−4=0⇒a= ±2
Asperthequestion,a>0,soa=2
So,equationofplaneP⇒2x −2y +4z =0
P⇒x−y+2z =0
CoordinatesofD
x−0
1=y−4
−1=z−5
2=−(0−4+10)
[12+ (−1)2+22]
If(x,y,z)bethefootofperpendiculardrawnfrom(x1,y1,z1)totheplaneax +by +cz +d=0.
Then,
x−x1
a=y−y1
b=z−z1
c=−(ax1+by1+cz1+d)
a2+b2+c2
Here,(x,y,z) = (0,4,5)
⇒x−0= −(y−4) = z−5
2=−6
6
∴x= −1,y=5,z=3
C= (0, −2, −1) ⇒ D= (−1,5,3)
∴CD = (0+1)2+ (−2−5)2+ (−1−3)2
= √1+49 +16
CD = √66
√
EquationofplanepassingthroughA(42,0,0), B(0,42,0)andC(0,0,42)willbe
x
42 +y
42 +z
42 =1
⇒x+y+z=42
(x−11) + (y−19) + (z−12) = 0
Now,
3+x−11
(y−19)2(z−12)2+z−12
(x−11)2(y−19)2
+y−19
(x−11)2(z−12)2−x+y+z
14(x−11)(y−19)(z−12)
-------------------------------------------------------------------------------------------------
Question143
Iftheequationoftheplanepassingthroughthelineofintersectionof
theplanes
2x −7y +4z −3=0,3x −5y +4z +11 =0
andthepoint(−2,1,3)isax +by +cz −7=0,thenthevalueof
2a +b+c−7is..........
[17Mar2021Shift1]
Answer:4
Solution:
Solution:
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Question144
Letthemirrorimageofthepoint(1,3,a)withrespecttotheplane
r⋅2∧
i−∧
j+∧
k−b=0be(−3,5,2).Thenthevalueof|a+b|isequalto
[18Mar2021Shift2]
Answer:1
( )
⇒3(x−11)2(y−19)2(z−12)2+ (x−11)3+ (y−19)3+ (z−12)3
(x−11)2(y−19)2(z−12)2−42
14(x−11)(y−19)(z−12)
⇒(x−11)3+ (y−19)3+ (z−12)3−3(x−11)(y−19)(z−12) + 3(x−11)2(y−19)2(z−12)2
(x−11)2(y−19)2(z−12)2
⇒IfA+B+C=0
Then,A3+B3+C3=3ABC
⇒(x−11)3+ (y−19)3+ (z−12)3=3(x−11)(y−19)(z−12)
⇒3(x−11)(y−19)(z−12) − 3(x−11)(y−19)(z−12) + 3(x−11)2(y−19)2(z−12)2
(x−11)2(y−19)2(z−12)2
⇒3(x−11)2(y−19)2(z−12)2
(x−11)2(y−19)2(z−12)=3
Equationoftheplanepassingthroughthelineofintersectionsofplanes2x −7y +4z −3=0and3x −5y +4z +11 =0
is
(2x −7y +4z −3) + λ(3x −5y +4z +11) = 0
Sincethisplanepassesthoughtthepoint(−2,1,3)
∴(−4−7+12 −3) + λ(−6−5+12 +11) = 0
−2+12λ =0⇒λ=1/6
∴Equationofplaneis
(2x −7y +4z −3) + 1
6(3x −5y +4z +11) = 0
15x −47y +28z −7=0
∴a=15,b= −47,c=28
2a +b+c−7=30 −47 +28 −7=4
Solution:
Solution:
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Question145
LetPbeaplanecontainingtheline x−1
3=y+6
4=z+5
2andparalleltothe
line x−3
4=y−2
−3=z+5
7.Ifthepoint(1, −1,α)liesontheplaneP,thenthe
valueof|5α|isequalto............
[18Mar2021Shift2]
Answer:38
Solution:
Solution:
Givenequationofplaneinvectorformisr⋅2
∧
i−∧
j+∧
k−b=0
ItsCartesianformwillbe
2x −y+z=b...(i)
∵Risthemid-pointofPQ.
∴R≡P+Q
2⇒R≡ −1,4,a+2
2
∵Rliesontheplane(i).
∴ − 2−4+a+2
2=b⇒a+2=2b +12
⇒a=2b +10...(ii)
∵Directionratio'sofQPis(1− (−3), 3−5,a−2)
i.e.(4, −2,a−2)
anddirectionratiosofnormaltothegivenplaneare(2, −1,1)
∵nandQPareparallel.
2
4=−1
−2=1
a−2
∴a−2=2⇒a=4
FromEq.(ii),b= −3
∴ | a+b|=|4−3|=|1| = 1
( )
( )
Equationofrequiredplaneis
x−1 y +6 z +5
3 4 2
4−37
=0
Since,(1, −1,∞)liesonit,
So,replacexby1,yby(−1)andzandα
0 5 α+5
3 4 2
4−37
=0
⇒5α +38 =0⇒5α = −38
| |
| |
-------------------------------------------------------------------------------------------------
Question146
LetPbeaplanel x +my +nz =0containingtheline, 1−x
1=y+4
2=z+2
3.If
planePdividesthelinesegmentABjoiningpointsA(−3, −6,1)and
B(2,4, −3)inratiok :1,thenthevalueofkisequalto
[16Mar2021Shift1]
Options:
A.1.5
B.3
C.2
D.4
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question147
Ifthedistanceofthepoint(1, −2,3)fromtheplanex +2y −3z +10 =0
measuredparalleltotheline, x−1
3=2−y
m=z+3
1is 7
2,thenthevalueof
|m|isequalto.........
[16Mar2021Shift2]
√
∴ | 5α | = | −38 | = 38
P⇒l x +my +nz =0
PcontainsL1
L1⇒x−1
−1=y+4
2=z+2
3
So,(1, −4, −2)liesonplane.
1−4m −2n =0...(i)
And(−1,2,3)willbeperpendicularto(I,m,n).
−1+2m +3n =0....(ii)
AddingEqs.(i)and(ii),
−2m +n=0
n=2m
l−4m −4m =0
l=8m
So,l=8mandn=2m
Plane⇒8x +y+2z =0
Now,A(−3, −6,1)andB(2,4, −3)
PlanePdividesABintheratioofk:1.
LetplanePintersectthelineABatpointO.
So,O=2k −3
k+1,4k −6
k+1,−3k +1
k+1
AndOliesonplaneP,
So,8(2k −3) + (4k −6) + 2(−3k +1) = 0
⇒14k −28 =0
∴k=2
( )
Answer:2
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question148
Ifthelinesx−k
1=y−2
2=z−3
3andx+1
3=y+2
2=z+3
1areco-planar,thenthe
valueofkis_______.
[25Jul2021Shift2]
Answer:1
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question149
Given,pointA⇒ (1, −2,3)
Plane⇒x+2y −3z +10 =0
Distanceofpointfromplanealongthevector 3
∧
i−m
∧
j+∧
kis√7/2.
Linepassingthrough(1, −2,3)inthedirectionof
3
∧
i−m
∧
j+∧
kis x−1
3=y+2
−m=z−3
1=λ
AnygeneralpointBwillbe(3λ +1, −mλ −2,λ+3)
Now,thispointBliesonplane
So,x+2y −3z +10 =0
(3λ +1) + 2(−mλ −2) − 3(λ+3) + 10 =0
= (3−2m −3)λ=2
⇒λ= −1/m
Now,A= (1, −2,3)
B= (3λ +1, −mλ −2,λ+3)
|AB|2= (3λ +1−1)2+ (−mλ −2+2)2+ (λ+3−3)2
⇒7/2=9λ2+m2λ2+λ2
⇒7/2=10λ2+1 [∵mλ = −1]
⇒10λ2=5/2⇒λ2=1/4
⇒=±1/2⇒m= −1/λ
∴m= ±2
and|m| = 2
( )
( )
k+14 6
1 2 3
3 2 1
=0
(k+1)[2−6] − 4[1−9] + 6[2−6] = 0
k=1
| |
Iftheshortestdistancebetweenthestraightlines
3(x−1) = 6(y−2) = 2(z−1)and4(x−2) = 2(y−λ) = (z−3), λ∈Ris 1
√38 ,
thentheintegralvalueofλisequalto:
[22Jul2021Shift2]
Options:
A.3
B.2
C.5
D.−1
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question150
Iftheshortestdistancebetweenthelines
→
r1=α^
i+2^
j+2^
k+λ^
i−2^
j+2^
k,λ ∈R,α>0and
overrightarrow mathrm r2= −4^
i−^
k+µ 3^
i−2^
j−2^
k,µ∈Ris9,thenα
isequalto________.
[20Jul2021Shift1]
( )
( )
L1:(x−1)
2=(y−2)
1=(z−1)
3→
r1=2^
i+^
j+3^
k
L2:(x−2)
1=y−λ
2=z−3
4→
r2=^
i+2^
j+4^
k
Shortestdistance=Projectionof→
aon→
r1×→
r2
=
→
a.→
r1×→
r2
→
r1×→
r2
→
a.→
r1×→
r2=
1λ−22
2 1 3
1 2 4
= |14 −5λ|
→
r1×→
r2= √38
∴1
√38 =|14 −5λ|
√38
⇒ | 14 −5λ | = 1
⇒14 −5λ =1or14 −5λ = −1
⇒λ=13
5or3
∴Integralvalueofλ=3
| ( ) |
| |
| ( ) | | |
| |
Answer:6
Solution:
Solution:
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Question151
Thelinesx =ay −1=z−2andx =3y −2=bz −2, (ab ≠0)arecoplanar,
if:
[20Jul2021Shift2]
Options:
A.b =1,a∈R− {0}
B.a =1,b∈R− {0}
C.a =2,b=2
D.a =2,b=3
Answer:A
Solution:
Solution:
If→
r=→
a+λ→
band→
r=→
c+λ→
d
thenshortestdistancebetweentwolinesis
L=
→
a−→
c.→
b×→
d
|b×d|
∴→
a−→
c= (α+4)^
i+2^
j+3^
k
→
b×→
d
|b×d|=2^
i+2^
j+^
k
3
∴ (α+4)^
i+2^
j+3^
k.2^
i+2^
j+^
k
3 = 9
orα=6
( ) ( )
( )
( )
( ) ( )
x+1
a=y=z−1
a
x+2
3=y=z
3∕b
linesareCo-planar
a 1 a
3 1 3
b
−10−1
=0⇒− 3
b−a−1(a−3) = 0
| | ( )
-------------------------------------------------------------------------------------------------
Question152
Lettheplanepassingthroughthepoint(−1,0, −2)andperpendicular
toeachoftheplanes2x +y−z=2andx −y−z=3be
ax +by +cz +8=0.Thenthevalueofa +b+cisequalto:
[27Jul2021Shift1]
Options:
A.3
B.8
C.5
D.4
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question153
Leta,bandcbedistinctpositivenumbers.Ifthevectors
a^
i+a^
j+c^
k,^
i+^
kandc^
i+c^
j+b^
kareco-planar,thencisequalto:
[25Jul2021Shift2]
Options:
A. 2
1
a+1
b
B.a+b
2
C.1
a+1
b
D.√ab
Answer:D
a−3
b−a+3=0
~b=1,a∈R− {0}
Normalofreq.plane 2^
i+^
j−^
k×^
i−^
j−^
k
= −2^
i+^
j−3^
k
Equationofplane
−2(x+1) + 1(y−0) − 3(z+2) = 0
−2x +y−3z −8=0
2x −y+3z +8=0
a+b+c=4
( ) ( )
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question154
LetaplanePpassthroughthepoint(3,7, −7)andcontaintheline,
x−2
−3=y−3
2=z+2
1.IfdistanceoftheplanePfromtheoriginisd ,thend 2
isequalto_________.
[27Jul2021Shift1]
Answer:3
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question155
Forrealnumbersαandβ ≠0,ifthepointofintersectionofthestraight
linesx−α
1=y−1
2=z−1
3andx−4
β=y−6
3=z−7
3liesontheplanex +2y −z=8,
thenα −βisequalto:
[27Jul2021Shift2]
Options:
A.5
B.9
Becausevectorsarecoplanar
Hence
a a c
1 0 1
c c b
=0
⇒c2=ab ⇒c= √ab
| |
→
BA =^
i+4^
j−5^
k
→
BA =^
i+4^
j−5^
k
→
BA ×→
l=→
n=
^
i^
j^
k
−32 1
1 4 −5
a^
i+b^
j+c^
k= −14^
i−^
j(14) + ^
k(−14)
a=1,b=1,c=1
Planeis(x−2) + (y−3) + (z+2) = 0
x+y+z−3=0
d= √3⇒d2=3
( )
( )
| |
C.3
D.7
Answer:D
Solution:
-------------------------------------------------------------------------------------------------
Question156
ThedistanceofthepointP(3,4,4)fromthepointofintersectionofthe
linejoiningthepoints.Q(3, −4, −5)andR(2, −3,1)andtheplane
2x +y+z=7,isequalto______.
[27Jul2021Shift2]
Answer:7
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question157
LetthefootofperpendicularfromapointP(1,2, −1)tothestraight
lineL :x
1=y
0=z
−1beN .
LetalinebedrawnfromPparalleltotheplanex +y+2z =0which
meetsLatpointQ.IfαistheacuteanglebetweenthelinesPN andPQ,
thencos αisequalto_________.
[25Jul2021Shift1]
Firstlineis(ϕ+α,2ϕ +1,3ϕ +1)
andsecondlineis(qβ +4,3q +6,3q +7).
Forintersection
ϕ+α=qβ +4.....(i)
2ϕ +1=3q +6......(ii)
3ϕ +1=3q +7........(iii)
for(ii)&(iii)ϕ=1,q= −1
So,from(i)α+β=3
Now,pointofintersectionis(α+1,3,4)
Itliesontheplane.
Hence,α=5&β= −2
→
QR : −x−3
1=y+4
−1=z+5
−6=r
⇒(x,y,z)≡ (r+3, −r−4, −6r −5)
Now,satisfyingitinthegivenplane.
Wegetr= −2.
so,requiredpointofintersectionisT(1, −2,7).
Hence,PT =7.
Options:
A. 1
√5
B.√3
2
C. 1
√3
D. 1
2√3
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question158
LetLbethelineofintersectionofplanes→
r.^
i−^
j+2^
k=2and
→
r.2^
i+^
j−^
k=2.IfP(α,β,γ)isthefootofperpendicularonLfrom
thepoint(1,2,0),thenthevalueof35(α+β+γ)isequalto:
[22Jul2021Shift2]
( )
( )
→
PN .^
i−^
k=0
⇒N(1,0, −1)
Now,
→
PQ .^
i+^
j+2^
k=0
⇒µ= −1
⇒Q(−1,0,1)
→
PN =2^
jand →
PQ =2^
i+2^
j−2^
k
⇒cos α =1
√3
( )
( )
Options:
A.101
B.119
C.143
D.134
Answer:B
Solution:
Solution:
P1:x−y+2z =2
P2=2x +y−3=2
LetlineofIntersectionofplanesP1andP2cutsxyplaneinpointQ.
⇒z-coordinateofpointQiszero
⇒x−y=2
and2x +y=2⇒x=4
3,y=−2
3
⇒Q4
3,−2
3,0
Vectorparalleltothelineofintersection
→
a=
^
i^
j^
k
1−12
2 1 −1
= −^
i+5^
j+3^
k
EquationofLineofintersection
x−4
3
−1=
y+2
3
5=z−0
3=λ(say)
Letcoordinatesoffootofperpendicularbe
F−λ+4
3,5λ −2
3,3λ
→
PF = −λ+1
3
^
i+5λ −8
3
^
j+ (3λ)^
k
→
PF .→
a=0
⇒λ−1
3+25λ−40
3+9λ =0
⇒35λ =41
3⇒λ=41
105
Now,α= −λ+4
3,β=5λ −2
3,γ=3λ
⇒α+β+γ=7λ +2
3
=741
105 +2
3
=51
15
⇒35(α+β+γ) = 51
15 ×35 =119
}
( )
| |
( )
( ) ( )
( )
Question159
LetPbeaplanepassingthroughthepoints(1,0,1), (1, −2,1)and
(0,1, −2).Letavector→
a=α^
i+β^
j+γ^
kbesuchthat→
aisparallelto
theplaneP,perpendicularto ^
i+2^
j+3^
kand→
a.^
i+^
j+2^
k=2,then
(α−β+γ)2equals_______.
[20Jul2021Shift1]
Answer:81
Solution:
-------------------------------------------------------------------------------------------------
Question160
ConsiderthelineLgivenbytheequationx−3
2=y−1
1=z−2
1.LetQbethe
mirrorimageofthepoint(2,3, −1)withrespecttoL.LetaplanePbe
suchthatitpassesthroughQ,andthelineLisperpendiculartoP.Then
whichofthefollowingpointsisontheplaneP?
[20Jul2021Shift2]
Options:
A.(−1,1,2)
B.(1,1,1)
C.(1,1,2)
D.(1,2,2)
Answer:D
( ) ( )
Equationofplane:
x−1 y −0 z −1
1−121−1
1−0 0 −1 1 +2
=0
⇒3x −z−2=0
→
a=α^
i+β^
j+γ^
k∥to3x −z−2=0
⇒3α −8=0.......(1)
→
a⊥^
i+2^
j+3^
k
⇒α+2β +38 =0.......(2)
→
a.^
i+^
j+2^
k=0
⇒α+β+28 =2.......(3)
onsolving1,2&3
α=1,β= −5,8=3
So(α−β+8) = 81
| |
( )
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question161
Theanglebetweenthestraightlines,whosedirectioncosinesaregiven
bytheequations2I +2m −n=0andmn +nI +Im =0,is
[27Aug2021Shift2]
Options:
A.π
2
B.π −cos−14
9
C.cos−18
9
D.π
3
Answer:A
Solution:
( )
( )
Planemathrm pis⊥rtoline
x−3
2=y−1
1=z−2
1
&passesthroughpt.(2,3)equationofplanep
2(x−2) + 1(y−3) + 1(z+1) = 0
2x +y+z−6=0
pt(1,2,2)satisfiesaboveequation
Given,
2l+2m−n=0....(i)
mn+nl+lm=0...(ii)
FromEq.(i),weget
n=2l+2m...(iii)
Substituting,n=2l+2minEq.(ii),wehave
m(2l+2m)+l(2l+2m)+lm=0
⇒2 Im +2m2+2I2+2 Im +Im =0
⇒2I2+4 Im +Im +2m2=0
⇒2l(l+2m)+m(l+2m)=0
⇒(2l+m)(l+2m)=0
When
2l=−m
FromEq(iii),
n=m
⇒2I
−1=m
1=n
1
⇒I
−1
2
=m
1=n
1
or I
1=m
−2=n
−2
⇒(l,m,n)=(1,−2,−2)
When
l=−2m
FromEq.(iii),
-------------------------------------------------------------------------------------------------
Question162
Thesquareofthedistanceofthepointofintersectionoftheline
x−1
2=y−2
3=z+1
6andtheplane2x −y+z=6fromthepoint(−1, −1,2)is
[31Aug2021Shift1]
Answer:61
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question163
Thedistanceofthepoint(−1,2, −2)fromthelineofintersectionofthe
planes2x +3y +2z =0andx −2y +z=0is
[31Aug2021Shift2]
Options:
A. 1
√2
B.5
2
n=−2m
⇒l=−2m=n
⇒I
−2=m
1=n
−2
⇒(l,m,n)=(−2,1,−2)
∴Anglesbetweenstraightlines
cos θ =
^
i−2^
j−2^
k−2^
i+^
j−2^
k
^
i−2^
j−2^
k−2^
i+^
j−2^
k
cos θ =−2−2+4
9=0
⇒θ=π
2
( ) ( )
| | |
x−1
2=y−2
3=z+1
6=λ
x=2λ +1
y=3λ +2
z=6λ −1
Equationofplaneis2x −y+z=6
⇒2(2λ +1) − (3λ +2) + (6λ −1) = 6
7λ =7
λ=1
P(3,5,5)
(Distance)2=(3+1) + (5+1)2+ (5−2)2
=16 +36 +9=61
{
C.√42
2
D.√34
2
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question164
Iftheequationofplanepassingthroughthemirrorimageofapoint
(2,3,1)withrespecttoline x+1
2=y−3
1=z+2
−1andcontainingtheline
x−2
3=1−y
2=z+1
1isαx +βy +γz =24,thenα +β+γisequalto
[17Mar2021Shift2]
Options:
A.20
B.19
C.18
D.21
Answer:B
LetLbethelineofintersectionof2x +3y +2z =0andx−2y +z=0
Ifz=0,thenx=y=0
ThelineLisparalleltor1×r2,wherer1=2^
i+3^
j+2^
kandr2=^
i−2^
j+^
k
^
i^
j^
k
232
1−21
=7^
i−7^
k
DR'sofis(1,0, −1)
andequationofLisx
1=y
0=z
−1=λ
LetPQbethedistancefromthepointP(−1,2, −2)tothelineL.
DRsofPQ =λ+1, −2,2−λ
∴PQ⊥r
⇒(λ+1)(1) + (−2)(0) + (2−λ)(−1) = 0
⇒λ+1−2+λ=0
⇒λ=1
2
CoordinateofQis 1
2,0,−1
2
⇒PQ =−1−1
2
2+ (2−0)2+ −2+1
2
2
=9
4+4+9
4=√34
2
| |
( )
√( ) ( )
√
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question165
Supposethelinex−2
α+y−2
−5=z+2
2liesontheplanex +3y −2z +β=0.
Then,(α+β)isequalto
[31Aug2021Shift2]
Answer:7
Solution:
LetA= (2,3,1)
L1⇒x+1
2=y−3
1=z+2
−1
L2⇒x−2
3=y−1
−2=z+1
1
AnypointMtakenonL1is(2r −1,r+3, −r−2)
∴DirectionratiosofAMare(2r −3,r, −r−3)
∵AM ⟂L1
∴2(2r −3) + 1×r+ (−1)(−r−3) = 0
⇒4r −6+r+r+3=0
⇒6r =3⇒r=1
2
∴M=0,7
2,−5
2
∴M=0,7
2,−5
2
∴B≡ (2×0) − 2,2×7
2−3,2×−5
2−1
⇒B= (−2,4, −6)
Now,equationofplanecontainingB(−2,4, −6)andthelineL2is
x−2 y −1 z +1
3−2 1
4−3 5
=0
⇒(x−2)(−10 +3) − (y−1)(15 −4) + (z+1)(−9+8) = 0
⇒−7(x−2) − 11(y−1) − 1(z+1) = 0
⇒ − 7x −11y −z= −14 −11 +1
⇒7x +11y +z=24comparingthisto
αx +βy +γz =24
Weget,α=7,β=11,γ=1
∴α+β+γ=7+11 +1=19
( )
( )
( ( ) ( ( ) ) )
| |
-------------------------------------------------------------------------------------------------
Question166
Lettheequationoftheplane,thatpassesthroughthepoint(1,4, −3)
andcontainsthelineofintersectionoftheplanes3x −2y +4z −7=0
andx +5y −2z +9=0beαx +βy +γz +3=0,thenα +β+γisequalto
[31Aug2021Shift1]
Options:
A.-23
B.-15
C.23
D.15
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question167
Thedistanceofthepoint(1, −2,3)fromtheplanex −y+z=5
measuredparalleltoaline,whosedirectionratiosare2,3, −6is
[27Aug2021Shift1]
Options:
Givenequationofline
x−2
α=y−2
−5=z+2
2...(i)
andplanex+3y −2z +β=0...(ii)
Line(i)pasesthrough(2,2, −2)
whichliesonplane(ii).
∴2+6+4+β=0
⇒β=−12
Also,givenlineisperpendiculartonormaloftheplane
α(1) − 5(3) + 2(−2) = 0
⇒α=19
∴α+β=7
Equationofplaneis
(3x −2y +4z −7) + λ(x+5y −2z +9) = 0
(λ+3)x+ (5λ −2)y+ (4−2λ)z+9λ −7=0
Passingthrough(1,4,−3)
(λ+3) + 4(5λ −2) − 3(4−2λ) + 9λ −7=0
⇒36λ −24 =0
λ=2
3
⇒Equationofplane
2
3+3 x +10
3−2 y +4−4
3z+6−7=0
⇒11x +4y +8z −3=0
α= −11,β= −4,γ= −8
α+β+γ= −23
( ) ( ) ( )
A.3
B.5
C.2
D.1
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question168
Equationofaplaneatadistance 2
21 fromtheorigin,whichcontains
thelineofintersectionoftheplanesx −y−z−1=0and
2x +y−3z +4=0,is
[27Aug2021Shift1]
Options:
A.3x −y−5z +2=0
B.3x −4z +3=0
C.−x+2y +2z −3=0
D.4x −y−5z +2=0
Answer:D
Solution:
√
LetAbeanypointontheplanex−y+z=5andB(1, −2,3).
ThenequationofthelineABwhosedirectionratiosare2,3, −6
x−1
2=y+2
3=z−3
−6=λ(Let)
⇒x=1+2λ,y= −2+3λ,z=3−6λ
A(1+2λ, −2+3λ,3−6λ)
Aliesonplane.
Then,1+2λ − (−2+3λ) + 3−6λ =5
⇒1+2λ +2−3λ +3−6λ =5
⇒λ=1
7
∴A9
7,−11
7,15
7
Distance AB =1−9
7
2+ −2+11
7
2+3−15
7
2
=4
49 +9
49 +36
49 =1
( )
√( ) ( ) ( )
√
Givenplanes,
x−y−z−1=0...(i)
2x +y−3z +4=0...(ii)
Equationofplanepassingthroughlineofintersectionofplanes(i)and(ii)isgivenby
-------------------------------------------------------------------------------------------------
Question169
Theequationoftheplanepassingthroughthelineofintersectionofthe
planesr .^
i+^
j+^
k=1andr .2^
i+3^
j−^
k+4=0andparalleltothe
X-axisis
[27Aug2021Shift2]
Options:
A.r .^
j−3^
k+6=0
B.r .^
i+3^
k+6=0
C.r .^
i−3^
k+6=0
D.r .^
j−3^
k−6=0
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
( ) ( )
( )
( )
( )
( )
(x−y−z−1) + λ(2x +y−3z +4) = 0
⇒(2λ +1)x+ (λ−1)y+ (−3λ −1)z+ (4λ −1) = 0...(iii)
Distanceofplane(iii)fromorigin = 2
21(given)
⇒|4λ −1|
(2λ +1)2+ (λ−1)2+ (−3λ +1)2 = 2
21
Squaringbothsides
(4λ −1)2
(2λ +1)2+ (λ−1)2+ (3λ +1)2=2
21
⇒21(16λ2−8λ +1) = 2(14λ2+8λ +3)
⇒308λ2−184λ +15 =0
308λ2−154λ −30λ +15 =0
(2λ −1)(154λ −15) = 0
⇒λ=1
2orλ=15
154
Puttingλ=1
2inEq.(iii),wehave
4x −y−5z +2=0
√√
Given,equationofplanes
r.^
i+^
j+^
k=1...(i)
r.2^
i+3^
j−^
k+4=0...(ii)
EquationofplanepassingthroughtheintersectionoftheplanesEqs.(i)and(ii)isgivenby
(x+y+z−1)+λ(2x+3y–z+4)=0
or(1+2λ)x+(1+3λ)y+(1−λ)z+(−1+4λ)=0...(iii)
Plane(iii)inparalleltoX-axis
1+2λ=0[Coefficientofx=0]
⇒λ=−1
2
∴FromEq.(iii)becomes
y−3z+6=0
orr.^
j−3^
k+6=0
( )
( )
( )
Question170
LetSbethemirrorimageofthepointQ(1,3,4)withrespecttothe
plane2x −y+z+3=0andletR(3,5,γ)beapointofthisplane.Then
thesquareofthelengthofthelinesegmentSRis
[27Aug2021Shift2]
Answer:72
Solution:
-------------------------------------------------------------------------------------------------
Question171
AplanePcontainsthelinex +2y +3z +1=0=x−y−z−6andis
perpendiculartotheplane−2x +y+z+8=0.Thenwhichofthe
followingpointsliesonP?
[26Aug2021Shift1]
Options:
A.(−1,1,2)
B.(0,1,1)
C.(1,0,1)
D.(2, −1,1)
Answer:B
Solution:
LetpointS(a,b,c)
Then,
a−1
2=b−3
−1=c−4
1=−2(2−3+4+3)
4+1+1 = −2
⇒a=−3,b=5,c=2
∴S(−3,5,2)
andpointR(3,5,γ)liesontheplane2x−y+z+3=0
⇒6−5+γ+3=0
⇒γ=−4
∴R(3,5,−4)
Now,SR2=62+02+ (6)2
=36 +0+36 =72
Equationofplanecontainingthegivenplanesis
(x+2y +3z +1) + λ(x−y−z−6) = 0
(1+λ)x+ (2−λ)y+ (3−λ)z+ (1−6λ) = 0
Thisplaneisperpendiculartotheplane−2x +y+z+8=0
So,−2(1+λ) + (2−λ) + (3−λ) = 0
−2−2λ +2−λ+3−λ=0
-------------------------------------------------------------------------------------------------
Question172
LetPbetheplanepassingthroughthepoint(1,2,3)andthelineof
intersectionoftheplanesr .^
i+^
j+4^
kandr . −^
i+^
j+^
k=6Then
whichofthefollowingpointsdoesnotlieonP?
[26Aug2021Shift2]
Options:
A.(3,3,2)
B.(6, −6,2)
C.(4,2,2)
D.(−8,8,6)
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question173
LetthelineLbetheprojectionofthelinex−1
2=y−3
1=z−4
2intheplane
x−2y −z=3.Ifdisthedistanceofthepoint(0,0,6)fromL,thend 2is
equalto
[26Aug2021Shift1]
Answer:26
Solution:
( ) ( )
λ=3
4
So,equationofrequiredplaneis7x +5y +9z =14
Now,(0,1,1)satisfiestheaboveplane.
PisaplanepassingthroughtheintersectionofP1andP2.
EquationofP:P1+λP2=0
(x+y+4z −16) + λ(−x+y+z−6) = 0...(i)
SinceplanePpassesthrough(1,2,3),then
(1+2+12 −16) + λ(−1+2+3−6) = 0
⇒−1+λ(−2) = 0
⇒λ=−1
2
Onputtingλ=−1
2inEq.(i),weget
P:3x +y+7z −26 =0
Clearly(4,2,2)notlieontheplane.
Solution:
-------------------------------------------------------------------------------------------------
Question174
LetQbethefootoftheperpendicularfromthepointP(7, −2,13)onthe
planecontainingthelinesx+1
6=y−1
7=z−3
8andx+1
3=y−2
5=z−3
7.Then
(PQ)2isequalto
[26Aug2021Shift2]
Answer:96
Solution:
Givenline,x−1
2=y−3
1=z−4
2
Givenplane,x−2y −z=3
Tofindtheprojectionlet'sfindthefootofperpendicularfrom( 1,3,,4)toplanex−2y −z=3
x−1
1=y−3
−2=z−4
−1=λ1
(λ1+1) − 2(−2λ1+3) − (−λ1+4) = 3
⇒6λ1=12
⇒λ1=2
So,footofperpendicularfrom(1,3,4)toplanex−2y −z=3isA(3, −1,2).
Letusalsofindtheintersectionpointoftheplaneandline
x−1
2=y−3
1=z−4
2=λ2
(2λ2+1) − 2(λ2+3) − (2λ2+4) = 3−2λ2 = 12
⇒λ2= −6
TheintersectionpointoftheplaneandlineisB(−11, −3, −8)
LinepassingthroughAandBis
x−3
−14 =y+1
−2=z−2
−10 =μ
x−3
7=y+1
1=z−2
5=μ
Now,let'sfindthedistancefromO(0,0,6)tothislineL.
Let'ssayC(7μ +3,μ−1,5μ +2)isanypointonL.
Then,
{(7μ +3) − 0}.7 + {(μ−1) − 0}.1 + {(5μ +2) − 6}.5 =0
⇒49μ +21 +μ−1+25μ −20 =0
⇒μ=0
C(3, −1,2)
Distance = (3−0)2+ (−1−0)2+ (2−6)2= √26
d2=26
√
Planecontainingthelineswouldbe
x+1 y −1 z −3
6 7 8
3 5 7
=0
| |
-------------------------------------------------------------------------------------------------
Question175
Thedistanceofline3y −2z −1=0 =3x −z+4fromthepoint(2, −1,6)
is
[1Sep2021Shift2]
Options:
A.√26
B.2√5
C.C.2√6
D.4√2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question176
Lettheacuteanglebisectorofthetwoplanesx −2y −2z +1=0and
2x −3y −6z +1=0betheplaneP.Then,whichofthefollowingpoints
liesonP?
⇒(x+1)(49 −40) − (y−1)(42 −24) + (z−3)(30 −21) = 0
⇒9(x+1) − 18(y−1) + 9(z−3) = 0
⇒x−2y +z=0
Now,PQwillbeequaltotheperpendiculardistanceofthepointP(7, −2,13)fromtheplanex−2y +z=0
∴PQ =7−2(−2) + 13
12+ (−2)2+ (1)2
=7+4+13
√1+4+1=24
√6=4√6
PQ2= (4√6)2=16 ×6=96
|√|
| | | |
Equationofline
3y −2z −1=0=3x −z+4
⇒3y −1
2=z−0
1=3x +4
1
⇒
x+4
3
1∕3=
y−1
3
2∕3=z−0
1
PR = | PQ |cos θ = |PQ|PQ .P
|PQ||P| = PQ .PQ
|PR|
PR =
1
32+4
3+2
3−1−1
3+1(6−0)
1
9+4
9+1
=414
9
OR2=PQ2−PR2
=100
9+16
9+36 −224
9
=100
9+16
9+36 −224
9
QR = √24 =2√6
|( ) ( )
√|√
[1Sep2021Shift2]
Options:
A. 3,1, −1
2
B. −2,0, −1
2
C.(0,2, −4)
D.(4,0, −2)
Answer:B
Solution:
-------------------------------------------------------------------------------------------------
Question177
Theshortestdistancebetweenthelinesx−3
3=y−8
−1=z−3
1and
x+3
−3=y+7
2=z−6
4is:
[Jan.08,2020(I)]
Options:
A.2√30
B.7
2√30
C.3√30
D.3
Answer:C
Solution:
( )
( )
Equationofanglebisectors
x−2y −2z +1
√1+4+4= ±2x −3y −6z +1
√4+9+36
⇒x−5y +4z +4 = 0 and 13 x −23y −32z +10 = 0
Then,cos θ =1+10 −8
√1+4+4√1+25 +16 = 1
√42
⇒tan θ = √41 >1
⇒θ>45°
Then,acuteanglebisectorinplane
P:13x −23y −32z +10 =0
∴Point −2,0,−1
2liesontheplaneP.
( )
→
AB =6^
i+15^
j+3^
k
→
p=^
i+4^
j+22^
k
-------------------------------------------------------------------------------------------------
Question178
Ifthefootoftheperpendiculardrawnfromthepoint(1,0,3)onaline
passingthrough(α,7,1)is,thenαisequalto_______.
[NAJan.07,2020(II)]
Answer:4
Solution:
-------------------------------------------------------------------------------------------------
Question179
IfforsomeαandβinR,theintersectionofthefollowingthreeplanes
x+4y −2z =1
x+7y −5z =β
x+5y +αz =5
isalineinR3,thenα +βisequalto:
[Jan.9,2020(I)]
Options:
A.0
B.10
C.2
→
q=^
i+^
j+7^
k
→
p×→
q=
^
i^
j^
k
1 4 22
1 1 7
= 6^
i+15^
j−3^
k
Shortestdistancebetweenthelinesis
=
→
AB .→
p×→
q
→
p×→
q
= |36 +225 +9|
√36 +225 +9=3√30
| |
|( ) |
| |
Since,PQisperpendiculartoL
∴1−5
3α−5
3+−7
37−7
3+ 3−17
31−17
3=0
⇒−2α
3+10
9−98
9+112
9=0
⇒2α
3=24
9⇒α=4
( ) ( ) ( ) ( ) ( ) ( )
D.–10
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question180
Ifthedistancebetweentheplane,23x −10y −2z +48 =0andtheplane
containingthelinesx+1
2=y−3
4=z+1
3andx+3
2=y+2
6=z−1
λ(λ∈R)isequalto
k
√633 ,thenkisequalto________.
[NAJan.9,2020(II)]
Answer:3
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question181
Themirrorimageofthepoint(1,2,3)inaplaneis −7
3, −4
3, −1
3.Which
ofthefollowingpointsliesonthisplane?
[Jan.8,2020(II)]
( )
Δ=0⇒
14−2
17−5
1 5 α
=0
⇒(7α +25) − (4α +10)+(−20 +14) = 0
⇒3α +9=0⇒α= −3
Also,Dz=0⇒
141
1 7 β
155
=0
⇒1(35 −5β) − (15) + 1(4β −7) = 0⇒β=13
Hence,α+β= −3+13 =10
| |
| |
Since,thelinex+1
2=y−3
4=z+1
3containsthepoint(-1,3,-1)andlinex+3
2=y+2
6=z−1
λcontainsthepoint(-3,-2,1)
Then,thedistancebetweentheplane23x −10y −2z +48 =0andtheplanecontainingthelines=perpendiculardistance
ofplane
23x −10y −2z +48 =0eitherfrom(-1,3,-1)or(-3,-2,1)
=23(−1) − 10(3) − 2(−1)
(23)2+ (10)2+ (−2)2 = 3
√633
Itisgiventhatdistancebetweentheplanes
=k
√633 ⇒k
√633 = 3
√633 ⇒k=3
|√|
Options:
A.(1,1,1)
B.(1,–1,1)
C.(–1,–1,1)
D.(–1,–1,–1)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question182
LetPbeaplanepassingthroughthepoints(2,1,0),(4,1,1)and(5,0,
1)andRbeanypoint(2,1,6).ThentheimageofRintheplanePis:
[Jan.7,2020(I)]
Options:
A.(6,5,2)
B.(6,5,–2)
C.(4,3,2)
D.(3,4,–2)
Answer:B
Solution:
→
n=−7
3−1,−4
3−2,−1
3−3
→
n=10
3,10
3,10
3
D.rofnormaltotheplane(1,1,1)
MidpointofPandQis −2
3,1
3,4
3
∴EquationofrequiredplaneQ
→
r.^
n=^
a.^
n
→
r.^
i+^
j+^
k = −2
3+1
3+4
3
∴Equationofplaneisx+y+z=1
( )
( )
-------------------------------------------------------------------------------------------------
Question183
AplanePmeetsthecoordinateaxesatA,BandCrespectively.The
centroidofΔABCisgiventobe(1,1,2).Thentheequationoftheline
throughthiscentroidandperpendiculartotheplanePis:
[Sep.06,2020(II)]
Options:
A.x−1
2=y−1
1=z−2
1
B.x−1
1=y−1
1=z−2
2
C.x−1
2=y−1
2=z−2
1
D.x−1
1=y−1
2=z−2
2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question184
If(a,b,c)istheimageofthepoint(1,2,-3)intheline,x+1
2=y−3
−2=z
−1,
thena +b+cisequalsto:
[Sep.05,2020(I)]
Options:
Equationofplaneisx+y−2z =3
⇒x−2
1=y−1
1=z−6
−2 = −2(2+1−12 −3)
6
⇒(x,y,z) = (6,5, −2)
∴α=3,β=3andγ=6asGiscentroid.
∴Theequationofplaneis
x
α+y
β+z
γ=1
⇒x
3+y
3+z
6 = 1⇒2x +2y +z=6
∴Therequiredlineis,x−1
2=y−1
2=z−2
1
A.2
B.–1
C.3
D.1
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question185
Thelines→
r=^
i−^
j+l 2^
i+^
kand→
r=2^
i−^
j+m^
i+^
j−^
k
[Sep.03,2020(I)]
Options:
A.donotintersectforanyvaluesofl andm
B.intersectforallvaluesofl andm
C.intersectwhenl =2andm =1
2
D.intersectwhenl =1andm =2
Answer:A
Solution:
Solution:
( ) ( ) ( ) ( )
x+1
2=y−3
−2=z
−1=λ
Anypointonline = Q(2λ −1, −2λ +3, −λ)
∴D.r.ofPQ = [2λ −2, −2λ +1, −λ+3]
D.r.ofgivenline = [2, −2, −1]
∵PQisperpendiculartolineL
∴2(2λ −2) − 2(−2λ +1)−1(−λ+3) = 0
⇒4λ −4+4λ −2+λ−3=0
⇒9λ −9=0⇒λ=1
∵QismidpointofPR =Q= (1,1, −1)
∴CoordinateofimageR= (1,0,1) = (a,b,c)
∴a+b+c=2
L1≡→
r=^
i−^
j+l 2^
i+^
k
L2≡→
r=2^
i−^
j+m^
i+^
j−^
k
Equatingcoeff.of^
i,^
jand^
kofL1andL2
( ) ( )
( ) ( )
-------------------------------------------------------------------------------------------------
Question186
Theshortestdistancebetweenthelinesx−1
0=y+1
−1=z
1and
x+y+z+1=0,2x −y+z+3=0is:
[Sep.06,2020(I)]
Options:
A.1
B. 1
√3
C. 1
√2
D.1
2
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question187
Ifforsomeα ∈R,thelinesL1:x+1
2=y−2
−1=z−1
1andL2:x+2
α=y+1
5−α=z+1
1
arecoplanar,thenthelineL2passesthroughthepoint:
[Sep.05,2020(II)]
Options:
A.(10,2,2)
B.(2,-10,-2)
C.(10,-2,-2)
2l +1=m+2......(i)
−1= −1+m⇒m=0.......(ii)
l= −m.......(iii)
⇒m=l=0whichisnotsatisfyeqn.(i)hencelinesdonotintersectforanyvalueoflandm.
Forlineofintersectionofplanesx+y+z+1=0and2x −y+z+3=0:
→
b2=
^
i^
j^
k
111
2−11
= 2^
i+^
j−3^
k
Puty=0,wegetx= −2andz=1
L2:r= −2^
i+^
k+λ 2^
i+^
j−3^
kandL1:r=^
i−^
j+µ−^
j+^
kGiven)
Now,b1×b2 = −2^
i+^
j+^
kanda2−oc a1= −3^
i+^
j+^
k
∴Shortestdistance = 1
√3
| |
( ) ( ) ( ) ( ) (
[ ]
D.(-2,10,2)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question188
IftheequationofaplaneP,passingthroughtheintersectionofthe
planes,x +4y −z+7=0and3x +y+5z =8isax +by +6z =15forsome
a,b∈R,thenthedistanceofthepoint(3,2,-1)fromtheplanePis
_________.
[Sep.04,2020(I)]
Answer:3
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question189
Thedistanceofthepoint(1,-2,3)fromtheplanex −y+z=5measured
paralleltothelinex
2=y
3=z
−6is:
[NASep.04,2020(II)]
Options:
∴
1 3 2
2−11
α5−α1
=0
⇒1(−1−5+α) − 3(2−α)+2(10 −2α +α) = 0
∴α= −4
∴EquationofL2:x+2
−4=y+1
9=z+1
1
∴Point(2,-10,-2)liesonlineL2
| |
EquationofplanePis
(x+4y −z+7) + λ(3x +y+5z −8) = 0
⇒x(1+3λ) + y(4+λ) + z(−1+5λ)+(7−8λ) = 0
⇒1+3λ
a=4+λ
b=5λ −1
6 = 7−8λ
−15
Fromlasttworatios,λ= −1
⇒−2
a=3
b= −1
∴a=2,b= −3
∴Equationofplaneis,2x −3y +6z −15 =0
Distance = |6−6−6−15|
7=21
7=3.
A.7
5
B.1
C.1
7
D.7
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question190
Thefootoftheperpendiculardrawnfromthepoint(4,2,3)totheline
joiningthepoints(1,-2,3)and(1,1,0)liesontheplane:
[Sep.03,2020(I)]
Options:
A.2x +y−z=1
B.x −y−2z =1
C.x −2y +z=1
D.x +2y −z=1
Answer:A
Solution:
EquationoflinethroughpointP(1, −2,3)andparalleltothelinex
2=y
3=z
−6is
x−1
2=y+2
3=z−3
−6=λ
So,anypointonline = Q(2λ +1,3λ −2, −6λ +3)
Since,thispointliesonplanex−y+2=5
∴2λ +1−3λ +2−6λ +3=5⇒λ=1
7
∴Pointofintersectionlineandplane,Q=9
7,11
7,15
7
∴RequireddistancePQ
=9
7−12+ −11
7+22+15
7−32=1
( )
√( ) ( ) ( )
Equationoflinethroughpoints(1,-2,3)and(1,1,0)is
-------------------------------------------------------------------------------------------------
Question191
Theplanewhichbisectsthelinejoiningthepoints(4,–2,3)and(2,4,–
1)atrightanglesalsopassesthroughthepoint:
[Sep.03,2020(II)]
Options:
A.(4,0,1)
B.(0,–1,1)
C.(4,0,–1)
D.(0,1,–1)
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question192
LetaplanePcontaintwolines→
r=^
i+λ^
i+^
j,λ∈Rand
→
r= −^
j+µ^
j−^
k,µ∈RIfQ(α,β,γ)isthefootoftheperpendicular
drawnfromthepointM (1,0,1)toP,then3(α+β+γ)equals_______.
[NASep.03,2020(II)]
Answer:5
Solution:
( )
( )
x−1
0=y−1
−3=z−0
3−0( = λsay)
∴M(1, −λ+1,λ)
DirectionratiosofPM = [−3, −λ−1,λ−3]
∵PM ⊥AB
∴(−3) . 0+ (−1−λ)(−1)+(λ−3) . 1=0
∴λ=1
∴Footofperpendicular = (1,0,1)
Thispointsatisfytheplane2x +y−z=1
Directionratiosofnormaltotheplaneare<1, −3,2>.
Planepassesthrough(3,1,1).
Equationofplaneis,
1(x−3) − 3(y−1) + 2(z−1) = 0
⇒x−3y +2z −2=0
-------------------------------------------------------------------------------------------------
Question193
Theplanepassingthroughthepoints(1,2,1),(2,1,2)andparallelto
theline,2x=3y,z=1alsothroughthepoint:
[Sep.02,2020(I)]
Options:
A.(0,6,–2)
B.(–2,0,1)
C.(0,–6,2)
D.(2,0,–1)
Answer:B
Solution:
-------------------------------------------------------------------------------------------------
Question194
Normalofplane=
^
i^
j^
k
1 1 0
0 1 −1
→
n= −^
i+^
j+^
k
Directionratiosofnormaltotheplane= < −1,1,1>
Equationofplane
−1(x−1) + 1(y−0) + 1(z−0) = 0
⇒x−y−z−1=0
If(x,y,z)isfootofperpendicularofM(1,0,1)ontheplanethen
⇒x−1
1=y−0
−1=z−1
−1 = −(1−0−1−1)
3
∴x=4
3,y= −1
3,z=2
3
α+β+γ=4
3−1
3+2
3=5
3
∴3(α+β+γ) = 3×5
3=5
| |
Letplanepassesthrough(2,1,2)be
a(x−2) + b(y−1) + (z−2) = 0
Italsopassesthrough(1,2,1)
∴−a+b−c=0⇒a−b+c=0
Thegivenlineis
x
3=y
2=z−1
0isparalleltoplane
∴3a +2b +c(0) = 0
⇒a
0−2=b
3−0=c
2+3
⇒a
2=b
−3=c
2+3
⇒a
2=b
−3=c
−5
∴planeis2x −4−3y +3−5z+10 =0
⇒2x −3y −5z +9=0
Theplanesatisfiesthepoint(-2,0,1).
Aplanepassingthroughthepoint(3,1,1)containstwolineswhose
directionratiosare1,–2,2and2,3,–1respectively.Ifthisplanealso
passesthroughthepoint(α,-3,5),thenαisequalto:
[Sep.02,2020(II)]
Options:
A.5
B.–10
C.10
D.–5
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question195
LetSbethesetofallrealvaluesofλsuchthataplanepassingthrough
thepoints(−λ2,1,1), (1, −λ2,1)and(1,1, −λ2)alsopassesthroughthe
point-(-1,-1,1).ThenSisequalto:
[Jan.12,2019(II)]
Options:
A.{√3}
B.{√3, −√3}
Solution:
∵Planecontainstwolines
∴→
n=
^
i^
j^
k
1−22
2 3 −1
=^
i(2−6) − ^
j(−1−4) + ^
k(3+4)
= −4^
i+5^
j+7^
k
So,equationofplaneis
−4(x−3) + 5(y−1) + 7(z−1) = 0
⇒−4x +12 +5y −5+7z−7=0
⇒−4x +5y +7z =0
Thisalsopassesthrough(α, −3,5)
So,−4α −15 +35 =0
⇒−4α = −20 ⇒α=5
| |
C.{1, −1}
D.{3, −3}
Answer:B
:
-------------------------------------------------------------------------------------------------
Question196
Theplanecontainingthelinex−3
2=y+2
−1=z−1
3andalsocontainingits
projectionontheplane2x +3y −z=5,containswhichoneofthe
followingpoints?
[Jan.11,2019(I)]
Options:
A.(2,2,0)
Solution:
-------------------------------------------------------------------------------------------------
Question197
4withtheplaney −z+5=0are:
[Jan.11,2019(I)]
Options:
LetA(−λ2,1,1), B(1, −λ2,1),C(1,1, −λ2), D(−1, −1,1)
lieonsameplane,then
1−λ22 0
2 1 −λ20
2 2 −λ2−1
=0
⇒(λ2+1)((1−λ2)2−4) = 0
⇒(3−λ2)(λ2+1) = 0⇒λ2=3
λ= ±√3
Hence,S= {−√3, √3}
| |
Letnormaltotherequiredplaneis→
n
⇒→
nisperpendiculartobothvector2^
i−^
j+3^
kand2^
i+3^
j−3^
k
⇒→
n=
^
i^
j^
k
2−13
2 3 −1
= −8^
i+8^
j+8^
k
⇒Equationoftherequiredplaneis
⇒(x−3)(−8) + (y+2)×8+ (z−1) × 8=0
⇒(x−3)(−1) + (y+2) × 1+(z−1) × 1=0
⇒x−3−y−2−z+1=0
∵x−y−z=4passesthrough(2,0,-2)
∴planecontains(2,0,-2)
| |
Thedirectionratiosofnormaltotheplanethroughthepoints(0,-1,0)
and(0,0,1)andmakinganangleπ
B.(-2,2,2)
C.(0,-2,2)
D.(2,0,-2)
Answer:D
A.2,-1,1
B.2, √2, −√2
C.√2,1, −1
Solution:
-------------------------------------------------------------------------------------------------
Question198
Ifthepoint(2,α,β)liesontheplanewhichpassesthroughthepoints
(3,4,2)and(7,0,6)andisperpendiculartotheplane2x −5y =15,then
2α −3βisequalto:
[Jan.11,2019(II)]
Options:
A.12
B.7
C.5
D.17
Answer:B
Solution:
(b,c)Letthed.r'softhenormalbe‹a,b,c›
Equationoftheplaneis
a(x−0) + b(y+1) + c(z−0) = 0
∵Itpassesthrough(0,0,1)
∴b+c=0
Also 0.a+b−c
a2+b2+c2. √2
= cos π
4=1
√2
⇒b−c=a2+b2+c2
Andb+c=0
⇒b= ± 1
√2a
∴Thed.r'sare√2,1, −1or2, √2, −√2
√
√
Letthenormaltotherequiredplaneis→
n,then
→
n=
^
i^
j^
k
4−44
2−50
= 20^
i+8^
j−12^
k
∴Equationoftheplane
(x−3) × 20 + (y−4) × 8+ (z−2)×(−12) = 0
5x −15 +2y −8−3z +6=0
5x +2y −3z −17 =0.......(1)
| |
D.2√3,1, −1
Answer:
B & D
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Question199
Theplanewhichbisectsthelinesegmentjoiningthepoints(–3,–3,4)
and(3,7,6)atrightangles,passesthroughwhichoneofthefollowing
points?
[Jan.10,2019(II)]
Options:
A.(–2,3,5)
B.(4,–1,7)
C.(2,1,3)
D.(4,1,–2)
Answer:D
Solution:
Solution:
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Question200
Onwhichofthefollowinglinesliesthepointofin-ter-sectionofthe
line,x−4
2=y−5
2=z−3
1andtheplane,x +y+z=2?
[Jan.10,2019(II)]
Options:
A.x+3
3=4−y
3=z+1
−2
B.x−4
1=y−5
1=z−5
−1
Since,equationofplane(1)passesthrough(2,α,β),then10 +2α −3β −17 =0⇒2α −3β =7
Since,directionratiosofnormaltotheplaneis
→
n=6^
i+10^
j+2^
k
Then,equationoftheplaneis
(x−0)6+ (y−2)10 + (z−5)2=0
3x +5y −10 +z−5=0
3x +5y +z=15
Since,plane(1)satisfiesthepoint(4,1,-2)
Hence,requiredpointis(4,1,-2)
C.x−1
1=y−3
2=z+4
−5
D.x−2
2=y−3
2=z+3
3
Answer:C
Solution:
Solution:
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Question201
Thesystemoflinearequations
x+y+z=2
2x +3y +2z =5
2x +3y + (a2−1)z=a+1
[Jan092019I]
Options:
A.isinconsistentwhena =4
B.hasauniquesolutionfor|a| = √3
C.hasinfinitelymanysolutionsfora =4
D.isinconsistentwhen|a| = √3
Answer:D
Solution:
Solution:
Letanypointonthelinex−4
2=y−5
2=z−3
1beA(2λ +4,2λ +5,λ+3)whichliesontheplanex+y+z=2
⇒2λ +4+2λ +5+λ+3=2
⇒5λ = −10 ⇒λ= −2
Then,thepointofintersectionis(0,1,1)
whichliesonthelinex−1
1=y−3
2=z+4
−5
Sincethesystemoflinearequationsare
x+y+z=2.....(1)
2x +3y +2z =5.....(2)
2x +3y + (a2−1)z=a+1.....(1)
Now,Δ=
1 1 1
2 3 2
23a2−1
(ApplyingR3→R3−R2)
⇒Δ=
1 1 1
2 3 2
00a2−3
=a2−3
When,Δ=0⇒a2−3=0⇒ | a| = √3
Ifa2=3,thenplanerepresentedbyeqn(2)andeqn(3)areparallel.
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Question202
Theequationofthelinepassingthrough(−4,3,1),paralleltotheplane
x+2y −z−5=0andintersectingthelinex+1
−3=y−3
2=z−2
−1is:
[Jan092019I]
Options:
A.x−4
2=y+3
1=z+1
4
B.x+4
1=y−3
1=z−1
3
C.x+4
3=y−3
−1=z−1
1
D.x+4
−1=y−3
1=z−1
1
Answer:C
Solution:
Solution:
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Question203
Theplanethroughtheintersectionoftheplanesx +y+z=1and
2x +3y −z+4=0andparalleltoy-axisalsopassesthroughthepoint:
[Jan092019I]
Options:
A.(-3,0,-1)
B.(-3,1,1)
Hence,thegivensystemofequationisinconsistent.
Letanypointontheintersectingline
x+1
−3=y−3
2=z−2
−1=λ(say)
is(−3λ −1,2λ +3, −λ+2)
Since,theabovepointliesonalinewhichpassesthroughthepoint(-4,3,1)
Then,directionratiooftherequiredline
=<−3λ −1+4,2λ +3−3,−λ+2−1>
or<−3λ +3,2λ, −λ+1>
Since,lineisparalleltotheplane
x+2y −z−5=0
Then,perpendicularvectortothelineis^
i+2^
j−^
k
Now(−3λ +3)(1) + (2λ)(2)+(−λ+1)(−1) = 0
⇒λ= −1
Nowdirectionratiooftherequiredline= < 6, −2,2>or<3,−1,1>
Hencerequiredequationofthelineis
x+4
3=y−3
−1=z−1
1
C.(3,3,-1)
D.(3,2,1)
Answer:D
Solution:
Solution:
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Question204
Theequationoftheplanecontainingthestraightlinex
2=y
3=z
4and
perpendiculartotheplanecontainingthestraightlinesx
3=y
4=z
2and
x
4=y
2=z
3is:
[Jan.09,2019(II)]
Options:
A.x −2y +z=0
B.3x +2y −3z =0
C.x +2y −2z =0
D.5x +2y −4z =0
Answer:A
Solution:
Solution:
Since,equationofplanethroughintersectionofplanes
x+y+z=1and2x +3y −z+4=0is
(2x +3y −z+4) + λ(x+y+z−1) = 0
(2+λ)x+ (3+λ)y+ (−1+λ)z+(4−λ) = 0......(1)
But,theaboveplaneisparalleltoy-axisthen
(2+λ) × 0+ (3+λ) × 1+(−1+λ) × 0=0
⇒λ= −3
Hence,theequationofrequiredplaneis
−x−4z +7=0
⇒x+4z −7=0
Therefore,(3,2,1)thepassesthroughthepoint.
Letthedirectionratiosoftheplanecontaininglines
x
3=y
4=z
2andx
4=y
2=z
3is<a,b,c>
∴3a +4b +2c =0
4a +2b +3c =0
∴a
12 −4=b
8−9=c
6−16
a
8=b
−1=c
−10
∴Directionratioofplane= < −8,1,10>
Letthedirectionratioofrequiredplaneis<l,m,n>
Then−8l +m+10n =0.......(1)
and2l +3m +4n =0.........(2)
From(1)and(2),
l
−26 =m
52 =n
−26
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Question205
AtetrahedronhasverticesP(1,2,1), Q(2,1,3), R(−1,1,2)and
O(0,0,0).TheanglebetweenthefacesOPQandPQRis:
[Jan.12,2019(I)]
Options:
A.cos−117
31
B.cos−119
35
C.cos−19
35
D.cos−17
31
Answer:B
Solution:
Solution:
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Question206
Twolinesx−3
1=y+1
3=z−6
−1andx+5
7=y−2
−6=z−3
4intersectatthepointR.The
reflectionofRinthexy-planehascoordinates:
[Jan.11,2019(II)]
Options:
A.(2,–4,–7)
( )
( )
( )
( )
∴D.R.sare<1, −2,1>
∴Equationofplane:x−2y +z=0
Let→
v1and→
v2bethevectorsperpendiculartotheplaneOPQandPQRrespectively.
→
v1=PQ ×OQ=
^
i^
j^
k
121
213
= 5^
i−^
j−3^
k
→
v2=PQ ×PR=
^
i^
j^
k
1−12
−2−11
= ^
i−5^
j−3^
k
∵cos θ =
→
v1.→
v2
→
v1
→
v2
= 5+5+9
25 +1+9=19
35
∴θ=cos−119
35
| |
| |
| | | |
( )
B.(2,4,7)
C.(2,–4,7)
D.(–2,4,7)
Answer:A
Solution:
Solution:
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Question207
Ifthelinesx=ay+b,z=cy+dandx=a'z+b',y=c'z+d'are
perpendicular,then:
[Jan.09,2019(II)]
Options:
A.ab'+bc'+1=0
B.cc'+a+a'=0
C.bb'+cc'+1=0
D.aa'+c+c'=0
Answer:D
Solution:
Solution:
LetthecoordinateofPwithrespecttoline
x−3
1=y+1
3=z−6
−1=λ
x+5
7=y−2
−6=z−3
4=µ
L1= (λ+3,3λ −1, −λ+6)
andcoordinateofPw.r.t.
lineL2= (7µ −5, −6µ +2,4µ +3)
∴λ−7µ = −8,3λ +6µ =3,λ+4µ =3
Fromaboveequation:λ= −1,µ=1
∴CoordinateofpointofintersectionR= (2, −4,7)
ImageofRw.r.t.xyplane = (2, −4, −7)
Firstlineis:x=ay +b,z=cy +d
x−b
a=y
1=z−d
c
andanotherlineis:x=a′z+b′, y=c′z+d′
⇒x−b′
a′=y−d′
c′=z
1
∵Bothlinesareperpendiculartoeachother
∴aa′ + c′ + c=0
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Question208
Theperpendiculardistancefromtheorigintotheplanecontainingthe
twolines,x+2
3=y−2
5=z+5
7andx−1
1=y−4
4=z+4
7,is:
[Jan.12,2019(I)]
Options:
A.11√6
B.11 ∕ √6
C.11
D.6√11
Answer:B
Solution:
Solution:
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Question209
Ifananglebetweentheline,x+1
2=y−2
1=z−3
−2andtheplane,
x−2y −kx =3iscos−12√2
3,thenavalueofkis
[Jan.12,2019(II)]
Options:
A. 5
3
B. 3
5
C.−3
5
D.−5
3
( )
√
√
∵planecontainingbothlines.
D.R.ofplane=
^
i^
j^
k
357
147
= 7^
i−14^
j+7^
k
Now,equationofplaneis,
7(x−1) − 14(y−4) + 7(z+4) = 0
⇒x−1−2y +8+z+4=0
⇒x−2y +z+11 =0
Hence,distancefrom(0,0,0)totheplane,
=11
√1+4+1 = 11
√6
| |
Answer:A
Solution:
Solution:
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Question210
Ifthelengthoftheperpendicularfromthepoint(β,0,β)( β≠0)tothe
line,x
1=y−1
0=z+1
−1is 3
2,thenβisequalto:
[April10,2019(I)]
Options:
A.1
B.2
C.–1
D.–2
Answer:C
Solution:
Solution:
√
Letanglebetweenlineandplaneisθ,then
sin θ =
→
b.→
n
→
b.→
n
=2^
i+^
j−2^
k.^
i−2^
j−K^
k
√9.1+4+K2
=2−2−2K
35+K2=2|K|
3√4+K2
Since,cos θ =2√2
3⇒sin θ =1
3
Then, 2|K|
3√5+K2=1
3⇒4K 2=5+K2
3K 2=5⇒K= ± 5
3
|| | | | |
|( ) ( )
√|
|√|
√
Given,x
1=y−1
0=z+1
−1=p(let)andpointP(β,0,β)
AnypointonlineA= (p,1, −p−1)
Now,DRofAPa"<p−β,1−0, −p−1−β>
Whichisperpendiculartoline.
∴(p−β)1+0.1 −1(−p−1−β) = 0
⇒p−β+p+1+β=0⇒p=−1
2
∴PointA−1
2,1−1
2
GiventhatdistanceAP =3
2⇒AP2=3
2
⇒β+1
2
2+1+β+1
2
2=3
2or2 β +1
2
2=1
2
( )
√
( ) ( ) ( )
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Question211
TheverticesBandCofa"ABClieontheline,x+2
3=y−1
0=z
4suchthat
BC =5units.Thenthearea(insq.units)ofthistriangle,giventhatthe
pointA(1, −1,2),is:
[April09,2019(II)]
Options:
A.5√17
B.2√34
C.6
D.√34
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question212
IfapointR(4,y,z)liesonthelinesegmentjoiningthepointsP(2,–3,4)
⇒β+1
2
2=1
4⇒β=0, −1, (β≠0)
∴β= −1
( )
LetapointDonBC = (3λ −2,1,4λ)
AD = (3λ −3)^
i+2^
j+ (4λ −2)^
k
∵AD ⊥BC,
∴AD .BC =0
⇒(3λ −3) + 3+2(0) + (4λ −2)4=0
⇒λ=17
25
Hence,D=1
25,1,68
25
|AD| = 1
25 −12+ (2)2+68
25 −22
=(24)2+4(25)2+ (18)2
25 = 3400
25 =2√34
5
Areaoftriangle = 1
2×BC ×AD
=1
2×5×2√34
5= √34
( )
√( ) ( )
√√
| | | |
andQ(8,0,10),thendistanceofRfromtheoriginis:
[April08,2019(II)]
Options:
A.2√14
B.2√21
C.6
D.√53
Answer:A
Solution:
Solution:
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Question213
Aperpendicularisdrawnfromapointonthelinex−1
2=y+1
−1=z
1tothe
planex +y+z=3suchthatthefootoftheperpendicularQalsolieson
theplanex −y+z=3.Thentheco-ordinatesofQare:
[April10,2019(II)]
Options:
A.(1,0,2)
B.(2,0,1)
C.(–1,0,4)
D.(4,0,–1)
Answer:B
Solution:
Solution:
Here,P,Q,Rarecollinear
∴PR =λPQ
2^
i+ (y+3)^
j+ (z−4)^
k = λ 6^
i+3^
j+6^
k
⇒6λ =2,y+3=3λ,z−4=6λ
⇒λ=1
3,y= −2,z=6
∴PointR(4, −2,6)
Now,OR =(4)2+ (−2)2+ (6)2 = √56 =2√14
[ ]
√
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Question214
Thelengthoftheperpendicularfromthepoint(2,-1,4)onthestraight
line,x+3
10 =y−2
−7=z
1is:
[April08,2019(I)]
Options:
A.greaterthan3butlessthan4
B.lessthan2
C.greaterthan2butlessthan3
D.greaterthan4
Answer:A
Solution:
Solution:
Letco-ordinatesofQbe(α,β,γ),then
α+β+γ=3.....(i)
α−β+γ=3.....(ii)
⇒α+γ=3andβ=0
Equatingdirectionratio'sofPQ,weget
α−2λ −1
1=λ+1
1=γ−λ
1
⇒α=3λ +2,γ=2λ +1
Substitutingthevaluesofαandγinequation(i),weget
⇒5λ +3=3⇒λ=0
Hence,pointisQ(2,0,1)
LetPbethefootofperpendicularfrompointT(2, −1,4)onthegivenline.SoPcanbeassumedasP
(10λ −3, −7λ +2,λ)
DR'sofT P proptoto10λ −5, −7λ +3,λ−4
∵T Pandgivenlineareperpendicular,so10(10λ −5) − 7(−7λ +3)+1(λ−4) = 0
⇒λ=1
2
⇒T P =(10λ −5)2+ (−7λ +3)2+ (λ−4)2
=0+1
4+49
4= √12.5 =3.54
Hence,thelengthofperpendicularisgreaterthan3butlessthan4
√
√
-------------------------------------------------------------------------------------------------
Question215
Ifthelinex−2
3=y+1
2=z−1
−1intersectstheplane2x +3y −z+13 =0ata
pointPandtheplane3x +y+4z =16atapointQ,thenPQisequalto:
[April12,2019(I)]
Options:
A.14
B.√14
C.2√7
D.2√14
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question216
Aplanewhichbisectstheanglebetweenthetwogivenplanes2x–y+2z
–4=0andx+2y+2z–2=0,passesthroughthepoint:
[April12,2019(II)]
Options:
A.(1,–4,1)
B.(1,4,–1)
C.(2,4,1)
D.(2,–4,1)
Answer:D
Solution:
Solution:
LetpointsP(3λ +2,2λ −1, −λ+1)andQ(3µ +2,2µ −1, −µ+1)
∵Plieson2x +3y −z+13 =0
∴6λ +4+6λ −3+λ−1+13 =0
⇒13λ = −13 ⇒λ= −1
Hence,P(−1, −3,2)
Similarly,Qlieson3x +y+4z =16
∴9µ +6+2µ −1−4µ+4=16
⇒7µ =7⇒µ=1
Hence,Qis(5,1,0)
Now,PQ = √36 +16 +4 = √56 =2√14
Theequationsofanglebisectorsare,
-------------------------------------------------------------------------------------------------
Question217
Thelengthoftheperpendiculardrawnfromthepoint(2,1,4)tothe
planecontainingthelines→
r=^
i+^
j+λ^
i+2^
j−^
kand
→
r=^
i+^
j+µ−^
i+^
j−2^
kis:
[April12,2019(II)]
Options:
A.3
B.1
3
C.√3
D. 1
√3
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question218
IfQ(0, −1, −3)istheimageofthepointPintheplane3x−y +4z =2
andRisthepoint(3, −1, −2),thenthearea(insq.units)ofΔPQRis:
[April10,2019(I)]
Options:
A.2√13
B.√91
4
C.√91
2
( ) ( )
( ) ( )
x+2y +2z −23 = ±2x −y+2z −4
3
⇒x−3y −2=0
or3x +y+4z −6=0
(2,-4,1)liesonthesecondplane.
Theequationofplanecontainingtwogivenlinesis,
x−1 y −1z
1 2 −1
−11−2
=0
Onexpanding,wegetx−y−z=0
Now,thelengthofperpendicularfrom(2,1,4)tothisplane
=2−1−4
12+12+12 = √3
| |
|√|
D.√65
2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question219
Iftheplane2x −y+2z +3=0hasthedistances1
2and2
3unitsfromthe
planes4x −2y +4z +λ=0and2x −y+2z+µ =0,respectively,thenthe
maximumvalueofλ +µisequalto:
[April10,2019(II)]
Options:
A.9
B.15
C.5
D.13
Answer:D
Solution:
Solution:
ImageofQ(0, −1, −3)inplaneis,
(x−0)
3=(y+1)
−1=z+3
+4 = −2(1−12 −2)
9+1+16 =1
⇒x=3,y= −2,z=1
⇒P(3, −2,1), Q(0, −1, −3), R(3, −1, −2)
∴AreaofΔPQRis
1
2
→
QP ×→
QR = 1
2
^
i^
j^
k
3−14
301
=1
2
^
i(−1) − ^
j(3−12) + ^
k(3)
=1
2√(1+81 +9)=√91
2
| | | | | |
|{ } |
Let,P1:2x −y+2z +3=0
P2:2x −y+2z+λ
2=0
P3:2x −y+2z +µ=0
Given,distancebetweenP1andP2is1
3
1
3=
3−λ
2
√9⇒3−λ
2=1⇒λmax =8
AnddistancebetweenP1andP3is2
3
2
3=|µ−3|
√9⇒µmax =5
| | | |
-------------------------------------------------------------------------------------------------
Question220
Iftheline,x−1
2=y+1
3=z−2
4meetstheplane,x +2y +3z =15atapointP,
thenthedistanceofPfromtheoriginis:
[April092019I]
Options:
A.√5∕2
B.2√5
C.9 ∕2
D.7 ∕2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question221
Aplanepassingthroughthepoints(0,-1,0)and(0,0,1)andmakingan
angleπ
4withtheplaney −z+5=0,alsopassesthroughthepoint:
[April092019I]
Options:
A.(−√2,1, −4)
B.(√2, −1,4)
C.(−√2, −1, −4)
D.(√2,1,4)
Answer:D
Solution:
⇒(λ+µ)max =13
LetpointonlinebeP(2k +1,3k −1,4k +2)
Since,pointPliesontheplanex+2y +3z =15
∴2k +1+6k −2+12k+6=15
⇒k=1
2
∴P≡2,1
2,4
ThenthedistanceofthepointPfromtheoriginisOP =4+1
4+16 =9
2
( )
√
Solution:
-------------------------------------------------------------------------------------------------
Question222
LetPbetheplane,whichcontainsthelineofintersectionoftheplanes,
x+y+z−6=0and2x +3y +z+5=0anditisperpendiculartothexy-
plane.Thenthedistanceofthepoint(0,0,256)fromPisequalto:
[April09,2019(II)]
Options:
A.17 ∕ √5
B.63√5
C.205√5
D.11 ∕ √5
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question223
Theequationofaplanecontainingthelineofintersectionoftheplanes
2x −y−4=0andy +2z −4=0andpassingthroughthepoint(1,1,0)is:
[April082019I]
Options:
A.x −3y −2z = −2
B.2x −z=2
Lettherequiredplanepassingthroughthepoints(0,-1,0)and(0,0,1)bex
λ+y
−1+z
1=1andthegivenplaneis
y−z+5=0
∴cos π
4=−1−1
1
λ2+1+1√2
⇒λ2=1
2⇒1
λ= ±√2
Then,theequationofplaneis±√2x−y+z=1
Thenthepoint(√2,1,4)satisfiestheequationofplane
√( )
Lettheplanebe
P≡ (2x +3y +z+5)+λ(x+y+z−6) = 0
∵aboveplaneisperpendiculartoxyplane.
∴ (2+λ)^
i+ (3+λ)^
j+ (1+λ)^
k.^
k=0⇒λ= −1
Hence,theequationoftheplaneis,
P≡x+2y +11 =0
DistanceoftheplanePfrom(0,0,256)
0+0+11
√5=11
√5
( )
| |
C.x −y−z=0
D.x +3y +z=4
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question224
Thevectorequationoftheplanethroughthelineofintersectionofthe
planesx +y+z=1and2x +3y +4z =5whichisperpendiculartothe
planex −y+z=0is:
[April08,2019(II)]
Options:
A.→
r×^
i−^
k+2=0
B.→
r.^
i−^
k−2=0
C.→
r×^
i+^
k+2=0
D.→
r.^
i−^
k+2=0
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question225
Thelengthoftheprojectionofthelinesegmentjoiningthepoints(5,–
1,4)and(4,–1,3)ontheplane,x+y+z=7is:
( )
( )
( )
( )
Lettheequationofrequiredplanebe;
(2x −y−4) + λ(y+2z −4) = 0
∵Thisplanepassesthroughthepoint(1,1,0)then(2−1−4) + λ(1+0−4) = 0
⇒λ= −1
Then,equationofrequiredplaneis,
(2x −y−4) − (y+2z −4) = 0
⇒2x −2y −2z =0⇒x−y−z=0
Equationoftheplanepassingthroughthelineofintersectionofx+y+z=1and2x +3y +4z =5is
(2x +3y +4z −5)+λ(x+y+z−1) = 0
⇒(2+λ)x+ (3+λ)y+ (4+λ)z+(−5−λ) = 0.......(i)
∵plane(i)isperpendiculartotheplanex−y+z=0
∴(2+λ)(1) + (3+λ)(−1)+(4+λ)(1) = 0
2+λ−3−λ+4+λ=0⇒λ= −3
Hence,equationofrequiredplaneis
−x+z−2=0orx−z+2=0
⇒r.^
i−^
k+2=0
( )
[2018]
Options:
A.2
3
B.1
3
C. 2
3
D. 2
√3
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question226
Ananglebetweenthelineswhosedirectioncosinesaregivenbythe
equations,l +3m +5n =0and5l m −2mn +6nl =0,is
[OnlineApril15,2018]
Options:
A.cos−11
8
B.cos−11
6
C.cos−11
3
D.cos−11
4
Answer:B
√
( )
( )
( )
( )
AC =→
AB .^
AC = ^
i+^
k.
^
i+^
j+^
k
√3 = 2
√3
Now,A′B′ = BC =AB2−AC2 = 2−4
3=2
3
∴Lengthofprojection = 2
3
( ) ( )
√√ √
√
Solution:
-------------------------------------------------------------------------------------------------
Question227
Iftheanglebetweenthelines,x
2=y
2=z
1and5−x
−2=7y −14
P=z−3
4iscos−12
3,
thenPisequalto
[OnlineApril16,2018]
Options:
A.−7
4
B.2
7
C.−4
7
D.7
2
Answer:D
Solution:
Solution:
( )
Given
l+3m +5n =0.......(1)
and5l m −2mn +6nl =0.......(2)
Fromeq.(1)wehave
l= −3m −5n
Putthevalueoflineq.(2),weget;
Putthevalueoflineq.(2),weget;
5(−3m −5n)m−2mn +6n(−3m −5n) = 0
⇒15m2+45mn +30n2=0
⇒m2+3mn +2n2=0
⇒m2+2mn +mn +2n2=0
⇒(m+n)(m+2n) = 0
∴m= −norm= −2n
Form= −n,l= −2n
Andform= −2n,l=n
∴(l,m,n) = (−2n, −n,n)Or(l,m,n) = (n, −2n,n)
⇒(l,m,n) = (−2, −1,1)Or(l,m,n) = (1, −2,1)
Therefore,anglebetweenthelinesisgivenas:
cos(θ) = (−2)(1) + (−1) . (−2) + (1)(1)
√6. √6
⇒cos(θ) = 1
6⇒θ=cos−11
6
( )
Letθbetheanglebetweenthetwolines
Heredirectioncosinesofx
2=y
2=z
1are2,2,1
Alsosecondlinecanbewrittenas:
x−5
2=y−2
P
7
=z−3
4
∴itsdirectioncosinesare2,P
7,4
-------------------------------------------------------------------------------------------------
Question228
IfL1isthelineofintersectionoftheplanes
2x −2y +3z −2=0,x−y+z+1=0andL2isthelineofintersectionof
theplanesx +2y −z−3=0,3x −y+2z −1=0,thenthedistanceofthe
originfromtheplane,containingthelinesL1andL2,is:
[2018]
Options:
A. 1
3√2
B. 1
2√2
C. 1
√2
D. 1
4√2
Answer:A
Solution:
Solution:
Also,cos θ =2
3(Given)
∵cos θ =a1a2+b1b2+c1c2
a1
2+b1
2+c1
2a2
2+b2
2+c2
2
⇒2
3=
(2×2) + 2×P
7+ (1×4)
22+22+1222+P2
49 +42
=
4+2P
7+4
3×22+P2
49 +42
⇒4+P
7
2=20 +P2
49⇒16 +8P
7+P2
49 =20 +P2
49
⇒8P
7=4⇒P=7
2
|√ √ |
|( )
√√|
√
( )
Equationofplanepassingthroughthelineofintersectionoffirsttwoplanesis:
(2x −2y +3z −2)+λ(x−y+z+1) = 0
orx(λ+2) − y(2+λ) + z(λ+3)+(λ−2) = 0......(i)
ishavinginfinitenumberofsolutionwith
x+2y −z−3=0and3x −y+2z −1=0,then
(λ+2) −(λ+2) (λ+3)
1 2 −1
3−12
=0
Nowputλ=5in(i),weget
7x −7y +8z +3=0
Nowperpendiculardistancefrom(0,0,0)totheplacecontainingL1andL2=3
√162 =1
3√2
| |
-------------------------------------------------------------------------------------------------
Question229
Thesumoftheinterceptsonthecoordinateaxesoftheplanepassing
throughthepoint(-2,-2,2)andcontainingthelinejoiningthepoints
(1,-1,2)and(1,1,1)is
[OnlineApril16,2018]
Options:
A.12
B.-8
C.-4
D.4
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question230
Avariableplanepassesthroughafixedpoint(3,2,1)andmeetsx,yand
zaxesatA,BandCrespectively.Aplaneisdrawnparalleltoyz−plane
throughA,asecondplaneisdrawnparallelzx−planethroughBanda
thirdplaneisdrawnparalleltoxy−planethroughC.Thenthelocusof
Equationofplanepassingthroughthreegivenpointsis:
x−x1y−y1z−z1
x2−x1y2−y1z2−z1
x3−x1y3−y1z3−z1
=0
⇒
x+2 y +2 z −2
1+2−1+2 2 −2
1+2 1 +2 1 −2
=0
⇒
x+2 y +2 z −2
3 1 0
3 3 −1
=0
⇒−x+3y +6z −8=0
⇒x
8−3y
8−6z
8+8
8=0
⇒x
8−y
8
3
−z
8
6
= −1
⇒x
−8+y
8
3
+z
8
6
=1
∴Sumofintercepts = −8+8
3+8
6= −4
| |
| |
| |
thepointofintersectionofthesethreeplanes,is
[OnlineApril15,2018]
Options:
A.x +y+z=6
B.x
3+y
2+z
1=1
C.3
x+2
y+1
z=1
D.1
x+1
y+1
z=11
6
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question231
Ananglebetweentheplane,x +y+z=5andthelineofintersectionof
theplanes,3x +4y +z−1=0and5x +8y +2z +14 =0,is
[OnlineApril15,2018]
Options:
A.cos−13
√17
B.cos−13
17
C.sin−13
√17
D.sin−13
17
Answer:D
Solution:
Solution:
( )
(√)
( )
(√)
Ifa,b,caretheinterceptsofthevariableplaneonthex,y,zaxesrespectively,thentheequationoftheplaneis
x
a+y
b+z
c=1
Andthepointofintersectionoftheplanesparalleltothexy,yzandzxplanesis(a,b,c).
Asthepoint(3,2,1)liesonthevariableplane,so
3
a+2
b+1
c=1
Therefore,therequiredlocusis3
x+2
y+1
z=1
Normalto3x +4y +z=1is3^
i+4^
j+^
k.
-------------------------------------------------------------------------------------------------
Question232
Aplanebisectsthelinesegmentjoiningthepoints(1,2,3)and(–3,4,
5)atrightangles.Thenthisplanealsopassesthroughthepoint.
[OnlineApril15,2018]
Options:
A.(–3,2,1)
B.(3,2,1)
C.(1,2,–3)
D.(–1,2,3)
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question233
IftheimageofthepointP(1, −2,3)intheplane,2x +3y −4z +22 =0
measuredparalleltoline,x
1=y
4=z
5isQ,thenPQisequalto:
[2017]
Options:
A.6√5
B.3√5
C.2√42
Normalto5x +8y +2z = −14is5^
i+8^
j+2^
k
Thelineofintersectionoftheplanesisperpendiculartobothnormals,so,directionratiosoftheintersectionlineare
directlyproportionaltothecrossproductofthenormalvectors.
Thereforethedirectionratiosofthelineis−^
j+4^
k
Hencetheanglebetweentheplanex+y+z+5=0andtheintersectionlineissin−1−1+4
√17√3=sin−13
17
( ) (√)
Sincetheplanebisectsthelinejoiningthepoints(1,2,3)and(-3,4,5)thentheplanepassesthroughthemidpointofthe
linewhichis:
1−3
2,2+4
2,5+3
2≡−2
2,6
2,8
2≡(−1,3,4)
Asplanecutsthelinesegmentatrightangle,sothedirectioncosinesofthenormaloftheplaneare
(−3−1,4−2,5−3) = (−4,2,2)
Sotheequationoftheplaneis:−4x +2y +2z =λ
Asplanepassesthrough(-1,3,4)so
−4(−1) + 2(3) + 2(4) = λ⇒λ=18
Therefore,equationofplaneis:−4x +2y +2z =18
Now,only(-3,2,1)satisfiesthegivenplaneas
−4(−3) + 2(2) + 2(1) = 18
( ) ( )
D.√42
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question234
Thedistanceofthepoint(1,3,-7)fromtheplanepassingthroughthe
point(1, −1, −1),havingnormalperpendiculartoboththelines
x−1
1=y+2
−2=z−4
3andx−2
2=y+1
−1=z+7
−1,is:
[2017]
Options:
A. 10
√74
B. 20
√74
C. 10
√83
D. 5
√83
Answer:C
Solution:
Solution:
EquationoflinePQis
x−1
1=y+2
4=z−3
5
LetFbe(λ+1,4λ −2,5λ +3)
SinceFliesontheplane
∴2(λ+1) + 3(4λ −2) − 4(5λ +3)+22 =0
2λ +2+12λ −6−20λ −12+22 =0
⇒−6λ +6=0⇒λ=1
∴Fis(2,2,8)
PQ =2PF =2 12+42+52 = 2√42
√
Lettheplanebe
a(x−1) + b(y+1) + c(z+1) = 0
Normalvector
-------------------------------------------------------------------------------------------------
Question235
Ifx =a,y=b,z=cisasolutionofthesystemoflinearequations
x+8y +7z =0
9x +2y +3z =0
x+y+z=0
suchthatthepoint(a,b,c)liesontheplanex +2y +z=6,then
2a +b+cequals:
[OnlineApril9,2017]
Options:
A.-1
B.0
C.1
D.2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question236
Ifavariableplane,atadistanceof3unitsfromtheorigin,intersects
thecoordinateaxesatA,BandC,thenthelocusofthecentroidof
ΔABCis:
[OnlineApril9,2017]
^
i^
j^
k
1−2 3
2−1−1
= 5^
i+7^
j+3^
k
Soplaneis5(x−1) + 7(y+1) + 3(z+1) = 0
⇒5x +7y +3z +5=0
Distanceofpoint(1,3,-7)fromtheplaneis
5+21 −21 +5
√25 +49 +9 = 10
√83
| |
x+8y +7z =0
9x +2y +3z =0
x+y+z=0
x=λ y =6λ z = −7λ
x=λ y =6λ z = −7λ
Now,λ+12λ −7λ =6
6λ =6
λ=1
∴2λ +6λ −7λ
=λ
=1
Options:
A. 1
x2+1
y2+1
z2=1
B. 1
x2+1
y2+1
z2=3
C. 1
x2+1
y2+1
z2=1
9
D. 1
x2+1
y2+1
z2=9
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question237
Iftheline,x−3
1=y+2
−1=z+λ
−2liesintheplane,2x−4y +3z =2,thenthe
shortestdistancebetweenthislineandtheline,x−1
12 =y
9=z
4is:
[OnlineApril9,2017]
Options:
A.2
B.1
C.0
D.3
Answer:C
Solution:
Solution:
Supposecentroidbe(h,k,l)
∴x−int p =3h,y−int p =3k,z−int p =3l
Equation x
3h +y
3k +z
3l =1
∴Distancefrom(0,0,0)
−1
1
9h2+1
9k2+1
9l 2
=3
⇒1
x2+1
y2+1
z2=1
|√|
Point(3, −2, −λ)onpline2x −4y +3z −2=0 = 6+8−3λ −2 = 0=3λ =12
λ=4
Now,
x−3
1=y+2
−1=z+4
−2=k1......(i)
x−1
12 =y
9=z
4=k2.......(ii)
-------------------------------------------------------------------------------------------------
Question238
Thecoordinatesofthefootoftheperpendicularfromthepoint(1,-2,1)
ontheplanecontainingthelines,x+1
6=y−1
7=z−3
8andx−1
3=y−2
5=z−3
7,is:
[OnlineApril8,2017]
Options:
A.(2,-4,2)
B.(-1,2,-1)
C.(0,0,0)
D.(1,1,1)
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question239
Thelineofintersectionoftheplanes→
r.3^
i−^
j+^
k=1 and
→
r.^
i+4^
j−2^
k=2,is:
[OnlineApril8,2017]
Options:
A.
x−4
7
−2=y
7=
z−5
7
13
( )
( )
Pointonequation(i)P(k1+3, −k1−2, −2k1−4)
Pointonequation(ii)Q(12k2+1,9k2,4k2)
k1+3=12k2+1| −k1−2 = 9k2| −2k1−4=4k2
k2=0
k1= −2
p(1,0,0)lieonequationofaline1
givesshortestdistance = 0
→
n=→
n1×→
n2=
^
i^
j^
k
678
357
= (9, −18,9) = (1, −2,1)
∴Equationofplaneis
1(x+1) − 2(y−1) + (z−3) = 0
⇒x−2y +z=0
foottoz
x−1
1=y+2
−2=z−1
1 = −[1+4+1]
6
x=0,y=0,z=0
| |
B.
x−4
7
2=y
−7=
z+5
7
13
C.
x−6
13
2=
y−5
13
−7=z
−13
D.
x−6
13
2=
y−5
13
7=z
−13
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question240
ABCistriangleinaplanewithverticesA(2,3,5), B( −1,3,2)and
C(λ,5,µ).IfthemedianthroughAisequallyinclinedtothecoordinate
axes,thenthevalueof(λ3+µ3+5)is:
[OnlineApril10,2016]
Options:
A.1130
B.1348
C.1077
D.676
Answer:B
Solution:
Solution:
→
n=→
n1×→
n2⇒
^
i^
j^
k
3−11
1 4 −2
= ^
i(−2) − ^
j(−7) + ^
k(13)⇒→
n= −2^
i+7^
j+13^
k
Now,
3x −y+z=1
x+4y −2z =2
butz=0&solvingthegiven
x=6∕13&y =5∕13
∴requiredequationofalineis
x−6∕13
2=y−5∕13
−7=z
−13
| |
DR'sofADareλ−1
2−2,4−3,µ+22 −5
i.e.λ−5
2,1,µ−8
2
∵Thismedianismakingequalangleswithcoordinateaxes,therefore,
-------------------------------------------------------------------------------------------------
Question241
Thenumberofdistinctrealvaluesoflambdaforwhichthelines
x−1
1=y−2
2=z+3
λ2andx−3
1=y−2
λ2=z−1
2arecoplanaris:
[OnlineApril10,2016]
Options:
A.2
B.4
C.3
D.1
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question242
Theshortestdistancebetweenthelinesx
2=y
2=z
1andx+2
−1=y−4
8=z−5
4lies
intheinterval:
[OnlineApril9,2016]
Options:
λ−5
2=1=µ−8
2
⇒λ=7&µ =10
∴λ3+µ3=5=1348
Linesarecoplanar
3−1 2 −2 1 − (−3)
1 2 λ2
1λ22
=0
⇒
2 0 4
1 2 λ2
1λ22
=0
⇒2(4−λ4) + 4(λ2−2) = 0
⇒4−λ4+2λ2−4=0⇒λ2(λ2−2) = 0
⇒λ=0, √2, −√2
| |
| |
A.(3,4]
B.(2,3]
C.[1,2)
D.[0,1)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question243
Iftheline,x−3
2=y+2
−1=z+4
3liesintheplane,1x +my −z=9,thenl 2+m2
isequalto:
[2016]
Options:
A.5
B.2
C.26
D.18
Answer:B
Solution:
Solution:
Shortestdistancebetweentwolines
x−x1
a1
=y−y1
b1
=z−z1
c1
andx−x2
a2
=y−y2
b2
=z−z2
c2
isgivenby,
x2−x1y2−y1z2−z1
a1b1c1
a2b2c2
(b1c2−b2c1)2+ (c1a2−c2a1)2+ (a1b2−a2b1)2
∴Theshortestdistancebetweengivenlinesare
−24 5
2 2 1
−18 4
(8−8)2+ (−1−8)2+ (16 +2)2
=0−36 +90
√405 =54
20.1 =2.68
|| |
√|
|| |
√|
| |
Lineliesintheplane⇒(3, −2, −4)lieintheplane
⇒3l −2m +4=9or3l −2m =5.......(1)
Also,l,m, −1aredr'soflineperpendiculartoplaneand2.-1,3aredr'soflinelyingintheplane
⇒2l −m−3=0or2l −m=3.......(2)
Solving(1)and(2)wegetl=1andm= −1
-------------------------------------------------------------------------------------------------
Question244
Thedistanceofthepoint(1,-5,9)fromtheplanex −y+z=5measured
alongthelinex =y=zis:
[2016]
Options:
A.10
√3
B.20
3
C.3√10
D.10√3
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question245
Thedistanceofthepoint(1,-2,4)fromtheplanepassingthroughthe
point(1,2,2)andperpendiculartotheplanesx −y+2z =3and
2x −2y +z+12 =0,is
[OnlineApril9,2016]
Options:
A.2
⇒l2+m2=2
E qnofPO :x−1
1=y+5
1=z−9
1=λ
⇒x=λ+1;y=λ−5;z=λ+9
Puttingtheseineqnofplane:
λ+1−λ+5+λ+9=5
⇒λ= −10
⇒Ois(-9,-15,-1)
⇒distanceOP =10√3
B.√2
C.2√2
D. 1
√2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question246
Theequationoftheplanecontainingtheline2x −5y +z=
3;x+y+4z =5,andparalleltotheplane,x +3y +6z =1,is:
[2015]
Options:
A.x +3y +6z =7
B.2x +6y +12z = −13
C.2x +6y +12z =13
D.x +3y +6z = −7
Answer:A
Solution:
Solution:
Letequationofplanebe
a(x−1) + b(y−2) + c(z−2) = 0.......(1)
isperpendiculartogivenplanesthen
a−b+2c =0
2a −2b +c=0
Solvingaboveequationc=0anda=b
equationofplane(1)canbe
x+y−3=0
distancefrom(1,-2,4)willbe
D=|1−2−3|
√1+1=4
√2=2√2
Equationoftheplanecontainingthelines
2x −5y +z=3andx+y+4z =5is
2x −5y +z−3+λ(x+y+4z −5) = 0
⇒(2+λ)x+ (−5+λ)y+ (1+4λ)z+ (−3−5λ) = 0
Sincetheplane(i)paralleltothegivenplane
x+3y +6z =1
∴2+λ
1=−5+λ
3=1+4λ
6
⇒λ= −11
2
Henceequationoftherequiredplaneis
2−11
2x+ −5−11
2y+1−44
2z+ −3+55
2=0
⇒(4−11)x+ (−10 −11)y+ (2−44)z+(−6+55) = 0
⇒−7x −21y −42z +49 =0
( ) ( ) ( ) ( )
-------------------------------------------------------------------------------------------------
Question247
Thedistanceofthepoint(1,0,2)fromthepointofintersectionofthe
linex−2
3=y+1
4=z−2
12 andtheplanex −y+z=16,is
[2015]
Options:
A.3√21
B.13
C.2√14
D.8
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question248
Theshortestdistancebetweenthez-axisandthelinex+y+2z −3
=0=2x +3y +4z −4,is
[OnlineApril11,2015]
Options:
A.1
B.2
C.4
D.3
Answer:B
Solution:
⇒x+3y +6z −7=0
⇒x+3y +6z =7
Generalpointongivenline≡P(3r +2,4r −1,12r +2)
PointPmustsatisfyequationofplane
(3r +2) − (4r −1) + (12r +2) = 16
11r +5=16
r=1
P(3×1+2,4×1−1,12 ×1+2) = P(5,3,14)
distancebetweenPand(1,0,2)
D=(5−1)2+32+ (14 −2)2 = 13
√
-------------------------------------------------------------------------------------------------
Question249
Aplanecontainingthepoint(3,2,0)andthelinex−1
1=y−2
5=z−3
4also
containsthepoint:
[OnlineApril11,2015]
Options:
A.(0,3,1)
B.(0,7,-10)
C.(0,-3,1)
D.0,7,10
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question250
Ifthepoints(1,1,λ)and(-3,0,1)areequidistantfromtheplane,
3x +4y −12z +13 =0,thenλsatisfiestheequation:
[OnlineApril10,2015]
Theequationofanyplanepassingthroughgivenlineis
(x+y+2z −3) + λ(2x +3y+4z −4) = 0
⇒(1+2λ)x+ (1+3λ)y+ (2+4λ)z−(3+4λ) = 0
Ifthisplaneisparalleltoz-axisthennormaltotheplanewillbeperpendiculartoz-axis.
∴(1+2λ)(0) + (1+3λ)(0)+(2+4λ)(1) = 0
λ= −1
2
Thus,Requiredplaneis
(x+y+2z −3) − 1
2(2x +3y +4z −4) = 0⇒y+2=0
∴S.D=2
(1)2=2
√
Equationoftheplanecontainingthegivenline
x−1
1=y−2
5=z−3
4is
A(x−1) + B(y−2) + C(z−3) = 0......(i)
whereA+5B +4C =0.......(ii)
Sincethepoint(3,2,0)containsintheplane(i),therefore2A +0.B−3C =0.......(iii)
Fromequations(ii)and(iii),
A
−15 −0=B
6+3=C
0−10 =k(let)
⇒A= −15k,B=9kandC= −10k
PuttingthevalueofA,BandCinequation(i),weget
−15(x−1) + 9(y−2) − 10(z−3) = 0.......(iv)
Nowthecoordinatesofthepoint(0,-3,1)
satisfytheequationoftheplane(iv)as
−15(0−1) + 9(−3−2)−10(1−3)
=15 −45 +20 =0
Hencethepoint(0,–3,1)containsintheplane.
Options:
A.3x2+10x −13 =0
B.3x2−10x +21 =0
C.3x2−10x +7=0
D.3x2+10x −7=0
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question251
Iftheshortestdistancebetweenthelinesx−1
α=y+1
−1=z
1, (α≠ −1)and
x+y+z+1=0=2x −y+z+3is 1
√3,thenavalueαis:
[OnlineApril10,2015]
Options:
A.−16
19
B.−19
16
C.32
19
D.19
32
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
|3+4−12λ +13|= | −9+0−12 +13|
⇒ | −12λ +20 |=|8|⇒3λ −5| = 2
⇒9λ2+25 −30λ =4⇒9λ2−30λ +21 =0
⇒3λ2−10λ +7=0
Planepassingthroughx+y+z+1=0and2x −y+z+3=0isx+y+z+1+λ(2x −y+z+3) = 0
⇒(2λ +1)x+ (1−λ)y+ (1+λ)z+3λ +1=0
Paralleltothegivenlineif
α(2λ +1) − 1(1−λ) + 1(1+λ) = 0
⇒α=−2λ
2λ +1.......(i)
Also, 2λ +1− (1−λ) + 0+3λ +1
(2λ +1)2+ (1−λ)2+ (1+λ)2 = 1
√3
⇒λ=0,−32
102 ;α=0orα=32
19
|√|
Question252
Theanglebetweenthelineswhosedirectioncosinessatisfythe
equationsl +m+n=0andl 2+m2+n2is
[2014]
Options:
A.π
6
B.π
2
C.π
3
D.π
4
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question253
LetA(2,3,5), B(−1,3,2)andC(λ,5,µ)betheverticesofaΔABC.Ifthe
medianthroughAisequallyinclinedtothecoordinateaxes,then:
[OnlineApril11,2014]
Options:
A.5λ −8µ =0
B.8λ −5µ =0
C.10λ −7µ =0
Given,l+m+n=0andl2=m2+n2
Now,(−m−n)2=m2+n2
⇒mn =0⇒m=0orn=0
Ifm=0thenl= −n
Weknowl2+m2+n2=1⇒n= ± 1
√2
i.e.(l1,m1,n1) = − 1
√2,0,1
√2
Ifn=0thenl= −m
l2+m2+n2=1⇒2m2=1
⇒m= ± 1
√2
Letm=1
√2
⇒l= − 1
√2andn=0
(l2,m2,n2) = − 1
√2,1
√2,0
∴cos θ =1
2⇒θ=π
3
( )
( )
D.7λ −10µ =0
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question254
Alineinthe3-dimensionalspacemakesanangleθ 0 <θ≤π
2with
boththexandyaxes.Thenthesetofallvaluesofthetaistheinterval:
[OnlineApril9,2014]
Options:
A. 0,π
4
B. π
6,π
3
C. π
4,π
2
D. π
4,π
2
Answer:C
( )
( ]
[ ]
[ ]
( ]
IfDbethemid-pointofBC,then
D=λ−1
2,4,µ+2
2
DirectionratiosofADareλ−5
2,1,µ−8
2
SincemedianADisequallyinclinedwithcoordinateaxes,thereforedirectionratiosofADwillbeequal,i.e,
λ−5
2
2
λ−5
2
2+1+µ−8
2
2 = 1
λ−5
2
2+1+µ−8
2
2
=
µ−8
2
2
λ−5
2
2+1+µ−8
2
2
⇒λ−5
2
2=1 = µ−8
2
2
⇒λ=7,3andµ=10,6
Ifλ=7andµ=10
Thenλ
µ=7
10 ⇒10λ −7µ =0
( )
( )
( ) ( ) ( ) ( )
( )
( ) ( )
( ) ( )
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question255
Equationofthelineoftheshortestdistancebetweenthelinesx
1=y
−1=z
1
andx−1
0=y+1
−2=z
1is:
[OnlineApril19,2014]
Options:
A.x
1=y
−1=z
−2
B.x−1
1=y+1
−1=z
−2
C.x−1
1=y+1
−1=z
1
D. x
−2=y
1=z
2
Answer:B
Solution:
Solution:
Itmakesθwithxandy-axes.
l=cos θ,m=cos θ,n=cos(π−2θ)
wehavel2+m2+n2=1
⇒cos2θ+cos2θ+cos2(π−2θ) = 1
⇒2cos2θ+ (−cos 2 θ)2=1
⇒2cos2θ−1+cos22θ =0
⇒cos 2 θ − [1+cos 2 θ] = 0
⇒cos 2 θ =0orcos 2 θ = −1
⇒2θ =π∕2or2θ =π
⇒θ=π∕4orθ=π
2
⇒θ=π
4,π
2
[ ]
Letequationoftherequiredlinebe
x−x1
a=y−y1
b=z−z1
c......(i)
Giventwolines
x
1=y
−1=z
1.......(ii)
andx−1
0=y+1
0=z
1......(iii)
Sincetheline(i)isperpendiculartoboththelines(ii)and(iii),therefore
a−b+c=0......(iv)
−2b +c=0.....(v)
From(iv)and(v)c=2banda+b=0,whicharenotsatisfybyoptions(c)and(d).Henceoptions(c)and(d)are
rejected.
Thuspoint(x1,y1,z1)ontherequiredlinewillbeeither(0,0,0)or(1,-1,0).
Nowfootoftheperpendicularfrompoint(0,0,0)totheline(iii) = (1, −2r −1,r)
-------------------------------------------------------------------------------------------------
Question256
Theimageofthelinex−1
3=y−3
1=z−4
−5intheplane2x −y+z+3=0isthe
line:
[2014]
Options:
A.x−3
3=y+5
1=z−2
−5
B.x−3
−3=y+5
−1=z−2
5
C.x+3
3=y−5
1=z−2
−5
D.x+3
−3=y−5
−1=z+2
5
Answer:C
Solution:
Solution:
Thedirectionratiosofthelinejoiningthepoints(0,0,0)and(1, −2r −1,r)are1, −2r −1,r
Sincesumofthexandy-coordinateofdirectionratiooftherequiredlineis0.
∴1−2r −1=0, ⇒r=0
Hencedirectionratioare1,-1,0
Butthez-directionratiooftherequiredlineistwicethey-directionratiooftherequiredline
i.e.0=2(−1),whichisnottrue.
Hencetheshortestlinedoesnotpassthroughthepoint(0,0,0).Thereforeoption(a)isalsorejected.
a−1
2=b−3
−1=c−4
1=λ(let)
a=2λ +1
b=3−λ
c=4+λ
P=a+1
2,b+3
2,c+4
2 = λ+1,6−λ
2,λ+8
2
∴2(λ+1) − 6−λ
2+λ+8
2+3=0
3λ +6=0⇒λ= −2
a= −3,b=5,c=2
Requiredlineisx+3
3=y−5
1=z−2
−5
( ) ( )
-------------------------------------------------------------------------------------------------
Question257
Iftheanglebetweentheline2(x+1) = y=z+4andtheplane
2x −y+ √λz+4=0isπ
6,thenthevalueofλis:
[OnlineApril19,2014]
Options:
A.135
7
B.45
11
C.45
7
D.135
11
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question258
Ifthedistancebetweenplanes,4x −2y −4z +1=0and
4x −2y −4z +d=0is7,thend is:
[OnlineApril12,2014]
Options:
A.41or-42
B.42or-43
C.-41or43
D.-42or44
Givenequationoflinecanbewrittenas
x+1
1=y
2=z+4
2
Eqnofplaneis2x −y+ √λz +4=0
Since,anglebetweenthelineandtheplaneisπ
6
therefore
sin π
6=2(1) + 2(−1) + 2(√λ)
√1+4+4√4+1+λ
1
2=2−2+2√λ
√9√5+λ
⇒√λ
√5+λ=3
4⇒ λ
5+λ=9
16
⇒7λ =45 ⇒λ=45
7
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question259
Asymmetricalformofthelineofintersectionoftheplanesx =ay +b
andz =cy +d is
[OnlineApril12,2014]
Options:
A.x−b
a=y−1
1=z−d
c
B.x−b−a
a=y−1
1=z−d−c
c
C.x−a
b=y−0
1=z−c
d
D.x−b−a
b=y−1
0=z−d−c
d
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question260
Theplanecontainingthelinex−1
1=y−2
2=z−3
3andparalleltotheline
x
1=y
1=z
4passesthroughthepoint:
Givenplanesare
4x −2y −4z +1=0
and4x −2y −4z +d=0
Theyareparallel.
Distancebetweenthemis±7=d−1
√16 +4+16
⇒d−1
6= ±7⇒d=42 +1
or−42 +1i.e.d= −41or43.
Giventwoplanes:
x−ay −b=0andcy −z+d=0
Let,l,m,nbethedirectionratiooftherequiredline.Sincetherequiredlineisperpendiculartonormalofboththeplane,
thereforel−am =0andcm −n=0
⇒l−am +0.n=0and0.l+cm −n=0
∴l
a−0=m
0+1=n
c−0
Hence,d.Roftherequiredlinearea,1,c.
Hence,options(c)and(d)arerejected.
Now,thepoint(a+b,1,c+d)satisfytheequationofthetwogivenplanes.
∴Option(b)iscorrect.
[OnlineApril11,2014]
Options:
A.(1,-2,5)
B.(1,0,5)
C.(0,3,-5)
D.(-1,-3,0)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question261
Equationoftheplanewhichpassesthroughthepointofintersectionof
linesx−1
3=y−2
1=z−3
2andx−3
1=y−1
2=z−2
3andhasthelargestdistance
fromtheoriginis:
[OnlineApril9,2014]
Options:
A.7x +2y +4z =54
B.3x +4y +5z =49
C.4x +3y +5z =50
D.5x +4y +3z =57
Answer:C
Equationoftheplanecontainingtheline
x−1
1=y−2
2=z−3
3is
a(x−1) + b(y−2) + c(z−3) = 0......(i)
wherea.1+b.2 +c.3 =0
i.e.,a+2b +3c =0........(ii)
Sincetheplane(i)paralleltotheline
x
1=y
1=z
4
∴a.1+b.1+c.4=0
i.e.,a+b+4c =0......(iii)
From(ii)and(iii),
a
8−3=b
3−4=c
1−2=k(let)
∴a=5k,b= −k,c= −k
Onputtingthevalueofa,bandcinequation(i),
5(x−1) − (y−2) − (z−3) = 0
⇒5x −y−z=0........(iv)
whenx=1,y=0andz=5;then
L.H.S.ofequation(iv) = 5x −y−2
=5×1−0−5=0
=R.H.S.ofequation
Hencecoordinatesofthepoint(1,0,5)satisfytheequationplanerepresentedbyequations(iv),
Thereforetheplanepassesthroughthepoint(1,0,5)
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question262
LetABCbeatrianglewithverticesatpointsA(2,3,5),B(−1,3,2)and
C(λ,5,µ)inthreedimensionalspace.IfthemedianthroughAisequally
inclinedwiththeaxes,then(λ,µ)isequalto:
[OnlineApril25,2013]
Options:
A.(10,7)
B.(7,5)
C.(7,10)
D.(5,7)
Answer:C
Solution:
Solution:
Givenequationoflinesare
x−1
3=y−2
1=z−3
2.........(1)
andx−3
1=y−1
2=z−2
3......(2)
Anypointonline(1)isP(3λ +1,λ+2,2λ +3)andonline(2)isQ(µ+3,2µ +1,3µ +2)
Onsolving3λ +1=µ+3andλ+2=2µ +1
wegetλ=1,µ=1
∴PointofintersectionoftwolinesisR(4,3,5)
So,equationofplane⊥toORwhereOis(0,0,0)andpassingthroughRis
4x +3y +5z =50
SinceADisthemedian
∴D=λ−1
2,4,µ+2
2
Now,d R'sofADis
a=λ−1
2−2=λ−5
2
b=4−3=1,c=µ+2
2−5=µ−8
2
Also,a,b,caredR's
∴a=kl ,b=km,c=knwherel=m=n
( )
( )
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Question263
Iftheprotectionsofalinesegmentonthex,yandz-axesin3-
dimensionalspaceare2,3and6respectively,thenthelengthofthe
linesegmentis:
[OnlineApril23,2013]
Options:
A.12
B.7
C.9
D.6
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question264
Theacuteanglebetweentwolinessuchthatthedirectioncosines
l,m,n,ofeachofthemsatisfytheequationsl +m+n=0and
l2+m2−n2=0is
[OnlineApril22,2013]
Options:
A.15°
B.30°
C.60°
D.45°
Answer:C
Solution:
Solution:
andl2+m2+n2=1
⇒l=m=n=1
√3
Now,a=1,b=1andc=1
⇒λ=7andµ=10
Lengthofthelinesegment = (2)2+ (3)2+ (6)2=7
√
Letl1,m1,n1andl2,m2,n2bethed.cofline1and2respectively,thenasgiven
-------------------------------------------------------------------------------------------------
Question265
Ifthelinesx−2
1=y−3
1=z−4
−kandx−1
k=y−4
2=z−5
1arecoplanar,thenkcan
have
[2013]
Options:
A.anyvalue
B.exactlyonevalue
C.exactlytwovalues
D.exactlythreevalues
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question266
IftwolinesL1andL2inspace,aredefinedbyL1= { x= √λy+ (√λ−1),
l1+m1+n1=0
andl2+m2+n2=0
andl1
2+m1
2−n1
2=0and
l2
2+m2
2−n2
2=0
( ∵l+m+n=0andl2+m2−n2=0)
Anglebetweenlines,θis
cos theta =l1l2+m1m2+n1n2.....(1)
Asgivenl2+m2=n2andl+m= −n
⇒(−n)2−2l m =n2⇒2l m =0orl m =0
Sol1m1=0,l2m2=0
Ifl1=0,m1≠0thenl1m2=0
Ifm1=0,l1≠0thenl2m1=0
Ifl2=0,m2≠0thenl2m1=0
Ifm2=0,l2≠0thenl1m2=0
Alsol1l2=0andm1m2=0
l2+m2−n2=l2+m2+n2−2n2=0
⇒1−2n2=0⇒n= ± 1
√2
∴n1= ± 1
√2,n2= ± 1
√2
∴cos θ =1
2⇒θ=60°(acuteangle)
Givenlineswillbecoplanar
If
−11 1
1 1 −k
k 2 1
=0
⇒−1(1+2k) − (1+k2)+1(2−k) = 0
⇒k=0, −3
| |
z= (√λ−1)y+ √λ}andL2= { x= √µy+ (1− √µ),z = (1− √µ)y+ √µ}
thenL1isperpendiculartoL2,forallnon-negativerealsλandµ,such
that:
[OnlineApril23,2013]
Options:
A.√λ+ √µ=1
B.λ ≠µ
C.λ +µ=0
D.λ =µ
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question267
Ifthelinesx+1
2=y−1
1=z+1
3andx+2
2=y−k
3=z
4arecoplanar,thenthevalue
ofkis:
[OnlineApril9,2013]
Options:
A.11
2
B.−11
2
C.9
2
D.−9
2
Answer:A
ForL1,
x= √λy + (√λ−1)⇒y=x− (√λ−1)
√λ.......(i)
z= (√λ−1)y+ √λ⇒y=z− √λ
√λ−1.......(ii)
From(i)and(ii)
x− (√λ−1)
√λ=y−0
1=z− √λ
√λ−1.......(A)
Theequation(A)istheequationoflineL1.
SimilarlyequationoflineL2is
x− (1− √µ)
√µ=y−0
1=z− √µ
1− √µ......(B)
SinceL1⊥L2,therefore
√λ√µ+1×1+ (√λ−1)(1− √µ) = 0
⇒√λ+ √µ=0⇒ √λ= −√µ
⇒λ=µ
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question268
Distancebetweentwoparallelplanes2x +y+2z =8and
4x +2y +4z +5=0is
[2013]
Options:
A.3
2
B.5
2
C.7
2
D.9
2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question269
Theequationofaplanethroughthelineofintersectionoftheplanes
x+2y =3,y−2z +1=0,andperpendiculartothefirstplaneis:
[OnlineApril25,2013]
Options:
Twogivenplanesarecoplanar,if
−2− (−1)k−1 0 − (−1)
2 1 3
2 3 4
= 0
⇒
−1 k −1 1
2 1 3
2 3 4
=0
⇒(−1)(4−9) − (k−1)(8−6)+6−2=0
⇒k=11
2
| |
| |
2x +y+2z −8=0....(Plane1)
2x +y+2z +5
2=0....(Plane2)DistancebetweenPlane1and2
=
−8−5
2
22+12+22= −21
6=7
2
|√|| |
A.2x −y−10z =9
B.2x −y+7z =11
C.2x −y+10z =11
D.2x −y−9z =10
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question270
LetQbethefootofperpendicularfromtheorigintotheplane
4x −3y +z+13 =0andRbeapoint(-1,-6)ontheplane.Thenlength
QRis:
[OnlineApril22,2013]
Options:
A.√14
B. 19
2
C.3 7
2
D. 3
√2
Answer:C
Solution:
Solution:
√
√
Equationofaplanethroughthelineofintersectionoftheplanes
x+2y =3,y−2z +1=0is
(x+2y −3) + λ(y−2z +1) = 0
⇒x+ (2+λ)y−2λ(z) − 3+λ=0.......(i)
Now,plane(i)is⊥tox+2y =3
∴Theirdotproductiszero
i.e.1+2(2+λ) = 0⇒λ= −5
2
Thus,requiredplaneis
x+2−5
2y−2×−5
2(z) − 3−5
2=0
⇒x−y
2+5z −11
2=0
⇒2x −y+10z −11 =0
( )
LetPbetheimageofOinthegivenplane.
-------------------------------------------------------------------------------------------------
Question271
Avector→
nisinclinedtox-axisat45°,toy-axisat60°andatanacute
angletoz-axis.If→
nisanormaltoaplanepassingthroughthepoint
(√2, −1,1)thentheequationoftheplaneis:
[OnlineApril9,2013]
Options:
A.4√2x+7y +z−2
B.2x +y+2z =2√2+1
C.3√2x−4y −3z =7
D.√2x−y−z=2
Answer:B
Solution:
Solution:
Equationoftheplane,4x −3y +z+13 =0
OPisnormaltotheplane,thereforedirectionratioofOPareproportionalto4,-3,1
SinceOPpassesthrough(0,0,0)andhasdirectionratioproportionalto4, −3,1.ThereforeequationofOPis
x−0
4=y−0
−3=z−0
1=r(let)
∴x=4r,y= −3r,z=r
LetthecoordinateofPbe(4r, −3r,r)
SinceQbethemidpointofOP
∴Q=2r, −3
2r,r
2
SinceQliesinthegivenplane
4x −3y +z+13 =0
∴8r +9
2r+r
2+13 =0
⇒r=−13
8+9
2+1
2
=−26
26 = −1
∴Q= −2,3
2, −1
2
QR =(−1+2)2+1−3
2
2+ −6+1
2
2
=1+1
4+121
4=37
2
( )
( )
√( ) ( )
√ √
Directioncosinesof→
nare1
2,1
4,1
2.
Equationoftheplane,
-------------------------------------------------------------------------------------------------
Question272
Ifthelinex−1
2=y+1
3=z−1
4andx−3
1=y−k
2=z
1intersect,thenkisequalto:
[2012]
Options:
A.-1
B.2
9
C.9
2
D.0
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question273
Thedistanceofthepoint−^
i+2^
j+6^
kfromthestraightlinethatpasses
throughthepoint2^
i+3^
j−4^
kandisparalleltothevector6^
i+3^
j−4^
kis
[OnlineMay26,2012]
Options:
1
2(x− √2) + 1
4(y+1) + 1
2(z−1) = 0
⇒2(x− √2) + (y+1) + 2(z−1) = 0
⇒2x +y+2z =2√2−1+2
⇒2x +y+2z =2√2+1
Givenlinesarex−1
2=y+1
3=z−1
4
andx−3
1=y−k
2=z
1
∴→
a1=^
i−^
j+^
k,→
b1 = 2^
i+3^
j+4^
k
→
a2=3^
i+k^
j,→
b2=^
i+2^
j+^
k
Givenliesareintersectif
→
a2−→
a1.→
b1×→
b2
→
b1
→
b2
=0
⇒→
a2−→
a1.→
b1×→
b2=0
⇒
2k+1−1
2 3 4
1 2 1
=0
⇒2(3−8) − (k+1)(2−4) − 1(4−3) = 0
⇒2(−5) − (k+1)(−2) − 1(1) = 0
⇒−10 +2k +2−1=0⇒k=9
2
( ) ( )
| | | |
( ) ( )
| |
A.9
B.8
C.7
D.10
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question274
Statement1:Theshortestdistancebetweenthelinesx
2=y
−1=z
2and
x−1
4=y−1
−2=z−1
4is√2.
Statement2:Theshortestdistancebetweentwoparallellinesisthe
perpendiculardistancefromanypointononeofthelinestotheother
line.
[OnlineMay19,2012]
Options:
A.Statement1istrue,Statement2isfalse.
B.Statement1istrue,Statement2istrue,Statement2isacorrectexplanationforStatement
1.
C.Statement1isfalse,Statement2istrue.
D.Statement1istrue,Statement2istrue,,Statement2isnotacorrectexplanationfor
Statement1
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Pointis(–1,2,6)
Linepassesthroughthepoint(2,3,–4)paralleltovectorwhosedirectionratiosis6,3,–4.
Equationisx−2
6=y−3
3=z+4
−4=λ
Anypointonthislineisgivenbyx=6λ +2,y=3λ +3,z= −4λ −4
Now,d.Rsoflinepassingthrough(-1,2,6)and⊥tothislineis{(x+1), (y−2), (z−6)}
So,6(x+1) + 3(y−2) − 4(z−6) = 0
⇒6x +3y −4z +24 =0
Now,6(6λ +2) + 3(3λ +3)+4(4λ +4) + 24 =0
⇒61λ +61 =0⇒λ= −1
So,x= −4,y=0,z=0
Now,distancebetween(-1,2,6)and(-4,0,0)is√9+4+36 = √49 =7
Onsolvingwewillgetshortestdistance≠√2
Question275
Thecoordinatesofthefootofperpendicularfromthepoint(1,0,0)to
thelinex−1
2=y+1
−3=z+10
8are
[OnlineMay12,2012]
Options:
A.(2,–3,8)
B.(1,–1,–10)
C.(5,–8,–4)
D.(3,–4,–2)
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question276
Aequationofaplaneparalleltotheplanex −2y +2z −5=0andata
unitdistancefromtheoriginis:
[2012]
Options:
A.x −2y +2z −3=0
B.x −2y +2z +1=0
C.x −2y +2z −1=0
D.x −2y +2z +5=0
LettheequationofABis
x−1
2=y− (−1)
−3=z− (−10)
8=k
LetLbethefootoftheperpendiculardrawnfromP(1,0,0)
Now,directionratioofPL = (2k, −3k −1,8k −10)anddirectionratioofAB = (2, −3,8)
Since,PLisperpendiculartoAB
∴2(2k) − 3(−3k −1)+8(8k −10) = 0
Now,k=2(1−1) + (−3)(0+1) + 8(0+10)
(2)2+ (−3)2+ (8)2
=0−3+80
4+9+64 =77
77 =1
∴Requiredco-ordinate = L= (2+1, −3−1,8−10)
= (3, −4, −2)
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question277
Theequationofaplanecontainingthelinex+1
−3=y−3
2=z+2
1andthepoint
(0,7,-7)is
[OnlineMay26,2012]
Options:
A.x +y+z=0
B.x +2y +z=21
C.3x −2y +5z +35 =0
D.3x +2y +5z +21 =0
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question278
Considerthefollowingplanes
P:x+y−2z +7=0
Q:x+y+2z +2=0
R:3x +3y −6z −11 =0
Giventhat,equationofaplaneis
x−2y +2z −5=0
So,Equationofparallelplaneis
x−2y +2z +d=0
Now,itisgiventhatdistancefromorigintotheparallelplaneis1.
∴d
12+22+22=1⇒d= ±3
Soequationofrequiredplanex−2y +2z ±3=0
|√|
Theequationoftheplanecontainingtheline
x+1
−3=y−3
2=z+2
1isa(x+1) + b(y−3) + c(z+2) = 0
where−3a +2b +c=0.......(A)
Thispassesthrough(0,7,-7)
∴a(0+1) + b(7−3) + c(−7+2) = 0
⇒a+4b −5c =0.......(B)
Onsolvingequation(A)and(B)weget
a=1,b=1,c=1
∴Requiredplaneis
x+1+y−3+z+2=0
⇒x+y+z=0
[OnlineMay26,2012]
Options:
A.PandRareperpendicular
B.QandRareperpendicular
C.PandQareparallel
D.PandRareparallel
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question279
Ifthethreeplanesx =5,2x −5ay +3z −2=0and3bx +y−3z =0
containacommonline,then(a,b)isequalto
[OnlineMay19,2012]
Options:
A. 8
15, −1
5
B. 1
5, − 8
15
C. −8
15,1
5
D. −1
5,8
15
Answer:B
Solution:
Solution:
( )
( )
( )
( )
Givenplanesare
P:x+y−2z +7=0
Q:x+y+2z +2=0
andR:3x +3y −6z −11 =0
ConsiderPlanePandR.
Herea1=1,b1=1,c1= −2
anda2=3,b2=3,c2= −6
Since,a1
a2
=b1
b2
=c1
c2
=1
3
thereforePandRareparallel.
Letthedirectionratiosofthecommonlinebel,mandn.
∴l×1+m×0+n×0=0⇒l=0
2l −5ma +3n =0⇒5ma −3n =0
3l b +m−3n =0⇒m−3n =0........(3)
Subtracting(3)from(1),wegetm(5a −1) = 0
-------------------------------------------------------------------------------------------------
Question280
Alinewithpositivedirectioncosinespassesthroughthepoint
P(2, −1,2)andmakesequalangleswiththecoordinateaxes.Iftheline
meetstheplane2x +y+z=9atpointQ,thenthelengthPQequals
[OnlineMay7,2012]
Options:
A.√2
B.2
C.√3
D.1
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question281
Thevaluesofaforwhichthetwopoints(1,a,1)and(−3,0,a)lieonthe
oppositesidesoftheplane3x +4y−12z +13 =0,satisfy
[OnlineMay7,2012]
Options:
A.0 <a<1
3
B.−1<a<0
Now,valueofmcannotbezerobecauseifm=0thenn=0
⇒l=m=n=0whichisnotpossible.
Hence,5a −1=0⇒a=1
5
Thus,option(b)iscorrect.
PointPis(2,-1,2)
LetthislinemeetatQ(h,k,w)
Directionratioofthislineis
(h−2,k+1,w−2)
Since,d csareequal&d rsarealsoequal,
So,h−2=k+1+w−2
⇒k=h−3andw=h
Thislinemeetstheplane
2x +y+z=9atQ,so,
2h +k+w=9or2h +h−3+h=9
⇒4h −3=9⇒h=3
andk=0andw=3
DistancePQ = (3−2)2+ (0− (−1))2+ (3−2)2
=12+12+12= √3
√
√
C.a < −1ora <1
3
D.a =0
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question282
Thelengthoftheperpendiculardrawnfromthepoint(3,-1,11)tothe
linex
2=y−2
3=z−3
4is:
[2011RS]
Options:
A.√29
B.√33
C.√53
D.√66
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question283
Statement-1:ThepointA(1,0,7) )isthemirrorimageofthepoint
Givenequationofplaneis
3x +4y −12z +13 =0
(1,a,1)and(−3,0,a)satisfytheequationofplane.
∴Wehave
3+4(a) − 12 +13 =0and3(−3) − 12(a) + 13 =0
⇒4+4a =0and4−12a =0
⇒a= −1anda=1
3
Since,(1,a,1)and(−3,0,a)lieontheoppositesidesoftheplane∴a=0
Anypointonlinex
2=y−2
3=z−3
4=αis(2α,3α +2,4α +3)
⇒Directionratioofthe⊥lineis
2α −3,3α +3,4α −8.
andDirectionratioofthegivenlineare2,3,4
⇒2(2α −3) + 3(3α +3) + 4(4α −8) = 0
⇒29α −29 =0
⇒α=1
⇒Footof⊥is(2,5,7)
⇒Length⊥is 12+62+42= √53
√
B(1,6,3)intheline:x
1=y−1
2=z−2
3
Statement-2:Thelinex
1=y−1
2=z−2
3bisectsthelinesegmentjoining
A(1,0,7)andB(1,6,3).
[2011]
Options:
A.Statement-1istrue,Statement-2istrue;Statement-2isnotacorrectexplanationfor
Statement-1.
B.Statement-1istrue,Statement-2isfalse.
C.Statement-1isfalse,Statement-2istrue.
D.Statement-1istrue,Statement-2istrue;Statement-2isacorrectexplanationforStatement-
1.
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question284
Thedistanceofthepoint(1,-5,9)fromtheplanex −y+z =5measured
alongastraightx =y=zis
[2011RS]
Options:
A.10√3
B.5√3
C.3√10
D.3√5
Answer:A
Solution:
Solution:
ThedirectionratioofthelinesegmentABis0,6,-4andthedirectionratioofthegivenlineis1,2,3.
Clearly1 times 0 +2×6+3× (−4) = 0
So,thegivenlineisperpendiculartolineAB.
Also,themidpointofAandBis(1,3,5)whichsatisfythegivenline.
So,theimageofBinthegivenlineisAstatement-1and2bothtruebut2isnotcorrectexplanation.of1.
EquationoflinethroughP(1, −5,9)andparalleltothelinex=y=zis
x−1
1=y+5
1=z−9
1=λ(say)
Q= (x=1+λ,y= −5+λ,z=9+λ)
SinceQliesonplanex−y+z=5
∴1+λ+5−λ+9+λ=5
⇒λ= −10
-------------------------------------------------------------------------------------------------
Question285
Iftheanglebetweenthelinex =y−1
2=z−3
λandtheplanex +2y +3z =4
iscos−15
14 ,thenλequals
[2011]
Options:
A.3
2
B.2
5
C.5
3
D.2
3
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question286
AlineABinthree-dimensionalspacemakesangles45°and120°with
thepositivex-axisandthepositivey-axisrespectively.IfABmakesan
acuteangleθwiththepositivez-axis,thenθequals
[2010]
Options:
A.45°
(√)
∴Q= (−9, −15, −1)
∴PQ = (1+9)2+ (15 −5)2+ (9+1)2
= √300 =10√3
√
Letθbetheanglebetweenthegivenlineandplane,then
sin θ =1×1+2×2+λ×3
12+22+λ2.12+22+32 = 5+3λ
√14 .5+λ2
⇒cos θ =1−(5+3λ)2
14(5+λ2)
⇒5
14 =1−(5+3λ)2
14(5+λ2)
Squaringbothsides,weget
5
14 =5λ2−30λ +45
14(5+λ2)
⇒λ=2
3
√ √ √
√
√√
B.60°
C.75°
D.30°
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question287
ThelineLgivenbyx
5+y
b=1passesthroughthepoint(13,32).Theline
K isparalleltoLandhastheequationx
c+y
3=1.Thenthedistance
betweenLandK is
[2010]
Options:
A.√17
B. 17
√15
C. 23
√17
D. 23
√15
Answer:C
Solution:
Solution:
Asperquestion,directioncosinesoftheline:l=cos 45° = 1
√2,m=cos 120° = −1
2,n=cos θ
wherethetaistheangle,whichlinemakeswithpositivez-axis.
Weknowthat,l2+m2+n2=1
⇒1
2+1
4+cos2θ=1
cos2θ=1
4
⇒cos θ =1
2=cos π
2( θbeingacute)
⇒θ=π
3
SlopeoflineL= −b
5
SlopeoflineK= −3
c
LineLisparalleltolinek.
⇒b
5=3
c⇒bc =15
(13,32)isapointonL.
∴13
5+32
b=1⇒32
b= −8
5
-------------------------------------------------------------------------------------------------
Question288
Statement-1:ThepointA(3,1,6)isthemirrorimageofthepointB(1,
3,4)intheplanex–y+z=5.
Statement-2:Theplanex–y+z=5bisectsthelinesegmentjoining
A(3,1,6)andB(1,3,4).
[2010]
Options:
A.Statement-1istrue,Statement-2istrue;Statement-2isnotacorrectexplanationfor
Statement-1.
B.Statement-1istrue,Statement-2isfalse.
C.Statement-1isfalse,Statement-2istrue.
D.Statement-1istrue,Statement2istrue;Statement-2isacorrectexplanationfor
Statement-1.
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question289
Theprojectionsofavectoronthethreecoordinateaxisare6,-3,2
respectively.Thedirectioncosinesofthevectorare
[2009]
Options:
A.6
5,−3
5,2
5
B.6
7,−3
7,2
7
⇒b= −20 ⇒c= − 3
4
EquationofK:
y−4x =3⇒4x −y+3=0
DistancebetweenLandK=|52 −32 +3|
√17 =23
√17
A(3,1,6); B= (1,3,4)
Puttingcoordinateofmid-pointofAB = (2,2,5)inplanex−y+z=5then2−2+5=5,satisfy
So,mid-pointofAB = (2,2,5)liesontheplane.
d.r'sofAB = (2, −2,2)
d.r'sofnormaltoplane = (1, −1,1)
DirectionratioofABandnormaltotheplaneareproportionaltherefore,
ABisperpendiculartothenormalofplane
∴AisimageofB
Statement-1iscorrect.
Statement-2isalsocorrectbutitisnotcorrectexplanation.
C.−6
7,−3
7,2
7
D.6, −3,2
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question290
Letthelinex−2
3=y−1
−5=z+2
2lieintheplanex +3y −αz +β=0.Then
(α,β)equals
[2009]
Options:
A.(-6,7)
B.(5,-15)
C.(-5,5)
D.(6,-17)
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question291
Ifthestraightlinesx−1
k=y−2
2=z−3
3andx−2
3=y−3
k=z−1
2intersectata
point,thentheintegerkisequalto
LetP(x1,y1,z1)andQ(x2,y2,z2)betheinitialandfinalpointsofthevectorwhoseprojectionsonthethreecoordinate
axesare6,-3,2then
x2−x1, = 6;y2−y1= −3;z2−z1=2
Sothatdirectionratiosof →
PQare6,-3,2
∴Directioncosinesof →
PQare 6
62+ (−3)2+22,−3
62+ (−3)2+222
62+ (−3)2+22=6
7,−3
7,2
7
√ √ √
Giventhat,thelinex−2
3=y−1
−5=z+2
2lieintheplanex+3y −αz +β=0
∴Pt(2,1, −2)liesontheplane
i.e.2+3+2α +β=0
or2α +β+5=0.......(i)
Alsonormaltoplanewillbeperpendiculartoline,
∴3×1−5×3+2× (−α) = 0
⇒α= −6
Fromequation(i)then,β=7
∴(α,β) = (−6,7)
[2008]
Options:
A.-5
B.5
C.2
D.-2
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question292
Thelinepassingthroughthepoints(5,1,a)and(3,b,1)crossestheyz-
planeatthepoint 0,17
2,−13
2.Then
[2008]
Options:
A.a =2,b=8
B.a =4,b=6
C.a =6,b=4
D.a =8,b=2
Answer:C
Solution:
( )
henthetwolinesintersectthenshortestdistancebetweenthemiszeroi.e.
→
a2−→
a1.→
b1×→
b2
→
b1×→
b2
=0
⇒→
a2−→
a1.→
b1×→
b2=0
where→
a1=^
i+2^
j+3^
k,→
b1=k^
i+2^
j+3^
k
→
a2=2^
i+3^
j+^
k,→
b2=3^
i+k^
j+2^
k
⇒
11−2
k 2 3
3 k 2
=0
⇒1(4−3k) − 1(2k −9)−2(k2−6) = 0
⇒−2k2−5k +25 =0⇒k= −5or5
2
∵kisaninteger,thereforek= −5
( )
| |
( )
| |
-------------------------------------------------------------------------------------------------
Question293
Ifalinemakesanangleofπ ∕4withthepositivedirectionsofeachofx
-axisandy-axis,thentheanglethatthelinemakeswiththepositive
directionofthez-axisis
[2007]
Options:
A.π
4
B.π
2
C.π
6
D.π
3
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question294
Equationoflinethrough(5,1,a)and(3,b,1)isx−5
−2=y−1
b−1=z−a
1−a=λ
x= −2λ +5
y= (b−1)λ+1
z= (1−a)λ+a
∴Anypointonthislineisa
[−2λ +5, (b−1)λ+1, (1−a)λ+a]
Giventhatitcrossesyzplane∴−2λ +5=0
λ=5
2
∴0, (b−1)5
2+1, (1−a)5
2+a = 0,17
2,−13
2
⇒(b−1)5
2+1=17
2
and(1−a)5
2+a= −13
2
⇒b=4anda=6
( ) ( )
Letthelinemakesananglethetawiththepositivedirectionofz-axis.Giventhatlinesmakesangleπ
4withxaxisandy-
axis.
∴l=cos π
4,m=cos π
4,n=cos θ
Weknowthat,l2+m2+n2=1
∴cos2π
4+cos2π
4+cos2θ=1
⇒1
2+1
2+cos2θ=1
⇒cos2θ=0⇒θ=π
2
Hence,anglewithpositivedirectionofthez-axisisπ
2.
LetLbethelineofintersectionoftheplanes2x +3y +z=1and
x+3y +2z =2.IfLmakesanangleαwiththepositivex-axis,thencos α
equals
[2007]
Options:
A.1
B. 1
√2
C. 1
√3
D.1
2.
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question295
TOPIC4-SphereandMiscellaneousProblemsonSphere
If(2,3,5)isoneendofadiameterofthespherex2+y2+z2
−6x −12y −2z +20 =0,thenthecooordinatesoftheotherendofthe
diameterare
[2007]
Options:
A.(4,3,5)
B.(4,3,-3)
C.(4,9,-3)
D.(4,-3,3)
LetthedirectioncosinesoflineLbel,m,n.Sinceline
Lliesonbothplanes.
∴2l +3m +n=0.......(i)
andl+3m +2n =0........(ii)
onsolvingequation(i)and(ii),weget
l
6−3=m
1−4=n
6−3⇒l
3=m
−3=n
3
Nowl
3=m
−3=n
3 = l2+m2+n2
32+ (−3)2+32
∵l2+m2+n2=1
∴l
3=m
−3=n
3=1
√27
⇒l=3
√27 =1
√3,m= − 1
√3,n=1
√3
LineL,makesanangleαwith+vex-axis
∴l=cos α ⇒cos α =1
√3
√
√
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question296
Theimageofthepoint(-1,3,4)intheplanex −2y =0is
[2006]
Options:
A. −17
3, −19
3,4
B.(15,11,4)
C. −17
3, −19
3,1
D.Noneofthese
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question297
Ifnonzeronumbersa,b,careinH.P.,thenthestraightlinex
a+y
b+1
c=0
alwayspassesthroughafixedpoint.Thatpointis
( )
( )
Weknowthatcentreofsphere
x2+y2+z2+2ux +2vy+2wz +d=0
is(−u, −v, −w)
Giventhat,x2+y2+z2−6x −12y −2z+20 =0
∴Centre ≡ (3,6,1)
Coordinatesofoneendofdiameterofthesphereare(2,3,5)
Letthecoordinatesoftheotherendofdiameterare(α,β,γ)
∴α+2
2=3,β+3
2=6,γ+5
2=1
⇒α=4,β=9andγ= −3
∴Coordinateofotherendofdiameterare(4,9,-3)
Let(α,β,γ)betheimage,thenmidpointof( α,β,γ)and(-1,3,4)mustlieonx−2y =0
∴α−1
2−2β+3
2=0
∴α−1−2β −6=0⇒α−2β =7.......(1)
Alsolinejoining(α,β,γ)and(-1,3,4)shouldbeparalleltothenormaloftheplanex−2y =0
∴α+1
1=β−3
−2=γ−4
0=λ
⇒α=λ−1,β= −2λ +3,γ=4........(2)
From(1)and(2)
α=9
5,β= −13
5,γ=4
Noneoftheoptionmatches.
( )
[2005]
Options:
A.(-1,2)
B.(-1,-2)
C.(1,-2)
D. 1, −1
2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question298
Theanglebetweenthelines2x =3y = −zand6x = −y= −4zis
[2005]
Options:
A.0°
B.90°
C.45°
D.30°
Answer:B
Solution:
Solution:
( )
a,b,careinH.P.⇒1
a,1
b,1
careinA.P.
⇒2
b=1
a+1
c⇒1
a−2
b+1
c=0
∴x
a+y
a+1
c=0passesthrough(1,-2)
Thegivenlinesare2x =3y = −z
orx
3=y
2=z
−6Dividingby6]
and6x = −y= −4z
orx
2=y
−12 =z
−3[Dividingby12]
∴Anglebetweentwolinesis
cos θ =a1a2+b1b2+c1c2
a1
2+b1
2+c1
2a2
2+b2
2+c2
2
cos θ =3.2 +2. (−12) + (−6) . (−3)
32+22+ (−6)222+ (−12)2+ (−3)2
=6−24 +18√49√157 = 0⇒θ=90°
[
√ √
√ √
Question299
Thedistancebetweentheline→
r=2^
i−2^
j+3^
k+λ(i−j+4k)andthe
plane→
r.^
i+5^
j+^
k=5is
[2005]
Options:
A.10
9
B. 10
3√3
C. 3
10
D.10
3
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question300
Iftheanglethetabetweenthelinex+1
1=y−1
2=z−2
2andtheplane
2x −y+ √λz+4=0issuchthatsin θ =1
3thenthevalueofλis
[2005]
Options:
A.5
3
B.−3
5
C.3
4
D.−4
3
Answer:A
( )
Thegivenlineis→
r=2^
i−2^
j+3^
k+λ(i−j+4k)
andtheplaneis→
r.^
i+5^
j+^
k=5
⇒x+5y +z=5
Requireddistance= ax1+by1+cz1+d
a2+b2+c2
=2−10 +3−5
√1+25 +1=10
3√3
( )
|√|
| |
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question301
Theplanex +2y −z=4cutsthespherex2+y2+z2−x+z −2=0ina
circleofradius
[2005]
Options:
A.3
B.1
C.2
D.√2
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question302
Letθistheanglebetweenlineandplanethen
sin θ =
→
b.→
n
→
b→
n
=
^
i+2^
j+2^
k.2^
i−^
j+ √λ^
k
√1+4+4√4+1+λ = 2−2+2√λ
3× √5+λ
⇒sin θ =2√λ
3√5+λ=1
3⇒4λ =5+λ
⇒λ=5
3
| | | |
( ) ( )
Centreofsphere = 1
2,0, −1
2andradiusofsphere
=1
4+1
4+2=5
2
PerpendiculardistanceOAofcentrefromx+2y −z=4isgivenby
1
2+1
2−4
√6=3
2
∴radiusofcircleAB =OB2−OA2 = 5
2−3
2=1
( )
√ √
| | √
√√
Iftheplane2ax −3ay +4az +6=0passesthroughthemidpointofthe
linejoiningthecentresofthespheresx2+y2+z2+6x−8y −2z =13and
x2+y2+z2−10x+4y −2z =8thenaequals
[2005]
Options:
A.-1
B.1
C.-2
D.2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question303
Alinemakesthesameangleθ,witheachofthexandzaxis.Ifthe
angleβ,whichitmakeswithy-axis,issuchthatsin2β=3sin2θ,then
cos2θequals
[2004]
Options:
A.2
5
B.1
5
C.3
5
D.2
3
Answer:C
Solution:
Solution:
Plane2ax −3ay +4az +6=0passesthroughthemidpointofthelinejoiningthecentresofspheres
x2+y2+z2+6x −8y−2z =13and
x2+y2+z2−10x +4y−2z =8
respectivelycentreofspheresarec1(−3,4,1)andc2(5, −2,1).Midpointofc1c2is(1,1,1)
Satisfyingthisintheequationofplane,weget
2a −3a +4a +6=0
⇒a= −2.
Asperquestionthedirectioncosinesofthelinearecos θ,cos β,cos θ
∴cos2θ+cos2β+cos2θ=1
∵2cos2θ=1−cos2θ
-------------------------------------------------------------------------------------------------
Question304
Ifthestraightlinesx =1+s,y= −3−λs,z=1+λsand
x=t
2,y=1+t,z=2−t,withparameterssandtrespectively,areco-
planar,thenλequals.
[2004]
Options:
A.0
B.-1
C.−1
2
D.-2
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question305
Alinewithdirectioncosinesproportionalto2,1,2meetseachofthe
linesx =y+a=zandx +a=2y =2z.Theco-ordinatesofeachofthe
⇒2cos2θ=sin2β=3sin2θ(given)
⇒2cos2θ=3−3cos2θ
∴cos2θ=3
5
Thegivenlinesare
x−1=y+3
−λ=z−1
λ=s......(1)
and2x =y−1=z−2
−1=t.....(2)
Thelinesarecoplanar,if
0−1 1 − (−3)2−1
1−λλ
1
21−1
= 0
⇒
−14 1
1−λλ
1
21−1
=0
Applyc2→c2+c3;
−15 1
10λ
1
20−1
=0
⇒−5−1−λ
2=0⇒λ= −2
| |
| |
| |
( )
pointsofintersectionaregivenby
[2004]
Options:
A.(2a,3a,3a), (2a,a,a)
B.(3a,2a,3a), (a,a,a)
C.(3a,2a,3a), (a,a,2a)
D.(3a,3a,3a), (a,a,a)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question306
Distancebetweentwoparallelplanes2x +y+2z =8and
4x +2y +4z +5=0is
[2004]
Options:
A.9
2
B.5
2
C.7
2
D.3
2
Answer:C
Solution:
Solution:
Letapointonthelinex=y+a=z=λis
(λ,λ−a,λ)andapointontheline
x+a=2y =2z =µis µ−a,µ
2,µ
2,thenDirectionratioofthelinejoiningthesepointsare
λ−µ+a,λ−a−µ
2,λ−µ
2
Ifitrespresentstherequiredlinewhosed.rbe2,1,2,then
λ−µ+a
2=
λ−a−µ
2
1 =
λ−µ
2
2
onsolvingwegetλ=3a,µ=2a
∴Therequiredpointsofintersectionare
(3a,3a −a,3a)and 2a −a,2a
2,2a
2
or(3a,2a,3a)and(a,a,a)
( )
( )
Theplanesare2x +y+2z −8=0........(1)
-------------------------------------------------------------------------------------------------
Question307
Theintersectionofthespheresx2+y2+z2+7x −2y −z=13and
x2+y2+z2−3x +3y +4z =8isthesameastheintersectionofoneof
thesphereandtheplane
[2004]
Options:
A.2x −y−z=1
B.x −2y −z=1
C.x −y−2z =1
D.x −y−z=1
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question308
Thelinesx−2
1=y−3
1=z−4
−kandx−1
k=y−4
2=z−5
1arecoplanarif
[2003]
Options:
A.k =3or-2
B.k =0or-1
C.k =1or-1
D.k =0or-3.
Answer:D
and4x +2y +4z +5=0
or2x +y+2z +5
2=0........(2)
Since,bothplanesareparallel
∴Distancebetween(1)and(2)
=
5
2+8
22+12+22= 21
2√9=7
2
|√|| |
Giventhat,theequationsofspheresare
S1:x2+y2+z2+7x −2y −z−13 =0and
S2:x2+y2+z2−3x +3y+4z −8=0
Weknowthateqn.ofintersectionplanebe
S1−S2=0⇒10x −5y −5z −5=0
⇒2x −y−z=1
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question309
Thetwolinesx=ay+b,z=cy+dandx=a'y+b',z=c'y+d'willbe
perpendicular,ifandonlyif
[2003]
Options:
A.aa'+cc'+1=0
B.aa'+bb'+cc'+1=0
C.aa'+bb'+cc'=0
D.(a+a')(b+b')+(c+c')=0.
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question310
Twosystemofrectangularaxeshavethesameorigin.Ifaplanecuts
thematdistancesa,b,canda′, b′, c′fromtheoriginthen
[2003]
Twoplanesarecoplanarif
x2−x1y2−y1z2−z1
l1m1n1
l2m2n2
= 0
1−1−1
1 1 −k
k 2 1
=0
ApplyingC2→C2+C1,C3→C3+C1
⇒
1 0 0
121−k
k k +2 1 +k
=0
⇒1[2+2k − (k+2)(1−k)] = 0
⇒2+2k − (−k2−k+2) = 0
k2+3k =0⇒k(k+3) = 0
ork=0or-3
| |
| |
| |
x−b
a=y
1=z−d
c;x−b′
a′ = y
1=z−d′
c′.
Forperpendicularityoflines,
aa′ + 1+cc′ = 0
Options:
A. 1
a2+1
b2+1
c2−1
a′2−1
b′2−1
c′2=0
B. 1
a2+1
b2+1
c2+1
a′2+1
b′2+1
c′2=0
C. 1
a2+1
b2−1
c2+1
a′2+1
b′2−1
c′2=0
D. 1
a2−1
b2−1
c2+1
a′2−1
b′2−1
c′2=0
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question311
Theradiusofthecircleinwhichthespherex2+y2+z2+2x −2y
−4z −19 =0iscutbytheplanex +2y +2z +7=0is
[2003]
Options:
A.4
B.1
C.2
D.3
Answer:D
Solution:
Solution:
Equationofplanesininterceptformbex
a+y
b+z
c=1&x
a′+y
b′+z
c′=1
( ⊥rdistanceonplanefromoriginissame.)
−1
1
a2+1
b2+1
c2
= −1
1
a′2+1
b′2+1
c′2
1
a2+1
b2+1
c2−1
a′2−1
b′2− 1
c′2=0
|√| | √|
Centreofsphere = (−1,1,2)
Radiusofsphere√1+1+4+19 =5
Perpendiculardistancefromcentretotheplane
-------------------------------------------------------------------------------------------------
Question312
Theshortestdistancefromtheplane12x +4y +3z =327tothesphere
x2+y2+z2+4x −2y−6z =155is
[2003]
Options:
A.39
B.26
C.11 4
13
D.13.
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question313
Thed.r.ofnormaltotheplanethrough(1,0,0),(0,1,0)whichmakesan
angleπ ∕4withplanex +y=3are
[2002]
Options:
A.1, √2,1
B.1,1, √2
C.1,1,2
D.√2,1,1
Answer:B
Solution:
OC =d=−1+2+4+7
√1+4+4=12
3=4
Inright,ΔAOC
AC2=AO2−OC2 = 52−42=9
⇒AC =3
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Centreofspherebe(-2,1,3)andradius13
Weknowthat,
Shortestdistance=perpendiculardistancebetweentheplaneandsphere=distanceofplanefromcentreofsphere-
radius
=−2×12 +4×1+3×3−327
√144 +9+16 −13
=26 −13 =13
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Solution:
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Question314
Aplanewhichpassesthroughthepoint(3,2,0)andtheline
x−4
1=y−7
5=z−4
4is
[2002]
Options:
A.x −y+z=1
B.x +y+z=5
C.x +2y −z=1
D.2x −y+z=5
Answer:A
Solution:
Solution:
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Equationofplanethrough(1,0,0)is
a(x−1) + by +cz =0.......(i)
Itisalsopassesthrough(0,1,0).
∴−a+b=0⇒b=a
cos 45° = a+a
2(2a2+c2)
⇒2a =2a2+c2⇒2a2=c2⇒c= √2a
Sod.rofnormalarea,a√2ai.e.1,1, √2.
√
√
Sincethepoint(3,2,0)liesonthegivenline
x−4
1=y−7
5=z−4
4
∴Therecanbeinfinitemanyplanespassingthroughthisline.Weobservedthatonlyoption(a)issatisfiedbythe
coordinatesofboththepoints(3,2,0)and(4,7,4)
∴x−y+z=1istherequiredplane.