VectorAlgebra
Question1
Let→
a=^
i+2^
j+k,→
b=3^
i−^
j+k .Let→
cbethevectorsuchthat→
a×→
c=→
b
and→
a⋅→
c=3.Then→
a⋅→
c×→
b−→
b−→
cisequalto:
[27-Jan-2024Shift1]
Options:
A.32
B.24
C.20
D.36
Answer:B
Solution:
Solution:
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Question2
Theleastpositiveintegralvalueofα,forwhichtheanglebetweenthe
vectorsα^
i−2^
j+2kandα^
i+2α^
j−2kisacute,is___
[27-Jan-2024Shift1]
( )
( ( ) )
Answer:5
Solution:
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Question3
Let→
a,→
band→
cbethreenon-zerovectorssuchthat→
band→
carenon-
collinear.if→
a+5→
biscollinearwith→
c,→
b+6→
ciscollinearwith→
aand
→
a+α→
b+β→
c=→
0,thenα +βisequalto
[29-Jan-2024Shift1]
Options:
A.35
B.30
C.-30
D.-25
Answer:A
Solution:
Solution:
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Question4
Let →
OA =→
a,→
OB =12→
a+4→
band →
OC =→
b,whereOistheorigin.IfSisthe
parallelogramwithadjacentsidesOAandOC,then areaofthequadrilateral OABC
areaofS
isequalto
[29-Jan-2024Shift2]
Options:
A.6
B.10
C.7
D.8
Answer:D
Solution:
Solution:
Question5
Let →
OA =→
a,→
OB =12→
a+4→
band →
OC =→
b,whereOistheorigin.IfSisthe
parallelogramwithadjacentsidesOAandOC,then areaofthequadrilateral OABC
areaofS
isequalto
[29-Jan-2024Shift2]
Options:
A.6
B.10
C.7
D.8
Answer:D
Solution:
Solution:
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Question6
Let→
a=ai
^
i+a2
^
j+a3
^
kand→
b=b1
^
i+b2
^
j+b3
^
kbetwovectorssuchthat
→
a=1;→
a⋅→
b=2and →
b=4.If→
c=2→
a×→
b−3→
b,thentheanglebetween
→
band→
cisequalto:
[30-Jan-2024Shift1]
Options:
A.cos−12
√3
B.cos−1−1
√3
C.cos−1−√3
2
D.cos−12
3
Answer:C
Solution:
| | | | ( )
( )
( )
( )
( )
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Question7
Letaunitvector ^
u=x˙^
i+y^
j+z^
kmakeangles π
2,π
3and 2π
3withthe
vectors 1
√2
^
i+1
√2
^
k,1
√2
^
j+1
√2
^
kand 1
√2
^
i+1
√2
^
jrespectively.If
→
v=1
√2
^
i+1
√2
^
j+1
√2
^
k,then ^
u−→
v2isequalto
[29-Jan-2024Shift2]
Options:
A. 11
2
B. 5
2
C.9
D.7
| |
Answer:B
Solution:
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Question8
LetA(2,3,5)andC(−3,4, −2)beoppositeverticesofaparallelogram
ABCDifthediagonal →
BD =^
i+2^
j+3^
kthentheareaoftheparallelogram
isequalto
[30-Jan-2024Shift1]
Options:
A. 1
2√410
B. 1
2√474
C. 1
2√586
D. 1
2√306
Answer:B
Solution:
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Question9
Let→
a=^
i+α^
j+β^
k,α,β∈R.Letavector→
bbesuchthattheangle
between→
aand→
bis π
4and →
b
2=6,If→
a⋅→
b=3√2,thenthevalueof
(α2+β2)→
a×→
b|2isequalto
[30-Jan-2024Shift2]
Options:
A.90
B.75
C.95
D.85
Answer:A
Solution:
Solution:
| |
|
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Question10
Let→
aand→
bbetwovectorssuchthat →
b=1and →
b×→
a=2.Then
→
b×→
a−→
b
2isequalto
[30-Jan-2024Shift2]
Options:
A.3
B.5
C.1
D.4
Answer:B
Solution:
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Question11
Let →
a=3^
i+^
j−2^
k,→
b=4^
i+^
j+7^
kand→
c=^
i−3^
j+4^
kbethreevectors.
Ifavectors→
psatisfies→
p×→
b=→
c×→
band→
p⋅→
a=0,then→
p⋅^
i−^
j−^
kis
| | | |
| ( ) |
( )
equalto
[31-Jan-2024Shift1]
Options:
A.24
B.36
C.28
D.32
Answer:D
Solution:
Solution:
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Question12
Let→
aand→
bbetwovectorssuchthat →
a=1,→
b=4and→
a⋅→
b=2.If
→
c=2→
a×→
b−3→
bandtheanglebetween→
band→
cisα,then192sin 2αis
equalto____
[31-Jan-2024Shift1]
Answer:48
Solution:
| | | |
( )
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Question13
Let→
a=3^
i+2^
j+^
k,→
b=2^
i−^
j+3^
kand→
cbeavectorsuchthat
→
a+→
b×→
c=2→
a×→
b+24^
j−6^
kand →
a−→
b+^
i⋅→
c= −3.Then →
c2is
equalto____
[31-Jan-2024Shift2]
Answer:38
Solution:
( ) ( ) ( ) | |
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Question14
Let→
a= −5^
i+^
j−3^
k,→
b=^
i+2^
j−4^
kand→
c=→
a×→
b×^
i×^
i×^
i.Then
→
c⋅ −^
i+^
j+^
kisequalto
[1-Feb-2024Shift1]
Options:
A.-12
B.-10
C.-13
D.-15
Answer:A
Solution:
Solution:
( ( ( ) ) )
( )
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Question15
Considera△ABCwhereA(1,2,3),, B(−2,8,0)andC(3,6,7).Ifthe
anglebisectorof∠BACmeetsthelineBCatD,thenthelengthofthe
projectionofthevector →
ADonthevector →
ACis:
[1-Feb-2024Shift2]
Options:
A. 37
2√38
B. √38
2
C. 39
2√38
D.√19
Answer:A
Solution:
Solution:
A(1,3,2); B(−2,8,0); C(3,6,7);
→
AC =2^
i+3^
j+5^
k
AB = √9+25 +4= √38
AC = √4+9−25 − √38
→
AD =1
2
^
i−4^
j−3
2
^
k = 1
2
^
i−8^
j−3^
k
( )
-------------------------------------------------------------------------------------------------
Question16
Let →
a=^
i+^
j+^
k,→
b= −^
i−8^
j+2^
kand→
c=4^
i+c2
^
j+c3
^
kbethree
vectorssuchthat→
b×→
a=→
c×→
a.Iftheanglebetweenthevector→
candthe
vector3^
i+4^
j+^
kisθ,thenthegreatestintegerlessthanorequalto
tan2θis:
[1-Feb-2024Shift2]
Answer:38
Solution:
Solution:
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Question17
Let→
a,→
band→
cbethreenon-zeronon-coplanarvectors.Lettheposition
vectorsoffourpointsA,B,CandDbe→
a−→
b+→
c,λ→
a−3→
b+4→
c,
−→
a+2→
b−3→
cand2→
a−4→
b+6→
crespectively.If →
AB, →
ACand →
ADarecoplanar,
thenλis:
OfficialAns.byNTA
[29-Jan-2023Shift1]
Lengthofprojectionof →
ADon →
AC
=
→
AD ⋅→
AC
→
AC
=37
2√38
|| | |
→
a=^
i+^
j+^
k
→
b= −^
i−8^
j+2^
k
→
c=4^
i+c2
^
j+c3
^
k
→
b×→
a=→
c×→
a
→
b−→
c×→
a=0
→
b−→
c=λ→
α
→
b=→
c+λ→
α
−^
i−8^
j+2^
k=4^
i+c2
^
j+c3
^
k+λ^
i+^
j+^
k
λ+4= −1⇒λ= −5
λ+c2= −8⇒c2= −3
λ+c3=2⇒c3=7
→
c=4^
i−3^
j+7k
cos θ =12 −12 +7
√26 ⋅ √74 = 7
√26 ⋅ √74 =7
2√481
tan2θ=625 ×3
49
[tan2θ] = 38
( )
( ) ( )
Answer:2
Solution:
Solution:
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Question18
Let →
u=∧
i−∧
j−2k,→
v=2∧
i+∧
j−k,→
v⋅→
w=2 and→
v×→
w=→
u+λ→
v.Then→
u⋅→
wis
equalto
[24-Jan-2023Shift1]
Options:
A.1
B. 3
2
C.2
D.−2
3
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
AB = (λ−1)a−2b +3c
AC =2a +3b −4c
AD =a−3b +5c
λ−1−23
−23−4
1−35
=0
⇒ (λ−1)(15 −12) + 2(−10 +4) + 3(6−3) = 0
⇒ (λ−1) = 1⇒λ=2
| |
→
u= (1, −1, −2), →
v= (2,1, −1), →
v⋅→
w=2
→
v×→
w=→
u+λ→
v............... ⋅ (1)
Takingdotwith →
w in (1)
→
W⋅→
v×→
w=→
u⋅→
w+λ→
v⋅→
w
⇒0=→
u⋅→
w+2λ
Takingdotwith →
v in (1)
→
v⋅→
v×→
w=→
u⋅→
v+λ→
v⋅→
v
⇒0= (2−1+2) + λ(6)
λ= − 1
2
⇒→
u⋅→
w= −2λ =1
( )
( )
Question19
Let→
α=4^
i+3^
j+5^
kand→
β=^
i+2^
j−4^
k.Let→
β1beparallelto→
αand→
β2be
perpendicularto→
α.If→
β=→
β1+→
β2,thenthevalueof5→
β2⋅^
i+^
j+^
kis
[24-Jan-2023Shift2]
Options:
A.6
B.11
C.7
D.9
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question20
Let→
a=^
i+2^
j+λ^
k,→
b=3^
i−5^
j−λ^
k,→
a⋅→
c=7,
2→
b⋅→
c+43 =0,→
a×→
c=→
b×→
c.Then →
a⋅→
bisequalto
[24-Jan-2023Shift2]
Answer:8
Solution:
( )
| |
Let→
β1=λ→
α
Now→
β2=→
β−→
β1
=
⋀
i+2
⋀
j−4
⋀
k−λ 4
⋀
i+3
⋀
j+5
⋀
k
= (1−4λ)
⋀
i+ (2−3λ)
⋀
j− (5λ +4)
⋀
k
→
β2⋅→
α=0
⇒4(1−4λ) + 3(2−3λ) − 5(5λ +4) = 0
⇒4−16α +6−9λ −25λ −20 =0
⇒50λ = −10
⇒λ=−1
5
→
β2=1+4
5
^
i+2+3
5
^
j− (−1+4)^
k
→
β2=9
5
⋀
i+13
5
⋀
j−3
⋀
k
5→
β2=9^
i+13 ^
j−15 ^
k
5→
β2⋅
⋀
i+
⋀
j+
⋀
k=9+13 −15 =7
( ) ( )
( ) ( )
( )
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Question21
Thevector→
a= −^
i+2^
j+^
kisrotatedthrougharightangle,passing
throughthey-axisinitswayandtheresultingvectoris→
b.Thenthe
projectionof3→
a+ √2→
bon→
c=5^
i+4^
j+3^
kis
[25-Jan-2023Shift1]
Options:
A.3√2
B.1
C.√6
D.2√3
Answer:A
Solution:
Solution:
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Question22
Let→
a,→
band→
cbethreenonzerovectorssuchthat→
b⋅→
c=0and
→
a=^
i+2^
j+λ^
k,→
b=3^
i−5^
j−λ^
k,→
a⋅→
c=7
→
a×→
c−→
b×→
c=→
0
→
a−→
b×→
c=0⇒→
a−→
bisparalleledto→
c
→
a−→
b=µ→
c,whereµisascalar
−2^
i+7^
j+2λ^
k=µ⋅→
c
Now→
a⋅→
c=7gives2λ2+12 =7µ
And→
b⋅→
c= − 43
2gives4λ2+82 =43µ
µ=2andλ2=1
→
a⋅→
b=8
( ) ( )
| |
→
b=λ→
a×→
a×^
j
⇒→
b=λ−2^
i−2^
j+2^
k
→
b=→
a∴ √6= √12 λ ⇒λ= ± 1
√2
λ=1
√2rejected ∵→
b makesacuteanglewithyaxis
→
b= −√2−^
i−^
j+^
k
3→
a+ √2→
b⋅
→
c
→
c=3√2
( )
( )
|||| | |
()
( )
()| |
→
a×→
b×→
c=→
b−→
c
2.If→
dbeavectorsuchthat→
b⋅→
d=→
a⋅→
b,then
→
a×→
b⋅→
c×→
disequalto
[25-Jan-2023Shift1]
Options:
A. 3
4
B. 1
2
C.−1
4
D. 1
4
Answer:D
Solution:
Solution:
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Question23
Ifthefourpoints,whosepositionvectorsare
3^
i−4^
j+2^
k,^
i+2^
j−^
k, −2^
i−^
j+3^
kand5^
i−2α^
j+4^
karecoplanar,then
αisequalto
[25-Jan-2023Shift2]
Options:
A. 73
17
B.−107
17
C.−73
17
D. 107
17
Answer:A
()
( ) ( )
→
a⋅→
c→
b−→
a⋅→
b→
c=→
b−
→
c
2
→
a⋅→
c=1
2,→
a⋅→
b=1
2
∴→
b⋅→
d=1
2
→
a×→
b⋅→
c×→
d=→
a⋅→
b×→
c×→
d
=→
a⋅→
b⋅→
d→
c−→
b⋅→
c→
d
=→
a⋅→
c→
b⋅→
d=1
4
()()
( ) ( ) ( ( ) )
( ( ) ( ))
()()
Solution:
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Question24
Let→
a= −^
i−^
j+^
k,→
a⋅→
b=1and→
a×→
b=^
i−^
j.Then→
a−6→
bisequalto
[25-Jan-2023Shift2]
Options:
A.3 ∧
i−∧
j−∧
k
B.3 ∧
i+∧
j+∧
k
C.3 ∧
i−∧
j+∧
k
D.3 ∧
i+∧
j−∧
k
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question25
Ifthevectors→
a=λ^
i+µ^
j+4^
k,→
b= −2^
i+4^
j−2^
kand→
c=2^
i+3^
j+^
kare
coplanarandtheprojectionof→
aonthevector→
bis√54units,thenthe
sumofallpossiblevaluesofλ +µisequalto
[29-Jan-2023Shift1]
( )
( )
( )
( )
LetA: (3, −4,2)
C: (−2, −1,3)
B: (1,2, −1)D: (5, −2α,4)
A,B,C,Darecoplanarpoints,then
⇒
1−3 2 +4−1−2
−2−3−1+4 3 −2
5−3−2α +4 4 −2
=0
⇒α=73
17
| |
→
a×→
b=^
i−^
j
Takingcrossproductwith→
a
⇒→
a×→
a×→
b=→
a×^
i−^
j
⇒→
a⋅→
b→
a−→
a⋅→
a→
b=^
i+^
j+2^
k
⇒→
a−3→
b=^
i+^
j+2^
k
⇒2→
a−6→
b=2^
i+2^
j+4^
k
⇒→
a−6→
b=3^
i+3^
j+3^
k
( )
( ) ( )
( ) ( )
Options:
A.0
B.6
C.24
D.18
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question26
Let→
a,→
band→
cbethreenon-zeronon-coplanarvectors.Lettheposition
vectorsoffourpointsA,B,CandDbe→
a−→
b+→
c,λ→
a−3→
b+4→
c,
−→
a+2→
b−3→
cand2→
a−4→
b+6→
crespectively.If →
AB, →
ACand →
ADarecoplanar,
thenλis:
OfficialAns.byNTA
[29-Jan-2023Shift1]
Answer:2
Solution:
λ µ 4
−24−2
2 3 1
=0
λ(10) − µ(2) + 4(−14) = 0
10λ −2µ =56
5λ −µ=28...(1)
→
a⋅→
b
→
b
= √54
−2λ +4µ −8∕ √24 = √54
−2λ +4µ −8= √54 ×24...(2)
Bysolvingequation(1)&(2)
⇒λ+µ=24
| |
| |
AB = (λ−1)a−2b +3c
AC =2a +3b −4c
AD =a−3b +5c
λ−1−23
−23−4
1−35
=0
⇒ (λ−1)(15 −12) + 2(−10 +4) + 3(6−3) = 0
⇒ (λ−1) = 1⇒λ=2
| |
-------------------------------------------------------------------------------------------------
Question27
Let→
a=4^
i+3^
jand→
b=3^
i−4^
j+5^
kand→
cisavectorsuchthat
→
c⋅→
a×→
b+25 =0,→
c⋅^
i+^
j+^
k=4andprojectionof→
con→
ais1,then
theprojectionof→
con→
bequals:
[29-Jan-2023Shift2]
Options:
A. 5
√2
B. 1
5
C. 1
√2
D. 3
√2
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question28
If→
a=^
i+2^
k,→
b=^
i+^
j+^
k,→
c=7^
i−3^
k+4^
k→
r×→
b+→
b×→
c=→
0and→
r⋅→
a=0
then→
r⋅→
cisequalto:
[29-Jan-2023Shift2]
Options:
A.34
B.12
C.36
D.30
( ) ( )
→
a×→
b=15 ^
i−20 ^
j−25 ^
k
Let →
c=x^
i+y^
j+z^
k
⇒15x −20y −25z +25 =0
⇒3x −4y −5z = −5
Alsox+y+z=4
and
→
c⋅→
a
→
a
=1⇒4x +3y =5
⇒→
c=2^
i−^
j+3^
k
Projectionof→
cor→
b=25
5√2=5
√2
| |
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question29
If→
a,→
b,→
carethreenon-zerovectorsand ^
nisaunitvectorperpendicular
to→
csuchthat→
a=α→
b−^
n, (α≠0)and→
b⋅→
c=12,then →
c×→
a×→
bis
equalto:
[30-Jan-2023Shift1]
Options:
A.15
B.9
C.12
D.6
Answer:C
Solution:
| ( ) |
→
r×→
b−→
c×→
b=0
⇒→
r−→
c×→
b=0
⇒→
r−→
c=λ→
b
⇒→
r=→
c+λ→
b
Andgiventhat→
r⋅→
a=0
⇒→
c+λ→
b⋅→
a=0
⇒→
c⋅→
a+λ→
b⋅→
a=0
⇒λ= −→
c⋅
→
a
→
b⋅→
a
Now →
r⋅→
c=→
c+λ→
b⋅→
c
=→
c−
→
c⋅→
a
→
b⋅→
a
→
b⋅→
c
=→
c−
→
c⋅→
a
→
b⋅→
a
→
b⋅→
c
=74 −15
38
=74 −40 =34
()
( )
( )
( )
|| ( ) ( )
[ ]
^
n⟂→
c→
a=α→
b−→
n
→
b⋅→
c=12
→
a⋅→
c=α→
b⋅→
c−→
n⋅→
c
→
a⋅→
c=α→
b⋅c
→
c×→
a×→
b=→
c⋅→
b→
a−→
c⋅→
a→
b
=→
c⋅→
b→
a−α→
b⋅→
c→
b
()
( )
| ( ) | | ( ) ( )|
| ( ) ( )|
-------------------------------------------------------------------------------------------------
Question30
Letλ ∈ ℝ, →
a=λ^
i+2^
j−3^
k,→
b=^
i−λ^
j+2^
k.
If →
a+→
b×→
a×→
b×→
a−→
b=8^
i−40^
j−24^
k,then
λ→
a+→
b×→
a−→
b
2isequalto
[30-Jan-2023Shift2]
Options:
A.140
B.132
C.144
D.136
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question31
( ( ) ( ) ) ( )
| ( ) ( ) |
=→
c⋅→
b→
a−α→
b
=12 ×→
n
=12 ×1
=12
| ( ) | | |
( | | )
→
a=λ^
i+2^
j−3^
k
→
b=^
i−λ^
j+2^
k
⇒→
b−→
a×→
a+→
b×→
a×→
b=8^
i−40 ^
j−24 ^
k
⇒→
a−→
b⋅→
a+→
b→
a×→
b=8^
i−40j −24 ^
k
⇒8→
a×→
b=8^
i−40 ^
j−24 ^
k
Now,→
a×→
b=
^
i^
j^
k
λ 2 −3
1−λ2
= (4−3λ)^
i− (2λ +3)^
j+ (−λ2−2)^
k
⇒λ=1
∴→
a=^
i+2^
j−3^
k
→
b=^
i−^
j+2^
k
⇒→
a+→
b=2^
i+^
j−^
k,→
a−→
b=3^
j−5^
k
⇒→
a+→
b×→
a−→
b=
^
i^
j^
k
2 1 −1
0 3 −5
=2^
i+10 ^
j+6^
k
∴requiredanswer =4+100 +36 =140
()( ( ) ( ) )
( ( ) ( ) ) ( )
( )
| |
( ) ( ) | |
Let→
aand→
bbetwovectors.Let →
a=1,→
b=4and→
a⋅→
b=2.If
→
c=2→
a×→
b−3→
b,thenthevalueof→
b⋅→
cis
[30-Jan-2023Shift2]
Options:
A.−24
B.−48
C.−84
D.−60
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question32
Let→
a=2^
i+^
j+^
k,and→
band→
cbetwononzerovectorssuchthat
→
a+→
b+→
c=→
a+→
b−→
cand→
b.→
c=0.Considerthefollowingtwo
statement:
(A) →
a+λ→
c≥→
aforallλ ∈ ℝ.
(B)→
aand→
carealwaysparallel
[31-Jan-2023Shift1]
Options:
A.only(B)iscorrect
B.neither(A)nor(B)iscorrect
C.only(A)iscorrect
D.both(A)and(B)arecorrect.
Answer:C
Solution:
Solution:
| | | |
( )
| | | |
| | | |
→
c=2→
a×→
b−3→
b
→
b⋅→
c=→
b⋅2→
a×→
b−3→
b⋅→
b
= −3|b|2
= −48
( )
( )
→
a+→
b+→
c|2=→
a+→
b−→
c|2
2→
a⋅→
b+2→
b⋅→
c+2→
c⋅→
a=2→
a⋅→
b−2→
b⋅→
c−2→
c⋅→
a
4→
a⋅→
c=0
Bisincorrect
| |
-------------------------------------------------------------------------------------------------
Question33
Let→
aand→
bbetwovectorsuchthat →
a= √14, →
b= √6and →
a×→
b= √48.
Then →
a⋅→
b
2isequalto_______.
[31-Jan-2023Shift1]
Answer:36
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question34
Let:→
a=^
i+2^
j+3^
k,→
b=^
i−^
j+2^
kand→
c=5^
i−3^
j+3^
kbetherevectors.
If→
risavectorsuchthat,→
r×→
b=→
c×→
band→
r⋅→
a=0.Then25 →
r|2isequalto
[31-Jan-2023Shift2]
Options:
A.449
B.336
C.339
D.560
Answer:C
Solution:
Solution:
| | | | | |
( )
|
→
a+λ→
c|2≥→
a|2
λ2c2≥0
True∀λ∈R(A)iscorrect.
| |
→
a= √14,→
b= √6→
a×→
b= √48
→
a×→
b2+→
a⋅→
b2=→
a2×→
b2
⇒→
a⋅→
b2=84 −48 =36
||| | | |
| | | | | | | |
( )
Sol. →
a=i+2 j +3 k
→
b=i−j+2 k
→
c=^
5i −3^
j+3^
k
→
r−→
c×→
b=0,→
r⋅→
a=0
⇒→
r−→
c=λ→
b
()
-------------------------------------------------------------------------------------------------
Question35
ThefootofperpendicularfromtheoriginOtoaplanePwhichmeets
theco-ordinateaxesatthepointsA,B,Cis(2,a,4), a∈N .Ifthe
volumeofthetetrahedronOABCis144unit3,thenwhichofthe
followingpointsisNOTonP?
[31-Jan-2023Shift2]
Options:
A.(2,2,4)
B.(0,4,4)
C.(3,0,4)
D.(0,6,3)
Answer:C
Solution:
Solution:
Also, →
c+λ→
b⋅→
a=0
⇒→
a⋅→
c+λ→
a⋅→
b=0
∴λ=
→
a⋅→
c
→
a⋅→
b
=−8
5
→
r=5 5 ^
i−3^
i+3^
k−8^
i−^
j+2^
k
5
→
r=17 ^
i−7^
j+^
k
5
→
r|2=1
25(289 +50)
25 →
r|2=339
( )
( )
( ) ( )
|
|
EquationofPlane:
2^
i+a^
j+4^
k⋅ (x−2)^
i+ (y−a)^
j+ (z−4)^
k=0
⇒2x +ay +4z =20 +a2
⇒A≡20 +a2
2,0,0
B≡0,20 +a2
a,0
C≡0,0,20 +a2
4
⇒Volumeoftetrahedron
=1
6
→
a→
b→
c
=1
6
→
a.→
b×→
c
⇒1
6
20 +a2
2⋅20 +a2
a⋅20 +a2
4=144
⇒ (20 +a2)3=144 ×48 ×a
⇒a=2
⇒Equationofplaneis2x +2y +4z =24
Orx+y+2z =12
⇒(3,0,4)NotliesonthePlane
x+y+2z =12
( ) [ ]
( )
( )
( )
[]
()
( ) ( ) ( )
Question36
Let→
a,→
b,→
cbethreevectorssuchthat →
a= √31,4→
b=→
c=2and
2→
a×→
b=3→
c×→
a.
Iftheanglebetween→
band→
cis 2π
3,then →
a×→
c
→
a⋅→
b
2
isequalto_________.
[31-Jan-2023Shift2]
Answer:3
Solution:
-------------------------------------------------------------------------------------------------
Question37
Let→
v=α^
i+2j −3k,→
w=2α^
i+j−k,and→
ubeavectorsuchthat
→
u=α>0.Iftheminimumvalueofthescalartripleproduct →
u→
v→
wis
−α√3401,and →
u.^
i
2=m
nwheremandnarecoprimenaturalnumbers,
thenm +nisequalto_______.
[1-Feb-2023Shift1]
Answer:3501
Solution:
||| | | |
( ) ( )
( )
| | [ ]
| |
Sol. 2 →
a×→
b=3→
c×→
a
→
a×2→
b+3→
c=0
→
a=λ 2→
b+3→
c
→
a|2=λ22→
b+3→
c|2
→
a|2=λ24→
b|2+9→
c|2+12→
b⋅→
c
31 =31λ2⇒λ= ±1
→
a= ± 2→
b+3→
c
→
b×→
c|2=→
b|2→
c|2−→
b⋅→
c)2=3
4
→
a×→
c
→
a⋅→
b
2
=3
( ) ( )
()
()
| |
| ( | | )
()
| | | (
( )
-------------------------------------------------------------------------------------------------
Question38
A(2,6,2), B(−4,0,λ), C(2,3, −1)andD(4,5,0),|λ| ≤ 5arethevertices
ofaquadrilateralABCD.Ifitsareais18squareunits,then5 −6λis
equalto______.
[1-Feb-2023Shift1]
Answer:11
Solution:
Solution:
-------------------------------------------------------------------------------------------------
→
u→
v→
w=→
u⋅→
v×→
w
min . |u|→
v×→
w cos θ = −α√3401
⇒cos θ = −1
|u| = α (Given)
→
v×→
w= √3401
→
v×→
w=
i j k
α 2 −3
2α 1 −1
→
v×→
w=i−5αj −3αk
→
v×→
w=
^
^
1+25α2+9α2= √3401
34α2=3400
α2=100
α=10
(as α >0 )
So
→
u=λ(i−5αj −3αk)
→
u=^
λ2+25α2λ2+9α2λ
α2=λ2(1+25α2+9α2)
100 =λ2(1+34 ×100)
λ2=100
3401 =m
n
[ ] ( )
( | | )
| |
| |
||√
√
A(2,6,2)B(−4,0,λ), C(2,3, −1)D(4,5,0)
Area =1
2
→
BD ×→
AC =18
→
AC ×→
BD =
∧
i j k
0−3−3
8 5 −λ
= (3λ +15)^
i−j(−24) + k(−24)
→
AC ×→
BD = (3λ +15)^
i+24j −24k
=(3λ +15)2+ (24)2+ (24)2=36
=λ2+10λ +9=0
=λ= −1, −9
|λ| ≤5⇒λ= −1
5−6λ =5−6(−1) = 11
| |
[ ]
√
Question39
Let→
a=5^
i−^
j−3^
kand→
b=^
i+3^
j+5^
kbetwovectors.Thenwhichoneof
thefollowingstatementsisTRUE?
[1-Feb-2023Shift2]
Options:
A.Projectionof→
aon→
bis 17
√35 andthedirectionofthep?
B.Projectionof→
aon→
bis −17
√35 andthedirectionofthep?
C.Projectionof→
aon→
bis 17
√35 andthedirectionoftheprojectionvectorisoppositetothe
directionof→
b
D.Projectionof→
aon→
bis −13
√35 andthedirectionoftheprojectionvectorisoppositetothe
directionof→
b
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question40
Let→
a=2^
i−7^
j+5^
k,→
b=^
i+^
kand→
c=^
i+2^
j−3^
kbethreegiven
vectors.If→
risavectorsuchthat→
r×→
a=→
c×→
aand→
r⋅→
b=0,then →
ris
equalto:
[1-Feb-2023Shift2]
Options:
A. 11
7√2
B. 11
7
C. 11
5√2
D. √914
7
Answer:A
Solution:
| |
→
a=5^
i−^
j−3^
k
→
b=^
i−3^
j+5^
k
→
a⋅^
b=5−3−15
√35 = − −13
√35
-------------------------------------------------------------------------------------------------
Question41
LetthepositionvectorsofthepointsA,B,CandDbe
5^
i+5^
j+2λ^
k,^
i+2^
j+3^
k, −2^
i+λ^
j+4^
kand−^
i+5^
j+6^
k.Lettheset
S= { λ∈ ℝ:thepointsA,B,CandDarecoplanar}.Then ∑
λ∈S(λ+2)2is
equalto:
[6-Apr-2023shift1]
Options:
A. 37
2
B.13
C.25
D.41
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question42
Let→
a=2^
i+3^
j+4^
k,b=^
i−2^
j−2^
kand→
c= −^
i+4^
j+3^
k.If→
disavector
a=2 i −7^
j+5^
k
→
b=^
i+^
k
→
c=^
i+2^
j−3^
k
→
r×→
a=→
c×→
a⇒→
r−→
c×→
a=0
∴→
r=→
c+λ→
a
→
r⋅→
b=0⇒→
c⋅→
b+λ→
b⋅→
a=0
−2+λ(7) = 0⇒λ=2
7
∴→
r=→
c+2
→
a
7=1
711 ^
i−11 ^
k
→
r=11√2
7
( )
( )
| |
A,B,C,Darecoplanar
⇒→
AB →
AC →
AD =0⇒
−4−3 3 −2λ
−7 λ −5 4 −2λ
−606−2λ
=0
⇒ −6[6λ −12 − (λ−5)(3−2λ)] + 0[ ] + (6−2λ)[20 −4λ −21]
⇒ −6[6λ −12 +2λ2+15 −13λ] + (6−2λ)[−4λ −1] = 0
⇒ −12λ2+42λ −18 +8λ2−22λ −6=0
⇒ −4λ2+20λ −24 =0⇒λ2−5λ +6=0
Now ∑
λ∈s
(λ+2)2=16 +25 =41
[][ ]
perpendiculartoboth→
band→
c,and→
a⋅→
d=18,then →
a×→
d2isequalto:
[6-Apr-2023shift1]
Options:
A.760
B.640
C.720
D.680
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question43
Letthevectors→
a,→
b,→
crepresentthreecoterminousedgesofa
parallelepipedofvolumeV.Thenthevolumeoftheparallelepiped,
whosecoterminousedgesarerepresentedby→
a,→
b+→
cand→
a+2→
b+3→
cis
equalto:
[6-Apr-2023shift2]
Options:
A.2V
B.6V
C.3V
D.V
Answer:D
Solution:
[ ]
→
d=λ→
b×→
c
For λ :→
a⋅→
d=18 ⇒λ→
a→
b→
c=18
⇒λ
234
1−2−2
−1 4 3
=18
⇒λ(4−3+8) = 18 ⇒λ=2
⇒→
d=2 2^
i−^
j+2^
k
Hence →
a×→
d|2=a2d2−→
a⋅→
d2
=29 ⋅36 − (18)2=18(58 −18)
=18 ⋅40 =720
( )
[ ]
| |
( )
| ( )
-------------------------------------------------------------------------------------------------
Question44
Thesumofallvaluesofα,forwhichthepointswhosepositionvectors
are^
i−2^
j+3k,2^
i−3^
j+4k, (α+1)^
i+2kand9^
i+ (α−8)^
j+6^
kare
coplanar,isequalto:
[6-Apr-2023shift2]
Options:
A.-2
B.2
C.6
D.4
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question45
Let→
a=6^
i+9^
j+12^
k,→
b=α^
i+11^
j−2^
kand→
cbevectorssuchthat
→
a×→
c−=→
a×→
b.If→
a.→
c= −12,→
c⋅^
i−2^
j+^
k=5,then→
c.^
i+^
j+^
kis
equalto_______.
[8-Apr-2023shift1]
( ) ( )
v1=→
a→
b+→
c→
a+2→
b+3→
c
v1=
100
011
123
→
a→
b→
c
vl= (3−2)v
=v
Ans.Option4
[ ]
| | [ ]
A= (1, −2,3)
B= (2, −3,4)
C= (α+1,0,2)
D= (9,α−8,6)
→
AB →
AC →
AD =0
1−1 1
α 2 −1
8 α −6 3
=0
⇒ (6+α−6) + 1(3α +8) + (α2−6α −16) = 0
⇒α2−2α −8=0
⇒α=4, −2
⇒sumofallvaluesof α =2
Ans.option 2
[ ]
| |
Answer:11
Solution:
-------------------------------------------------------------------------------------------------
Question46
Letthevectors→
u1=^
i+^
j+a^
k,→
µ2=^
i+b^
j+^
kand→
u3=c^
i+^
j+^
kbe
coplanar.Ifthevectors→
v1= (a+b)^
i+cj +ck,→
v2=a^
i+ (b+c)^
j+ak ^
kand
→
v3=b^
i+b^
j+ (c+a)^
karealsocoplanar,then6(a+b+c)isequalto
[8-Apr-2023shift2]
Options:
A.4
B.12
C.6
D.0
Answer:B
Solution:
→
a×→
c=→
a×5
⇒→
a×→
c−→
b=0
→
a|rl →
c−→
b
∴→
a=λ→
c−→
b
(6,9,12) = λ[x−α,y−11,z+2]
x−α
2=y−11
3=z+2
4
4y −44 =3z +6
4y −3z =50
6x +9y +12z = −12
2x +3y +4z = −4
(∵x−2y +z=5)
2x −4y +2z =10
+ −
7y +2z = −14 .. . (2)
8y −6z =100
21y +6z = −42
29y =58
y=2,z= −14
∴x−4−14 =5
x=23
c= (23,2, −14)
c⋅ (1,1,1) = 23 +2−14 =11
( )
( )
( )
-------------------------------------------------------------------------------------------------
Question47
AnarcPQofacirclesubtendsarightangleatitscentreO.Themid
pointofthearcPQisR.If →
OP =→
u,→
OR =→
vand →
OQ =α→
u+β→
v,thenα,β2are
therootsoftheequation:
[10-Apr-2023shift1]
Options:
A.3x2−2x −1=0
B.3x2+2x −1=0
C.x2−x−2=0
D.x2+x−2=0
Answer:C
Solution:
→
u1
→
u2
→
u3=0∴
1 1 a
1 b 1
c 1 1
=0
⇒b−1+c−1+a(1−bc) = 0
∴abc =a+b+c−2
→
v1
→
v2
→
v3=0∴
a+b c c
a b +c a
b b c +a
=0
⇒ −2a(ac −bc −c2) + 2c(a2+ab −ac) = 0
⇒ −2a2c+2 abc +2ac2+2a2c+2 abc −2ac2=0
⇒4abc =0∴abc =0
∴a+b+c=2∴6(a+b+c) = 12 Ans.
[ ] | |
[ ] | |
-------------------------------------------------------------------------------------------------
Question48
Let→
a=2^
i+7^
j−^
k,→
b=3^
i+5^
kand→
C=^
i+^
j+2^
k,Let→
d-beavectorwhich
isperpendiculartoboth→
a,and→
b,and→
c⋅→
d=12.The
−^
i+^
j−^
k⋅→
c×→
d-isequalto
[10-Apr-2023shift2]
Options:
A.24
B.42
C.48
D.44
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question49
( ) ( )
Then →
OR =→
v=1
√2
^
i+1
√2
^
j
Now
→
OQ =α→
u+β→
v
^
j=α^
i+β1
√2
^
i+1
√2
^
j
β= √2,α+β
√2=0⇒α= −1
Nowequation
x2− (α+β2)x+αβ2=0
x2− (−1+2)x+ (−1)(2) = 0
x2−x−2=0
( )
→
a=2^
i+7^
j−^
k
→
b=3^
i+5^
k
→
c=^
i−^
j+2^
k
→
d=λ→
a×→
b=λ
^
i^
j^
k
2 7 −1
3 0 5
→
d=λ 35^
i−13^
j−21^
k
λ(35 +13 −42) = 12
λ=2
→
d=2 35^
i−13^
j−21^
k
^
i+^
j−^
k→
c×→
d
=
−11−1
1−12
70 −26 −42
=44
( ) | |
( )
( )
( ) ( )
| |
Foranyvector→
a=a1
^
i+a2
^
j+a3
^
k,with10|ai| < 1,i=1,2,3,considerthe
followingstatements:
(A):max{|a1|, |a2|, |a3|}≤ →
a
(B): →
a≤3 max{|a1|, |a2|, |a3|}
[11-Apr-2023shift1]
Options:
A.Only(B)istrue
B.Both(A)and(B)aretrue
C.Neither(A)nor(B)istrue
D.Only(A)istrue
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question50
Let→
abeanon-zerovectorparalleltothelineofintersectionofthetwo
placesdescribedby^
i+^
j,^
i+^
kand^
i−^
j,^
j−^
k.Ifθistheanglebetween
thevector→
aandthevector→
b=2^
i−2^
j+^
kand→
a⋅→
b=6,thenorderedpair
θ,→
a×→
bisequalto:
[11-Apr-2023shift1]
Options:
A. π
3,6
B. π
4,3√6
C. π
3,3√6
| |
| |
( | | )
( )
( )
( )
Withoutlossofgenerality
Let|a1| ≤ |a2| ≤ |a3|
→
a|2=→
a1
2+→
a2
2+→
a3
2≥ (a3)2
⇒→
a≥|a3| = max{|a1|, |a2|, |a3|}
Aistrue
→
a|2= |a1|2+ |a2|2+ |a3|2≤ |a3|2+ |a3|2+ |a3|2
⇒→
a|2≤3|a3|2
⇒→
a≤√3→
a3= √3 max{|a1|, |a2|, |a3|}
≤3 max{|a1|, |a2|, |a3|}
(2)istrue
|| | | | | |
||
|
|
|| | |
D. π
4,6
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question51
Iffourdistinctpointswithpositionvectors→
a,→
b,→
cand→
darecoplanar,
then →
a→
b→
cisequalto
[11-Apr-2023shift2]
Options:
A. →
d→
c→
a+→
b→
d→
a+→
c→
d→
b
B. →
d→
b→
a+→
a→
c→
d+→
d→
b→
c
C. →
a→
d→
b+→
d→
c→
a+→
d→
b→
c
D. →
b→
c→
d+→
d→
a→
c+→
d→
b→
a
( )
[ ]
[ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ] [ ]
→
n1and→
n2arenormalvectortotheplane^
i+^
j,^
i+^
kand^
i−^
j,^
i−^
krespectively
→
n1=
^
i^
j^
j
110
101
=^
i−^
j−^
k
→
n2=
^
i^
j^
j
1−10
1 0 −1
=^
i+^
j+^
k
→
a=λ→
n2×→
n2
=λ
^
i^
j^
j
1−1−1
1 1 −1
=λ−2^
j+2^
k
→
a⋅→
b=λ 0 +4+2=6
⇒λ=1
→
α= −2^
j+2^
k
cos θ =
→
a⋅→
b
|a||b|
cos θ =6
2√2×3=1
√2
θ=π
4
Now →
a⋅→
b|2+→
a×→
b|2=a|2b|2
36 +→
a×b2=8×9=72
→
a×b|2=36
→
a×→
b=6
| |
| |
| |
| | ( )
| |
| | | |
| |
|
| |
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question52
Let→
a=^
i+2^
j+3^
kand→
b=^
i+^
j−^
k.If→
cisavectorsuchthat
→
a⋅→
c=11,→
b⋅→
a×→
c=27and→
b⋅→
c= −√3→
b,then →
a×→
c2isequalto
_______.
[11-Apr-2023shift2]
Answer:285
Solution:
-------------------------------------------------------------------------------------------------
Question53
Leta,b,cbethreedistinctrealnumbers,noneequaltoone.Ifthe
vectorsa^
i+^
j+^
k,^
i+b^
j+^
kand^
i+^
j+c^
karecoplanar,then
1
1−a+1
1−b+1
1−cisequalto
[12-Apr-2023shift1]
( ) | | | |
→
a,→
b,→
c,→
d→coplanar
→
a→
b→
c= ?
→
b−→
a,→
c−→
b,→
d−→
c→coplanar
→
b−→
a→
c−→
b,→
d−→
c=0
⇒→
b−→
a⋅→
c−→
b×→
d−→
c=0
→
b−→
a⋅→
c×→
b−→
c×→
a−→
a×→
d=0
[bcd] − [bca] − [bad] − [acd] = 0
→
a→
b→
c=→
d→
c→
a+→
b→
d→
a+→
c→
d→
b
[ ]
[ ]
( ) ( ( ) ( ) )
( ) ( )
[ ] [ ][ ] [ ]
→
a=^
i+2^
j+3^
k,→
b=^
i+^
j−^
k
→
b⋅→
a×→
c=27,→
a⋅→
b=0
→
b×→
a×→
c= −3→
a
Letθbeanglebetween→
b,→
a×→
c
Then →
b⋅→
a×→
c sin θ =3√14
→
b⋅→
a×→
c cos θ =27
⇒sin θ =√14
√95
∴→
b×→
a×→
c=3√95
⇒→
a×→
c= √3× √95
( )
( )
| | | |
| | | |
| | | |
| |
Options:
A.1
B.2
C.-2
D.-1
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question54
Letλ ∈Z,→
a=λ^
i+^
j−^
kand→
b=3^
i−^
j+2^
k.Let→
cbeavectorsuchthat
→
a+→
b+→
c×→
c=→
0,→
a⋅→
c= −17and→
b⋅→
c= −20.Then →
c×λ^
i+^
j+^
k
2is
equalto
[12-Apr-2023shift1]
Options:
A.62
B.53
C.49
D.46
Answer:D
Solution:
( ) | ( ) |
a 1 1
1 b 1
1 1 c
=0
C2→C2−C1,C3→C3−C1
a1−a 1 −a
1b−10
1 0 c−1
=0
a(b−1)(c−1) − (1−a)(c−1) + (1−a)(1−b) = 0
a(1−b)(1−c) + (1−a)(1−c) + (1−a)(1−b) = 0
a
1−a+a
1−b+a
1−c=0
⇒ −1+1
1−a+1
1−b+1
1−c=0
⇒1
1−a+1
1−b+1
1−c=1
| |
| |
→
a+→
b+→
c×
→
c=0
( )
-------------------------------------------------------------------------------------------------
Question55
Let→
a=3^
i+^
j−^
kand→
c=2^
i−3^
j+3^
k.If→
bisavectorsuchthat→
a=→
b×→
c
and →
b
2=50,then|72−|→
b+→
c|2∣isequalto_______.
[13-Apr-2023shift1]
Answer:66
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question56
Let→
a=^
i+4^
j+2^
k,→
b=3^
i−2^
j+7^
kand→
c=2^
i−^
j+4^
k.Ifavector→
d
satisfies→
d×→
b=→
c×→
band→
d⋅→
a=24,then →
d
2isequalto-
[13-Apr-2023shift1]
| |
| |
→
a+→
b×→
c=0
→
c=α→
a+→
b=α(λ+3)^
i+α^
k
→
b⋅→
c= −20 ⇒3α(λ+3) + 2α = −20
→
a⋅→
c= −17 ⇒αλ(λ+3) − α= −17
⇒α(3λ +9+2) = −20
α(λ2+3λ −1) = −17
17(3λ +11) = 20(λ2+3λ −1)
20λ2+9λ −207 =0
λ=3(λ∈Z)
⇒α= −1⇒→
c= − 6^
i+^
k
→
v=→
c×3^
i+^
j+^
k
=
^
i^
j^
k
−6 0 −1
311
=^
i+3^
j−6^
k
→
v|2= (−1)2+32+62=46
( )
( )
( )
( )
| |
|
→
a= √11,→
c= √22
→
a=→
b×→
c=→
b→
c sin θ
√11 = √50√22sin θ
⇒sin θ =1
10
→
b+→
c|2=→
b|2+→
c|2+2→
b⋅→
c
=→
b|2+→
c|2+2→
b→
c cos θ
=50 +22 +2× √50 × √22×√99
10
=72 +66
|72−|→
b+→
c|2∣ = 66
| | | |
| | | | | | | |
| | |
| | | | | |
Options:
A.323
B.423
C.413
D.313
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question57
Let →
a=2,→
b=3andtheanglebetweenthevectors→
aand→
bbe π
4.Then
→
a+2→
b×2→
a−3→
b
2isequalto
[13-Apr-2023shift2]
Options:
A.482
B.841
C.882
D.441
Answer:C
Solution:
| | | |
| ( ) ( ) |
→
d×→
b=→
c×→
b
⇒→
d−→
c×→
b=0
⇒→
d=→
c+λ→
b
Also →
dä =24
⇒→
c+λ→
b⋅→
a=24
λ=24 −→
a⋅→
c
→
b⋅→
a
=24 −6
9=2
⇒→
d=→
c+2→
b
=8^
i−5^
j+18^
k
⇒→
d|2=64 +25 +324 =413
( )
( )
( )
|
cos π
4=
→
a⋅→
b
→
a→
b
⇒1
√2=
→
a⋅→
b
(2)(3)⇒→
a⋅→
b=3√2
Let→
p=→
a+2→
b
( ) | | | |
-------------------------------------------------------------------------------------------------
Question58
LetSbethesetofall(λ,µ)forwhichthevectorsλ^
i−^
j+^
k,^
i+2^
j+µ^
k
and3^
i−4^
j+5^
k,whereλ −µ=5,arecoplanar,then ∑
(λ,µ) ∈ S80(λ 2 +µ2)is
equalto
[15-Apr-2023shift1]
Options:
A.2130
B.2210
C.2290
D.2370
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question59
→
q=2→
a−3→
b
→
p|2=→
a|2+4→
b|2+4→
a⋅→
b
=4+36 +12√2
=40 +12√2
→
q|2=4→
a|2+9→
b|2−12 →
a−→
b
=16 +81 −36√2
=97 −36√2
→
p⋅→
q=2→
a|2−6→
b|2+→
a⋅→
b
=8−54 +3√2
= −46 +3√2
→
p×→
q=→
p→
q2−→
p⋅→
q2
= (40 +12√2(97 −36√2)) − (3√2−46)2
= (3016 −276√2) − (2134 −276√2)
=882
| | | ( )
| | | ( )
||
| | ( | | | | ) ( )
λ−11
1 2 µ
3−45
=0&λ−µ=5
λ(10 +4µ) + (5−3µ) + (−10) = 0
(µ+5)(4µ +10) + 5−3µ −10 =0
µ= −15;λ=5∕4
µ= −3;λ=2
Hence
2+µ2)
∑
(λ,µ) ∈ S80(λ
=80 250
16 +13
=1250 +1040
=2290
| |
( )
LetABCDbeaquadrilateral.IfEandFarethemidpointsofthe
diagonalsACandBDrespectivelyand →
AB −→
BC +→
AD −→
DC =k→
FE,thenk
isequalto
[15-Apr-2023shift1]
Options:
A.4
B.2
C.-2
D.-4
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question60
Let∧
aand∧
bbetwounitvectorssuchthat ∧
a+∧
b+2∧
a×∧
b=2.If
θ∈ (0,π)istheanglebetween∧
aand∧
b,thenamongthestatements:
(S1) : 2∧
a×∧
b=∧
a−∧
b
(S2):Theprojectionof∧
aon ∧
a+∧
bis 1
2
[24-Jun-2022-Shift-2]
Options:
A.Only(S1)istrue.
B.Only(S2)istrue.
( ) ( )
| ( ) ( ) |
| | | |
( )
→
AB −→
BC +→
AB −→
DC =k→
FE
→
b−→
a−→
c−→
b+→
d−→
a−→
c−→
d=k→
FE
2→
b+→
d−2→
a−→
c=k→
FE
2 2→
f−2 2→
e=k→
FE
4→
f−→
e=k→
FE
−4→
FE =k→
FE
k= −4
( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( )
C.Both(S1)and(S2)aretrue.
D.Both(S1)and(S2)arefalse.
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question61
Let→
a=a1
∧
i+a2
∧
j+a3widehat k ai>0,i=1,2,3beavectorwhichmakes
equalangleswiththecoordinateaxesOX,OYandOZ.Also,letthe
projectionof→
aonthevector3∧
i+4∧
jbe7.Let→
bbeavectorobtainedby
rotating→
awith90∘.If→
a,→
bandx-axisarecoplanar,thenprojectionofa
vector→
bon3∧
i+4∧
jisequalto:
[25-Jun-2022-Shift-1]
Options:
A.√7
B.√2
C.2
D.7
Answer:B
Solution:
-------------------------------------------------------------------------------------------------
Question62
Letθbetheanglebetweenthevectors→
aand→
b,where→
a∣=4,→
b∣=,3and
θ∈π
4,π
3.Then →
a−→
b×→
a+→
b
2+4→
a⋅→
b
2isequalto__
[25-Jun-2022-Shift-1]
Answer:576
Solution:
| |
( ) | ( ) ( ) | ( )
-------------------------------------------------------------------------------------------------
Question63
Let→
b=∧
i+∧
j+λ∧
k,λ∈R.If→
aisavectorsuchthat→
a×→
b=13∧
i−∧
j−4∧
kand
→
a⋅→
b+21 =0,then →
b−→
a⋅∧
k−∧
j+→
b+→
a⋅∧
i−∧
kisequalto____
[25-Jun-2022-Shift-2]
Answer:14
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question64
If→
a⋅→
b=1,→
b⋅→
c=2and→
c⋅→
a=3,thenthevalueof
→
a×→
b×→
c,→
b×→
c×→
a,→
c×→
b×→
ais:
[26-Jun-2022-Shift-1]
Options:
A.0
B.−→
6a ⋅→
b×→
c
( ) ( ) ( ) ( )
[ ( ) ( ) ( ) ]
( )
Let→
a=x
∧
i=y
∧
j+z
∧
k
So,
∧
i
∧
j
∧
k
x y z
11λ
=∧
i(λy −z) + ∧
j(z−λx) + ∧
k(x−y)
⇒λy −z=13,z−λx = −1,x−y= −4
andx+y+λz = −21
⇒Clearly,λ=3,x= −2,y=2andz= −7
So,→
b−→
a=3
∧
i−∧
j+10
∧
k
and→
b+→
a= −∧
i+3
∧
j−4
∧
k
⇒→
b−→
a⋅∧
k−∧
j+→
b+→
a⋅∧
i−∧
k=11 +3=14
| |
( ) ( ) ( ) ( )
C.−12→
c⋅→
a×→
b
D.−12→
b⋅→
c×→
a
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question65
Let→
a=∧
i+∧
j+2∧
k,→
b=2∧
i−3∧
j+∧
kand→
c=∧
i−∧
j+∧
kbethreegivenvectors.Let
→
vbeavectorintheplaneof→
aand→
bwhoseprojectionon→
cis 2
√3.If
→
v⋅∧
j=7,then→
v⋅∧
i+∧
kisequalto:
[26-Jun-2022-Shift-2]
Options:
A.6
B.7
C.8
D.9
Answer:D
Solution:
( )
( )
( )
-------------------------------------------------------------------------------------------------
Question66
Let→
a=∧
i+∧
j−∧
kand→
c=2∧
i−3∧
j+2∧
k.Thenthenumberofvectors→
bsuch
that→
b×→
c=→
aand →
b∈ {1,2, ......, 10}is:
[27-Jun-2022-Shift-1]
Options:
A.0
B.1
C.2
D.3
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question67
| |
→
a=∧
i+∧
j−∧
k
→
c=2
∧
i−3
∧
j+2
∧
k
Now,→
b×→
c=→
a
→
c⋅→
(b×→
c) = →
c⋅→
a
→
c⋅→
a=0
⇒∧
i+∧
j−∧
k2
∧
i−3
∧
j+2
∧
k=0
=2−3−2=0
⇒−3=0(Notpossible)
⇒Nopossiblevalueof→
bispossible.
( ) ( )
Let→
aand→
bbethevectorsalongthediagonalsofaparallelogramhaving
area2√2.Lettheanglebetween→
aand→
bbeacute, →
a=1,and
→
a⋅→
b=→
a×→
b.If→
c=2√2→
a×→
b−2→
b,thenananglebetween→
band→
cis
[27-Jun-2022-Shift-2]
Options:
A. π
4
B.−π
4
C. 5π
6
D. 3π
4
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question68
If→
a=2∧
i+∧
j+3∧
k,→
b=3∧
i+3∧
j+∧
kand→
c=c1
∧
i+c2
∧
j+c3
∧
karecoplanarvectors
and→
a⋅→
c=5,→
b⟂→
c,then122(c1+c2+c3)isequalto
[28-Jun-2022-Shift-1]
| |
| | | | ( )
∵→
aand→
bbethevectorsalongthediagonalsofaparallelogramhavingarea2√2.
∴1
2∣→
a×→
b∣ = 2√2
→
a→
b sin θ =4√2
⇒→
∣b∣sin θ =4√2.... (i)
and →
a⋅→
b=→
a×→
b∣
→
a→
b cos θ =→
a→
b sin θ
⇒tan θ =1
∴θ=π
4
By(i) →
|b| = 8
Now→
c=2√2→
(a×→
b−2→
b
⇒→
c⋅→
b= −2→
b|2= −128
and→
c⋅→
c=8→
a×→
b|2+4→
b|2
⇒ | c|2=8.32 +4.64
⇒→
c=16√2
From(ii)and(iii)
→
|c|→
b∣cos α = −128
⇒cos α =−1
√2
α=3π
4
| | | |
| |
| | | | | | | |
)
|
| |
| |
| |
Answer:150
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question69
Let→
a=α∧
i+2∧
j−∧
kand→
b= −2∧
i+α∧
j+∧
k,whereα ∈R.Iftheareaofthe
parallelogramwhoseadjacentsidesarerepresentedbythevectors→
aand
→
bis 15(α2+4),thenthevalueof2 →
a2+→
a⋅→
b→
b
2isequalto:
[28-Jun-2022-Shift-2]
Options:
A.10
B.7
C.9
D.14
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
√| | ( ) | |
2C1+C2+3C3=5....... (i)
3C1+3C2+C3=0
→
a→
b→
c=
213
331
C1C2C3
=2(3C3−C2) − 1(3C3−C1) + 3(3C2−3C1)
=3C3+7C2−8C1
⇒8C1−7C2−3C3=0...... (iii)
C1=10
122,C2=−85
122 ,C3=225
122
So122(C1+C2+C3) = 150
[ ] | |
→
a=α
∧
i+2
∧
j−∧
kand→
b= −2
∧
i+α
∧
j+∧
k
∴→
a×→
b=
∧
i
∧
j
∧
k
α 2 −1
−2α 1
= (2+α)∧
i− (α−2)∧
j+ (α2+4)∧
k
Now →
a×→
b=15(α2+4)
⇒(2+α)2+ (α−2)2+ (α2+4)2=15(α2+4)
⇒α4−5α2−36 =0
∴α= ±3
Now,2→
a|2+→
a−→
b→
b|−2=2.14 −14 =14
| |
| | √
|( ) |
Question70
Let→
abeavectorwhichisperpendiculartothevector3∧
i+1
2
∧
j+2∧
k.If
→
a×2∧
i+∧
k=2∧
i−13∧
j−4∧
k,thentheprojectionofthevector→
aonthe
vector2∧
i+2∧
j+∧
kis:
[28-Jun-2022-Shift-2]
Options:
A. 1
3
B.1
C. 5
3
D. 7
3
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question71
Let→
a=α∧
i+3∧
j−∧
k,→
b=3∧
i−β∧
j+4∧
kand→
c=∧
i+2∧
j−2∧
kwhereα,β∈R,be
threevectors.Iftheprojectionof→
aon→
cis 10
3and→
b×→
c= −6∧
i+10∧
j+7∧
k,
thenthevalueofα +βisequalto:
[29-Jun-2022-Shift-1]
Options:
A.3
B.4
C.5
( )
Let→
a=a1
∧
i+a2
∧
j+a3
∧
k
and→
a⋅3
∧
i−1
2
∧
j+2
∧
k=0⇒3a1+a2
2+2a3=0.....(i)
and→
a×2
∧
i+∧
k=2
∧
i−13
∧
j−4
∧
k
⇒a2
∧
i+ (2a3−a1)∧
j−2a2
∧
k=2
∧
i−13
∧
j−4
∧
k
∴a2=2.....(ii)
anda1−2a3=13.....(iii)
Fromeq.(i)and(iii):a1=3anda3= −5
∴→
a=3
∧
i+2
∧
j−5
∧
k
∴projectionof→
aon2
∧
i+2
∧
j+∧
k=6+4−5
3=5
3
( )
( )
D.6
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question72
LetA,B,Cbethreepointswhosepositionvectorsrespectivelyare
→
a=∧
i+4∧
j+3∧
k
→
b=2∧
i+α∧
j+4∧
k,α∈R
→
c=3∧
i−2∧
j+5∧
k
Ifαisthesmallestpositiveintegerforwhich→
a,→
b,→
carenoncollinear,
thenthelengthofthemedian,in∆ABC,throughAis:
[29-Jun-2022-Shift-2]
Options:
A. √82
2
B. √62
2
C. √69
2
D. √66
2
Answer:A
Solution:
→
a=α
∧
i+3
∧
j−∧
k
→
b=3
∧
i−β
∧
j+4
∧
k
→
c=∧
i+2
∧
j−2
∧
k
Projectionof→
aon→
cis
→
a⋅→
c
→
|b|
=10
3
α+6+2
12+22+ (−2)2=α+8
3=10
3
∴α=2
→
b×→
c= −6
∧
i+10
∧
j+7
∧
k
∧
i
∧
j
∧
k
3−β4
12−2
= (2β −8)∧
i+10
∧
j+ (6+β)∧
k= −6
∧
i+10
∧
j+7
∧
k
2β −8= −6&6+β=7
∴β=1
α+β=2+1=3
√
| |
Solution:
-------------------------------------------------------------------------------------------------
Question73
Let→
a=∧
i−2∧
j+3∧
k,→
b=∧
i+∧
j+∧
kand→
cbeavectorsuchthat→
a+→
b×→
c=→
0
and→
b⋅→
c=5.Thenthevalueof3 →
c⋅→
aisequalto____
[29-Jun-2022-Shift-2]
Answer:0
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question74
Let∧
a,∧
bbeunitvectors.If→
cbeavectorsuchthattheanglebetween∧
aand
→
cis π
12,and∧
b=→
c+2→
c×∧
a,then 6→
c|2isequalto:
[24-Jun-2022-Shift-1]
Options:
A.6(3− √3)
B.3 + √3
C.6(3+ √3)
( )
( )
( ) |
→
AB →
ACif 1
2=α−4
−6=1
2⇒α=1
→
a,→
b,→
carenon-collinearforα=2(smallestpositiveinteger)
Mid-pointofBC =M5
2,0,9
2
AM =9
4+16 +9
4=√82
2
|
( )
√
→
a+→
b×→
c=0
→
a×→
b+→
b|2→
c−5→
b=0
Itgives→
c=1
310
∧
i+3
∧
j+2
∧
k
so3→
a⋅→
c=10
Butitdoesnotsatisfy→
a+→
b×→
c=0.
Thisquestionhasdataerror.
Alternate(Explanation):
Accordingtogiven→
a&→
b→
a⋅→
b=1−2+3=2...(i)
butgivenequation
→
a= − →
b×→
c
⇒→
a⟂→
b⇒→
a⋅→
b=0
whichcontradicts.
|
( )
()
D.6(√3+1)
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question75
Iftheshortestdistancebetweenthelines
→
r= −∧
i+3∧
k+λ∧
i−a∧
j
and→
r= −∧
j+2∧
k+µ∧
i−∧
j+∧
kis 2
3,thentheintegralvalueofais
equalto
[24-Jun-2022-Shift-1]
Answer:2
Solution:
( ) ( )
( ) ( ) √
Question76
LetABCbeatrianglesuchthat
→
BC =→
a,→
CA =→
b,→
AB =→
c,a∣ = 6√2,overrightarrow ∣→
b∣ = 2√3and→
b⋅→
c=12.
Considerthestatements:
(S1) : ∣ →
a×→
b+→
c×→
b−→
c∣ = 6(2√2−1)
(S2) : ∠ACB =cos−12
3
Then
[25-Jul-2022-Shift-1]
Options:
A.both(S1)and(S2)aretrue
B.only(S1)istrue
C.only(S2)istrue
D.both(S1)and(S2)arefalse
Answer:D
Solution:
( ) ( ) | |
(√)
∵→
a+→
b+→
c=0
thena+→
c= −→
b
then →
a+→
c×→
b= −→
b×b
( )
-------------------------------------------------------------------------------------------------
Question77
Let→
a=∧
i−∧
j+2∧
kandlet→
bbeavectorsuchthat→
a×→
b=2∧
i−∧
kand→
a⋅→
b=3.
Thentheprojectionof→
bonthevector→
a−→
bis:
[25-Jul-2022-Shift-2]
Options:
A. 2
√21
B.2 3
7
C. 2
3
7
3
D. 2
3
Answer:A
Solution:
√
√
∴→
a×→
b+→
c×→
b=0.. . (i)
For(S1): →
a×→
b+→
c×→
b−→
c=6(2√2−1)
→
a+→
c×→
b−→
c=6(2√2−1)
→
c=6−12√2(notpossible)
For(S2):from(i)→
b+→
c= −→
a
⇒→
b⋅→
b+→
c⋅→
b= −→
a⋅→
b
⇒12 +12 = −6√2⋅2√3cos(π− ∠ACB)
∴cos(∠ACB) = 2
3
∴ ∠ACB =cos−12
3
∴S(2)iscorrect.
| | | |
|( ) | | |
| |
√
√
→
a=∧
i−∧
j+2
∧
k
→
a×→
b=2
∧
i−∧
k
→
a⋅→
b=3
→
a×→
b2+→
a⋅→
b|2=→
a|2⋅→
|b|2
⇒5+9=6→
b|2
⇒ | b|2=7
3
→
a−→
b=→
a2+ . →
|b|2−2→
a⋅→
b=7
3
projectionof→
bon→
a−→
b=
→
b⋅→
a−→
b
→
a−→
b
=
→
b⋅→
a−→
b|2
→
a−→
b∣
=
3−7
3
7
3
=2
√21
| | | |
|
| | √| | √
( )
| |
|
√
Question78
Let→
a=α∧
i+∧
j−∧
kand→
b=2∧
i+∧
j−α∧
k,α>0.Iftheprojectionof→
a×→
bonthe
vector−∧
i+2∧
j−2∧
kis30,thenαisequalto:
[26-Jul-2022-Shift-1]
Options:
A. 15
2
B.8
C. 13
2
D.7
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question79
Let→
a=α∧
i+∧
j+β∧
kand→
b=3∧
i−5∧
j+4∧
kbetwovectors,suchthat
→
a×→
b= −∧
i+9∧
i+12∧
k.Thentheprojectionof→
b−2→
aon→
b+→
aisequalto
[27-Jul-2022-Shift-1]
Options:
A.2
B. 39
5
C.9
Given:→
a= (α,1, −1)and→
b= (2,1, −α)
→
c=→
a×→
b=
∧
i
∧
j
∧
k
α 1 −1
2 1 −α
= (−α+1)∧
i+ (α2−2)∧
j+ (α−2)∧
k
Projectionof→
con→
d= −∧
i+2
∧
j−2
∧
k
. = →
c⋅
→
d
|d|=30 Given
⇒ = α−1−4+2α2−2α +4
√1+4+4=30
Onsolvingα=−13
2(Rejectedas.α>0)andα=7
| |
| | }
| |
D. 46
5
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question80
Let→
a=2∧
i−∧
j+5∧
kand→
b=α∧
i+β∧
j+2∧
k.If →
a×→
b×∧
i⋅∧
k=23
2,then →
b×2∧
j
isequalto
[27-Jul-2022-Shift-1]
Options:
A.4
B.5
C.√21
D.√17
Answer:B
Solution:
Solution:
(( ) ) | |
→
a=α
∧
i+∧
j+β
∧
k,→
b=3
∧
i−5
∧
j+4
∧
k
→
a×→
b= −∧
i+9
∧
j+12
∧
k
∧
i
∧
j
∧
k
α 1 β
3−54
= −∧
i+9
∧
j+12
∧
k
4+5β = −1⇒β= −1
−5α −3=12 ⇒α= −3
→
b−2→
a=3
∧
i−5
∧
j+4
∧
k−2−3
∧
i+∧
j−∧
k
→
b−2→
a=9
∧
i−7
∧
j+6
∧
k
→
b+→
a=3
∧
i−5
∧
j+4
∧
k+ −3
∧
i+∧
j−∧
k
→
b+→
a= −4
∧
j+3
∧
k
Projectionof→
b−2→
aon→
b+→
ais =
→
b−2→
a⋅→
b+→
a
→
b+→
a
=28 +18
5=46
5
| |
( )
( ) ( )
( ) ( )
| |
Given,→
a=2
∧
i−∧
j+5
∧
kand→
b=α
∧
i+β
∧
j+2
∧
k
Also, →
a×→
b×i⋅∧
k=23
2
⇒→
a⋅∧
i→
b−→
b⋅∧
i⋅a⋅∧
k=23
2
⇒2⋅→
b−α⋅→
a⋅∧
k=23
2
( ( ) )
( ( ) ( ) )
( )
-------------------------------------------------------------------------------------------------
Question81
Let→
a,→
b,→
cbethreenon-coplanarvectorssuchthat→
a×→
b=4→
c,→
b×→
c=9→
a
and→
c×→
a=α→
b,α>0
If →
a+→
b+→
c=1
36,thenαisequalto________.
[27-Jul-2022-Shift-2]
Answer:36
Solution:
-------------------------------------------------------------------------------------------------
Question82
Letthevectors→
a=(1+t)
∧
i+(1−t)
∧
j+∧
k,→
b=(1−t)
∧
i+ (1+t)∧
j+2∧
kand
→
c−t∧
i−t∧
j+∧
k,t∈Rbesuchthatfor
α,β,γ∈R,α→
a+β→
b+γ→
c=→
0⇒α=β=γ=0.Then,thesetofallvaluesoft
| | | | | |
⇒2⋅2−5α =23
2⇒α=−3
2
Now, →
b×2j =α
∧
i+β
∧
j+2
∧
k×2
∧
j
=2α
∧
k+0−4
∧
i
=4α2+16
=4−3
2
2+16 =5
| | | ( ) |
| |
√
√( )
Given,
→
a×→
b=4⋅→
c..... (i)
→
b×→
c=9⋅→
a..... (ii)
→
c×→
a=α⋅→
b... .(iii)
Takingdotproductswithc,→
a,
→
→
bweget
→
a⋅→
b=→
b⋅→
c=→
c⋅→
a=0
Hence,
(i) ⇒→
a⋅→
b=4.→
c∣..... (iv)
(ii)⇒ →
|b|⋅ →
c=9⋅→
a....(v)
(iii)⇒ →
c∣⋅ →
a=α⋅→
|b|...(vi)
Multiplying(iv),(v)and(vi)
⇒→
a⋅→
∣b⋅→
c∣ = 36α..... (vii)
Dividing(vii)by(iv)⇒ →
c|2=9α⇒ ∣ →
c=3√α....(viii)
Dividing(vii)by(v)⇒ →
a|2=4α⇒→
a=2√α
Dividing(viii)by(vi)⇒overrightarrow →
b2=36 ⇒→
b∣ = 6
Now,asgiven,3√α+2√α+6=1
36 ⇒ √α=−43
36
| | | |
| | | |
| |
|| | |
|
| | |
| |
is:
[28-Jul-2022-Shift-1]
Options:
A.anon-emptyfiniteset
B.equaltoN
C.equaltoR − {0}
D.equaltoR.
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question83
Letavector→
a,hasmagnitude9.Letavector→
bbesuchthatforevery
(x,y) ∈ R×R− {(0,0)},thevector x→
a+y→
bisperpendiculartothe
vector 6y→
a−18x→
b.Thenthevalueof →
a×→
bisequalto:
[28-Jul-2022-Shift-1]
Options:
A.9√3
B.27√3
C.9
D.81
Answer:B
Solution:
Solution:
( )
( ) | |
Clearly→
a,→
b,→
carenon-coplanar
1+t 1 −t1
1−t 1 +t2
t−t1
≠0
⇒(1+t)(1+t+2t) − (1−t)(1−t−2t) + 1(t2−t−t−t2) ≠ 0
⇒(3t2+4t +1) − (1−t)(1−3t) − 2t ≠0
⇒(3t2+4t +1) − (3t2−4t +1) − 2t ≠0
⇒t≠0
| |
→
xa +→
yb ⋅6→
ya −18x→
b=0
⇒6xy →
a|2−18xy →
b|2+ (6y2−18x2)→
a⋅→
b=0
Asgivenequationisidentity
Coefficientofx2= coefficientofy2= coefficientofxy =0
( ) ( )
(|| )
-------------------------------------------------------------------------------------------------
Question84
LetSbethesetofalla ∈Rforwhichtheanglebetweenthevectors
→
u=a(logeb)∧
i−6∧
j+3∧
kand→
v= (logeb)∧
i+2∧
j+2a(logeb)∧
k, (b>1)isacute.
ThenSisequalto:
[28-Jul-2022-Shift-2]
Options:
A. −∞, − 4
3
B.Φ
C. −4
3,0
D. 12
7,∞
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question85
Let→
a=3∧
i+∧
jand→
b=∧
i+2∧
j+∧
k.Let→
cbeavectorsatisfying
→
a×→
b×→
c=→
b+→
c.If→
band→
carenon-parallel,thenthevalueofλis:
[29-Jul-2022-Shift-1]
Options:
A.−5
B.5
( )
( )
( )
()
⇒→
a|2=.3 →
∣b|2⇒→
b∣ = 3√3
anda.
→
→
b=0
→
a×→
b=→
a→
b sinθ
=9.3√3.1 =27√3
|
| | | | | |
→
u=a(logeb)∧
i−6
∧
j+3
∧
k
→
v= (logeb)∧
i+2
∧
j+2a(logeb)∧
k
Foracuteangle→
u⋅→
v>0
⇒a(logeb)2−12 +6a(logeb) > 0
∵b>1
Letlogeb=t⇒t>0asb>1
at2+6at −12 >0∀t>0
⇒a∈φ
C.1
D.−1
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question86
Let∧
aand∧
bbetwounitvectorssuchthattheanglebetweenthemis π
4.If
θistheanglebetweenthevectors ∧
a+∧
band ∧
a+2∧
b+2∧
a×∧
b,then
thevalueof164cos2θisequalto:
[29-Jul-2022-Shift-1]
Options:
A.90 +27√2
B.45 +18√2
C.90 +3√2
D.54 +90√2
Answer:A
Solution:
Solution:
( ) ( ( ) )
→
a=3
∧
i+∧
j&→
b=∧
i+2
∧
j+∧
k
→
a×→
(b×→
c) = →
a⋅→
c)b−→
(a⋅→
b→
c=→
b+→
λc
If→
b&→
carenon-parallel
then→
a⋅→
c=1&→
a⋅→
b= −λ
but→
a⋅→
b=5⇒λ= −5
( )
∧
a⋅∧
b=1
√2and ∧
a×∧
b=1
√2
∧
a+∧
b⋅∧
a+2
∧
b+2∧
a×∧
b
∧
a+∧
b∧
a+2
∧
b+2∧
a×∧
b
=cos θ
⇒cos θ =1+3∧
a
∧
b+2
∧
a+∧
b∧
a+2
∧
b+2∧
a×∧
b
∧
a+∧
b
2
=2+ √2
∧
a+2
∧
b+2∧
a×∧
b
2
=1+4+4∧
a×∧
b|2+4∧
a
∧
b
=5+4⋅1
2+4
√2=7+2√2
So,cos2θ=
3+3
√2
2
(2+ √2)(7+2√2)=9√2(5√2+3)
164
⇒164cos2θ=90 +27√2
| |
( ) ( ( ) )
| | | ( ) |
| | | ( ) |
| |
| ( ) | |
( )
-------------------------------------------------------------------------------------------------
Question87
If(2,3,9), (5,2,1), (1,λ,8)and(λ,2,3)arecoplanar,thentheproduct
ofallpossiblevaluesofλis:
[29-Jul-2022-Shift-2]
Options:
A. 21
2
B. 59
8
C. 57
8
D. 95
8
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question88
Let→
a,→
b,→
cbethreecoplanarconcurrentvectorssuchthatangles
betweenanytwoofthemissame.Iftheproductoftheirmagnitudesis
14and →
a×→
b⋅→
b×→
c+→
b×→
c⋅→
c×→
a+→
c×→
a⋅→
a×→
b=168,then
→
a+→
b+→
cisequalto:
[29-Jul-2022-Shift-2]
Options:
A.10
B.14
C.16
D.18
( ) ( ) ( ) ( ) ( ) ( )
| | | | | |
∵A(2,3,9), B(5,2,1), C(1,λ,8)andD(λ,2,3)arecoplanar.
∴────────▶
ABACAD =0
3−1−8
−1 λ −3−1
λ−2−1−6
=0
⇒[−6(λ−3) − 1] − 8(1− (λ−3)(λ−2)) + ( 6+ (λ−2) = 0
⇒3(−6λ +17) − 8(−λ2+5λ −5) + (λ+4) = 8
⇒8λ2−57λ +95 =0
∴λ1λ2=95
8
[ ]
| |
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question89
Let→
aand→
bbetwovectorssuchthat →
a+→
b
2=→
a|2+2→
b|2,→
a⋅→
b=3and
→
a×→
b
2=75.Then →
a2isequalto_______.
[29-Jul-2022-Shift-2]
Answer:14
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question90
If(1,5,35), (7,5,5), (1,λ,7)and(2λ,1,2)arecoplanar,thenthesum
ofallpossiblevaluesofλis
[26Feb2021Shift1]
| | | |
| | | |
→
a→
b→
c=14
→
a∧→
b=→
b∧→
c=→
c∧→
a=θ=2π
3
So,→
a⋅→
b= − 1
2ab,→
b⋅→
c= − 1
2bc,→
a⋅→
c. = − 1
2ac
(let)
→
a×→
b⋅→
b×→
c=→
a⋅→
b→
b⋅→
c−→
a⋅→
c→
b⋅→
b
=1
4ab2c+1
2ab2c=3
4ab2c
Similarly
→
b×→
c⋅→
c×→
a
=3
4abc2→
c×→
a⋅→
a×→
b
=3
4a2bc168
=3
4abc(a+b+c)
So,(a+b+c) = 16
| | | | | |
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
∵→
a+ ˙b|2=→
a|2+2 b|2
or →
a2+→
b|2+2→
a⋅→
b=→
a|2+2→
b|2
∴→
b|2=6
Now →
a×→
b2=→
a|2→
b|2−→
a⋅→
b2
75 =→
a|2⋅6−9
∴→
a|2=14
| | |
| | | | |
|
| | | | ( )
|
|
Options:
A. 39
5
B.−39
5
C. 44
5
D.−44
5
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question91
Ifaandbareperpendicular,thena × (a× (a× (a×b)))isequalto
[26Feb2021Shift1]
Options:
A.0
B. 1
2a|4b
C.a ×b
D.|a|4b
Answer:D
Solution:
|
LetP(1,5,35), Q(7,5,5), R(1,λ,7), S(2λ,1,2)
GivenP,Q,R,Sarecoplanar.Then,PQ,PR,PSlieonthesameplane.
PQ = (7−1)∧
i+ (5−5)∧
j+ (5−35)∧
k=6
∧
i−30
∧
k
PR = (1−1)∧
i+ (λ−5)∧
j+ (7−35)∧
k= (λ−5)∧
j−28
∧
k
PS = (2λ −1)∧
i+ (1−5)∧
j+ (2−35)∧
k= (2λ −1)∧
i−4
∧
j−33
∧
k
∵PQ,PRandPSlieonsameplane,then
6 0 −30
0λ−5−28
2λ −1−4−33
=0
Expandalongfirstrow,
6[−33(λ−5) − 112] + 30[(2λ −1)(λ−5)] = 0
⇒6(−33λ +53) + 30(2λ2−11λ +5) = 0
⇒60λ2−528λ +468 =0
⇒10λ2−88λ +78 =0
⇒5λ2−44λ +39 =0...(i)
PossiblevalueofλarerootsofEq.(i).
Then,sumofallpossiblevaluesofλ= SumofrootsofEq.(i)
=−(−44)
5=44
5
[ ∵ax2+bx++c=0,sumofroots = −b∕a}
| |
-------------------------------------------------------------------------------------------------
Question92
Letthreevectors→
a,→
band→
cbesuchthat→
ciscoplanarwith→
aand
→
b,→
a⋅→
c=7and→
bisperpendicularto→
c,where→
a= −∧
i+∧
j+∧
kand→
b=2∧
i+∧
k,
thenthevalueof2 →
a+→
b+→
c|2is
24Feb2021Shift1
Answer:75
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question93
Ifvectorsa1=x∧
i−∧
j+∧
kanda2=∧
i+∧
yj+z∧
karecollinear,thenapossible
unitvectorparalleltothevectorx∧
i+∧
j+z∧
kis
[26Feb2021Shift2]
Options:
A. 1
√2−∧
j+∧
k
|
( )
a× [a× {a× (a×b)}]
=a× (a× [(a⋅b)a− (a⋅a)b])
[U sing,a× (b×c) = (a⋅cb − (a⋅b)c]
=a× [a× ((a⋅b)a− | a|2b)]
=a× [(a× (a⋅b)a)− | a|2(a×b)]
=a× [0− | a|2(a×b)]
=−|a|2[a× (a×b)]
=−|a|2[(a⋅b)a− (a⋅a)b]
= −(a⋅b)a|a|2+ | a|4b
=0+ | a|4b
= | a|4b
Let →
c=λ→
b×→
a×→
b
=λ→
b⋅→
b→
a−→
b⋅→
a→
b
=λ 5 −∧
i+∧
j+∧
k+2
∧
i+∧
k
=λ−3
∧
i+5
∧
j+6
∧
k
→
c⋅→
a=7⇒3λ +5λ +6λ =7
λ=1
2
∴2−3
2−1+2
∧
i+5
2+1
∧
j+ (3+1+1)∧
k
2
=21
4+49
4+25 =25 +50 =75
( ( ) )
( ( ) ( ) )
( ( ) )
( )
|( ) ( ) |
( )
B. 1
√2
∧
i−∧
j
C. 1
√3
∧
i+∧
j−∧
k
D. 1
√3
∧
i−∧
j+∧
k
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question94
Leta =∧
i+α∧
j+3∧
kandb =3∧
i−α∧
j+∧
k.Iftheareaoftheparallelogram
whoseadjacentsidesarerepresentedbythevectorsaandbis8√3
squareunits,thena ⋅bisequalto.............
[25Feb2021Shift2]
Answer:2
Solution:
Solution:
( )
( )
( )
Given,a1=x
∧
i−∧
j+∧
kanda2=∧
i+y
∧
j+z
∧
karecollinear,then x
1=−1
y=1
z=λ(Say)
Thisgivesx=λ,y= − 1
λ,z=1
λ
Then,unitvectorparalleltovectorx
∧
i+y
∧
j+zkwillbe
= ± (λ)∧
i−1
λ
∧
j+1
λ
∧
k
(λ)2+−1
λ
2+1
λ
2
= ± λ2∧
i−∧
j+
∧
kλ
λλ4+2
= ± λ2∧
i−∧
j+
∧
k
λ4+2
Take,λ=1= ± ∧
i−∧
j+
∧
k
√3
(( ) ( ) )
√( ) ( )
()
√()
√
()
Areaofparallelogram= | a×b|
=∧
i+α
∧
j+3
∧
k×3
∧
i−α
∧
j+∧
k
(64)(3) = 16α2+64 +16α2(given,area =8√3)
=∧
i+α
∧
j+3
∧
k×3
∧
i−α
∧
j+∧
k
(64)(3) = 16α2+64 +16α2(given,area =8√3)
(squaringonbothsides)
⇒α2=4
Now,a.b = 3−α2+3
=6−α2=6−4=2
| ( ) ( ) |
| ( ) ( ) |
-------------------------------------------------------------------------------------------------
Question95
Leta =∧
i+2∧
j−∧
k,b=∧
i−∧
jandc =∧
i−∧
j−∧
kbethreegivenvectors.Ifrisa
vectorsuchthatr ×a=c×aandr ⋅b=0,thenr ⋅aisequalto..........
[25Feb2021Shift1]
Answer:12
Solution:
-------------------------------------------------------------------------------------------------
Question96
Avectorahascomponents3pand1withrespecttorectangular
cartesiansystem.Thissystemisrotatedthroughacertainangleabout
theorigininthecounterclockwisesense.If,withrespecttonew
system,ahascomponentsp +1and√10,thenavalueofpisequalto
[18Mar2021Shift1]
Options:
A.1
B.−5
4
C. 4
5
D.−1
Answer:D
b=∧
i−∧
j
c=∧
i−∧
j−∧
k
r×a=c×a
⇒r×a−c×a=0
⇒ (r−c) × a=0
∴r−c=λa
⇒r=λa +c
⇒r⋅b=λa ⋅b+c⋅b(takingdotw)
⇒0=λa ⋅b+c⋅b
⇒λ
∧
i+2
∧
j−∧
k⋅∧
i−∧
j+∧
i−∧
j−∧
k⋅∧
i−∧
j=0
⇒λ(1−2) + 2=0
⇒λ=2
∴r=2a +c
⇒r⋅a=2a ⋅a+c⋅a[takingdotwitha]
=2|a|2+a⋅c
=2(1+4+1) + (1−2+1)
r⋅a=12
( ) ( ) ( ) ( )
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question97
Letavectorα∧
i+β∧
jbeobtainedbyrotatingthevector√3
∧
i+∧
jbyanangle
45∘abouttheoriginincounterclockwisedirectioninthefirstquadrant.
Thentheareaoftrianglehavingvertices(α,β), (0,β)and(0,0)isequal
to
[16Mar2021Shift1]
Options:
A. 1
2
B.1
C. 1
√2
D.2√2
Answer:A
Solution:
Solution:
Aftercounterclockwise(oranti-clockwise)rotation,thelengthofthevectoraremainsconstant.
i.e.|a|atoldposition= | a|atnewposition
⇒ (3p)2+ (1)2= (p+1)2+ (√10)2
⇒9p2+1=p2+1+2p +10
⇒8p2−2p −10 =0⇒4p2−p−5=0
⇒ (p+1)(4p −5) = 0
⇒p=5
4, −1
LetOAbe√3
∧
i+∧
jandOBbeα
∧
i+β
∧
j.
As,wecannoticeinOA,1
√3=tan 30∘.So,itmakesanangleof30∘withtheX-axis.
Now,whenOAisrotatedfurtherby45∘anticlockwise,theresultantvectorOBmakesanangleof75∘withtheX-axis.
| ( )
-------------------------------------------------------------------------------------------------
Question98
Let→
xbeavectorintheplanecontainingvectors→
a=2∧
i−∧
j+∧
kand
→
b=∧
i+2∧
j−∧
k.Ifthevector→
xisperpendicularto 3∧
i+2∧
j−∧
kandits
projectionon→
ais 17√6
2,thenthevalueof →
x2isequalto
[17Mar2021Shift2]
Answer:486
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question99
( )
| |
So,OB= | OA cos 75∘∧
i+sin 75∘∧
j
Let△OBCbetherequiredtrianglewhoseareawehavetodetermine.
Areaof△OBC = (1/2) × (Base) × (Height)
=1/2×β×α
=1
2(2 sin 75∘)(2 cos 75∘) = 2 sin 75∘cos 75∘
=sin 150∘=sin 30∘
=1/2
Hence,theareais1/2sq.unit.
| ( )
Letx=λa +µb,whereλandµarescalars.
⇒x=λ 2
∧
i−∧
j+∧
k+µ
∧
i+2
∧
j−∧
k
x=∧
i(2λ +µ) + ∧
j(2µ −λ) + ∧
k(λ−µ)
Since,xisperpendicularto 3
∧
i+2
∧
j−∧
k.
Then,x⋅3
∧
i+2
∧
j−∧
k=0
⇒3λ +8µ =0
Also,givenprojectionofxonais 17√6
2.....(i)
∴x⋅a
|a|=17√6
2
⇒2(2λ +µ) + (λ−2µ) + (λ−µ) = 51
6λ −µ=51...(ii)
FromEqs.(i)and(ii),
λ=8,µ= −3
∴x=13
∧
i−14
∧
j+11
∧
k
⇒ | x= (13)2+ (−14)2+ (11)2
∴ | x| = (13)2+ (−14)2+ (11)2=486
( ) ( )
( )
( )
|√
Letaandbbetwonon-zerovectorsperpendiculartoeachotherand
|a|= |b|.If|a×b|= |a|,thentheanglebetweenthevectors
[a+b+ (a×b)]andaisequalto
[18Mar2021Shift2]
Options:
A.sin−11
√3
B.cos−11
√3
C.cos−11
√2
D.sin−11
√6
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question100
Leta =2∧
i−3∧
j+4∧
kandb =7∧
i+∧
j−6∧
k.Ifr ×a=r×b,r⋅∧
i+2∧
j+∧
k= −3,
then.r ⋅2∧
i−3∧
j+∧
kisequalto
[17Mar2021Shift1]
Options:
A.12
B.8
C.13
D.10
( )
( )
( )
( )
( )
( )
Given,a⟂b...(i)
|a|= |b|...(ii)
and|a×b|= |a|
⇒ | a||b|sin 90∘= | a| [fromEq.(i)]
⇒ | b| = 1= | a|...(iii) [fromEq.(ii)]
FromEq.(iii),wecansaythat
a×baremutuallyperpendicularunitvectors.
Let a=∧
iandb=∧
j
∴a×b=∧
k
Now, [a+b+ (a×b)] = ∧
i+∧
j+∧
k
∴cos θ =∧
i+∧
j+∧
k⋅
∧
i
√3√1=1
√3
∴θ=cos−11
√3
( )
()
( )
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question101
Leta =∧
i+2∧
j−3∧
kandb =2∧
i−3∧
j+5∧
k.Ifr ×a=b×r,r⋅α∧
i+2∧
j+∧
k=3
andr. 2∧
i+5∧
j−a∧
k= −1,α∈R,thenthevalueofα+ | r|2isequalto
[16Mar2021Shift2]
Options:
A.9
B.15
C.13
D.11
Answer:B
Solution:
Solution:
( )
( )
a=2
∧
i−3
∧
j+4
∧
k
b=7
∧
i+∧
j−6
∧
k
Ifr×a=r×b
⇒r=λ(a−b) = λ 5
∧
i+4
∧
j−10k
Now,r⋅∧
i+2
∧
j+∧
k= −3
⇒λ 5
∧
i+4
∧
j+10
∧
k⋅∧
i+2
∧
j+∧
k= −3
⇒λ(5+8+10) = −3
⇒λ= +1
∴r= −5
∧
i−4
∧
j+10
∧
k
So,r⋅2
∧
i−3
∧
j+∧
k
= −5
∧
i−4
∧
j+10
∧
k2
∧
i−3
∧
j+∧
k
= −10 +12 +10
=12
( )
( )
( ) ( )
( )
( ) ( )
Given,a=∧
i+2
∧
j−3
∧
k
andb=2
∧
i−3
∧
j+5
∧
k
Ifr×a=b×r,r⋅α
∧
i+2
∧
j+∧
k=3
r⋅2
∧
i+5
∧
j−α
∧
k= −1
⇒r×a=b×r
(r×a) = −(r×b) ⇒ (r×a) + (r×b) = 0
⇒r× (a+b) = 0⇒r=λ(a+b)
⇒r=λ(1+2)∧
i+ (2−3)∧
j+ (−3+5)∧
k
⇒r=λ 3
∧
i−∧
j+2
∧
k
r⋅α
∧
i+2
∧
j+∧
k=3
( )
( )
( )
( )
[ ]
( )
( )
( ) ( )
-------------------------------------------------------------------------------------------------
Question102
Let→
cbeavectorperpendiculartothevectors→
a=∧
i+∧
j−∧
kand
→
b=∧
i+2∧
j+∧
kIf→
c⋅∧
i+∧
j+3∧
k=8thenthevalueof→
c⋅→
a×→
bisequalto
..........
[16Mar2021Shift2]
Answer:28
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question103
( ) ( )
⇒λ 3
∧
i−∧
j+2
∧
kα
∧
i+2
∧
j+∧
k=3
⇒λ(3α −2+2) = 3
⇒αλ =1
r⋅2
∧
i+5
∧
j−α
∧
k= −1
⇒λ 3
∧
i−∧
j+2
∧
k2
∧
i+5
∧
j−α
∧
k= −1
⇒λ(6−5−2α) = −1⇒λ(1−2α) = −1
⇒λ−2αλ = −1⇒λ−2= −1
⇒λ=1
So,α=1
r=3
∧
i−∧
j+2
∧
k
⇒ | r|2=9+1+4=14
∴α+ | r|2=1+14 =15
( ) ( )
( )
( ) ( )
( )
Since,cisperpendiculartoaandb.
So,c=λ(a×b)
a=∧
i+∧
j−∧
k
b=∧
i+2
∧
j+∧
k
Now,a×b=
∧
i
∧
j
∧
k
11−1
1 2 1
= (1+2)∧
i− (1+1)∧
j+ (2−1)∧
k
=3
∧
i−2
∧
j+∧
k
⇒c=λ 3
∧
i−2
∧
j+∧
k
Now,c⋅ ∧
i+∧
j+3
∧
k=8
λ 3
∧
i−2
∧
j+∧
k
∧
i+∧
j+3
∧
k=8
⇒4λ =8
∴λ=2
So,c=2 3
∧
i−2
∧
j+∧
k
So,c⋅ (a×b) = 2 3
∧
i−2
∧
j+∧
k3
∧
i−2
∧
j+∧
k
=2(9+4+1)
=28
| |
( )
( )
( ) ( )
( )
( ) ( )
Ifa =α∧
j+β∧
j+3∧
k,b= −β∧
i−α∧
j−∧
kandc =∧
i−2∧
j−∧
ksuchthata ⋅b=1and
b.c= −3,then 1
3[(a×b) ⋅ c]isequalto
[17Mar2021Shift1]
Answer:2
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question104
LetObetheorigin.LetOP =x∧
i+y∧
j−∧
kand
OQ = −∧
i+2∧
j+3x∧
k,x,y∈R,x>0,besuchthat|PQ| = √20andthevector
OPisperpendiculartoOQ.IfOR =3∧
i+2∧
j−7∧
k,z ∈R,iscoplanarwith
OPandOQ,thenthevalueofx2+y2+z2isequalto
[17Mar2021Shift2]
Options:
A.7
B.9
C.2
a= ⟨α,β,3⟩
b= ⟨−β, −α, −1⟩
c= ⟨1, −2, −1⟩
a⋅b=1
−αβ −αβ −3=1
αβ = −2
b⋅c= −3
−β+2α +1= −3
β−2α =4
⇒β−2−2
β=4
⇒β2+4=4β ⇒β2−4β +4=0
⇒ (β−2)2=0⇒β=2
αβ = −2⇒α⋅2= −2
⇒α= −1
Hence, 1
3[(a×b) ⋅ c]
a= ⟨−1,2,3⟩
b= ⟨−2,1, −1⟩
c= ⟨1, −2, −1⟩
(a×b) =
∧
i
∧
j
∧
k
−12 3
−21−1
= −5
∧
i−7
∧
j+3
∧
k
(a×b) ⋅ c= −5
∧
i−7
∧
j+3
∧
k
∧
i−2
∧
j−∧
k
= −5+14 −3=6
∴1
3[(a×b) ⋅ c] = 1
3×6=2
( )
| |
( ) ( )
D.1
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question105
If →
a+3→
bisperpendicularto 7→
a−5→
band →
a−4→
bisperpendicularto
7→
a−2→
b,thentheanglebetweenaand→
b(indegrees)is________.
[25Jul2021Shift2]
Answer:60
Solution:
( ) ( ) ( )
( )
Given,OP =x
∧
i+y
∧
j−∧
k
OQ = −∧
i+2
∧
j+3x
∧
k
OR =3
∧
i+z
∧
j−7
∧
k
and | PQ | = √20
Now,|PQ|= |OQ −OP |=|OP −OQ|
= (x+1)∧
i+ (y−2)∧
j− (1+3x)∧
k
⇒PQ |2= (√20)2=20
⇒ (x+1)2+ (y−2)2+ (1+3x)2=20
⇒(x+1)2+ (2x −2)2+ (1+3x)2=20
∵OP⟂OQ
∴OP ⋅OQ =0
⇒−x+2y −3x =0
⇒y=2x
⇒x2+1+2x +4x2+4−8x +1+9x2+6x =20
⇒14x2+6=20 ⇒14x2=14
⇒x2=1⇒x= ±1butxmustbepositiveasinquestionconditionsi.e.x>0.
∴x= −1 (Rejected)
Hence,x=1
∴y=2x =2×1=2
Now,OP,OQandORarecoplanar.
∴ [OP OQ OR] = 0
⇒
x y −1
−12 3x
3 z −7
=0⇒
1 2 −1
−12 3
3 z −7
=0
⇒1(−14 −3z) − 2(7−9) − 1(−z−6) = 0⇒z= −2
∴x2+y2+z2=1+4+4=9
[ ]
| | | |
→
a+3→
b⊥7→
a−5→
b
→
a+3→
b.7→
a−5→
b=0
7→
a2−15 →
b2+16→
a.→
b=0......(1)
→
a−4→
b.7→
a−2→
b=0
( ) ( )
( ) ( )
| | | |
( ) ( )
-------------------------------------------------------------------------------------------------
Question106
Let→
a,→
b,→
cbethreemutuallyperpendicularvectorsofthesame
magnitudeandequallyinclinedatanangleθ,withthevector→
a+→
b+→
c.
Then36cos22θisequalto_______.
[20Jul2021Shift1]
Answer:4
Solution:
-------------------------------------------------------------------------------------------------
Question107
Forp >0,avector→
v2=2^
i+ (p+1)^
jisobtainedbyrotatingthevector
→
v1= √3p^
i+^
jbyanangleθaboutoriginincounterclockwisedirection.
Iftan θ =(α√3−2)
(4√3+3),thenthevalueofαisequalto_________.
[20Jul2021Shift2]
Answer:6
Solution:
7→
a2+8→
b2−30→
a.→
b=0......(2)
from(1)&(2)
→
a=→
b
cos θ =
→
b
2→
a
∴θ=60°
| | | |
| | | |
| |
| |
→
a+→
b+→
c2=→
a2+→
b2+→
c2+2→
a.→
b+→
a.→
c+→
b.→
c=3
⇒→
a+→
b+→
c= √3
→
a→
a+→
b+→
c=→
a+→
a+→
b+→
ccos θ
⇒1= √3 cos θ
⇒cos 2 θ = − 1
3
⇒36cos22θ =4
| | | | | | | | ( )
| |
( ) | | |
-------------------------------------------------------------------------------------------------
Question108
InatriangleABC,if →
BC =3,→
CA =5and →
BA =7,thentheprojectionof
thevector →
BAon →
BCisequalto
[20Jul2021Shift2]
Options:
A.19
2
B.13
2
C.11
2
D.15
2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
| | | | | |
Projectionof →
BAon →
BCisequalto
=B→
A cos ∠ABC
=772+32−52
2×7×3=11
2
| |
| |
Question109
Let→
p=2^
i+3^
j+^
kand→
q=^
i+2^
j+^
kbetwovectors.
Ifavector→
r=α^
i+β^
j+γ^
kisperpendiculartoeachofthevectors
→
p+→
qand →
p−→
q,and →
r= √3,then|α|+ | β|+|γ|isequalto_______.
[25Jul2021Shift1]
Answer:3
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question110
If →
a=2,→
b=5and →
a×→
b=8,then →
a.→
bisequalto:
[25Jul2021Shift2]
Options:
A.6
B.4
C.3
D.5
Answer:A
Solution:
( )
( ) ( ) | |
| | | | | | | |
→
p=2^
i+3^
j+^
k(Given)
→
q=^
i+2^
j+^
k
Now →
p+→
q× →
p−→
q=
^
i^
j^
k
351
110
= −2^
i−2^
j−2^
k
→
r= ±√3
→
p+→
q×→
p−→
q
→
p+→
q×→
p−→
q
±√3−2^
i−2^
j−2^
k
22+22+22
→
r= ± −^
i−^
j−^
k
Accordingtoquestion
→
r=α^
i+β^
j+γ^
k
So|α| = 1, | β| = 1, | γ| = 1
⇒ | α|+|β|+|γ| = 3
( ) ( ) | |
( ( ) ( ) )
| ( ) ( ) |
( )
√
( )
-------------------------------------------------------------------------------------------------
Question111
Let→
a=^
i+^
j+2^
kand→
b= −^
i+2^
j+3^
k.Thenthevectorproduct
→
a+→
b×→
a×→
a−→
b×→
b×→
bisequalto:
[27Jul2021Shift1]
Options:
A.5 34^
i−5^
j+3^
k
B.7 34^
i−5^
j+3^
k
C.7 30^
i−5^
j+7^
k
D.5 30^
i−5^
j+7^
k
Answer:B
Solution:
Solution:
( ) ( ( ( ( ) ) ) )
( )
( )
( )
( )
→
a=2,→
b=5
→
a×→
b=→
a→
b sin θ = ±8
sin θ = ±4
5∴→
a.bv =→
a→
b cos θ
=10 . ±3
5= ±6
→
a.→
b=6
| | | |
| | | | | |
| | | |
( )
| |
→
a=^
i+^
j+2^
k
→
b= −^
i+2^
j+3^
k.
→
a+→
b=3^
j+5^
k;→
a.→
b= −1+2+6=7
→
a×→
a−→
b×→
b×→
b
→
a×→
a×→
b−→
b×→
b×→
b
→
a×→
a×→
b−0×→
b
→
a×→
a×→
b×→
b
→
a.→
b→
a−→
a.→
a→
b×→
b
→
a.→
b→
a×→
b−→
a.→
a→
b×→
b
→
a.→
b→
a×→
b
→
a×→
b=
^
i^
j^
k
1 1 2
−12 3
= −^
i−5^
j+3^
k
∴7−^
i−5^
j+3^
k
→
a+→
b×7−^
i−5^
j+3^
k
7 0^
i+3^
j+5^
k× −^
i−5^
j+3^
k
^
i^
j^
k
0 3 5
−1−53
( ( ( ( ) ) ) )
( ( ( ) ) )
( ( ) )
( ( ) )
( ( ) ( ) )
( ) ( ) ( )
( ) ( )
| |
( )
( ) ( ( ) )
( ) ( )
| |
( )
-------------------------------------------------------------------------------------------------
Question112
Let→
a=^
i+^
j+^
k,→
band→
c=^
j−^
kbethreevectorssuchthat→
a×→
b=→
cand
→
a.→
b=1.Ifthelengthofprojectionvectorofthevector→
bonthevector
→
a×→
cisl ,thenthevalueof3l 2isequalto_______.
[27Jul2021Shift1]
Answer:2
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question113
Let→
a,→
band→
cbethreevectorssuchthat→
a=→
b×→
b×→
c.Ifmagnitudesof
thevectors→
a,→
band→
care√2,1and2respectivelyandtheanglebetween
→
band→
cisθ 0 <θ<π
2,thenthevalueof1 +tan θisequalto:
[27Jul2021Shift2]
Options:
A.√3+1
B.2
C.1
D.√3+1
√3
Answer:B
( )
( )
⇒34^
i− (5)^
j+3^
k
⇒34^
i−5^
j+3^
k
∴7 34^
i−5^
j+3^
k
( )
( )
→
a×→
b=→
c
TakeDotwith→
c
→
a×→
b→
c=→
c|2=2
Projectionof→
bor→
a×→
c=l
→
b.→
a×→
c
→
a×→
c
=l
∴l=2
√6⇒l2=4
6
3l2=2
( ) |
|( ) |
| |
Solution:
-------------------------------------------------------------------------------------------------
Question114
Letthevectors(2+a+b)^
i+ (a+2b +c)^
j− (b+c)^
k,(1+b)^
i+2b −b^
kand
(2+b)^
i+2b^
j+ (1−b)^
k a,b,c, ∈Rbeco-planar.Thenwhichofthe
followingistrue?
[25Jul2021Shift1]
Options:
A.2b =a+c
B.3c =a+b
C.a =b+2c
D.2a =b+c
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
→
a=→
b.→
c→
b−→
b.→
b→
c
=1.2 cos θ →
b−→
c
⇒→
a=2 cos θ →
b−→
c
→
a2= (2 cos θ)2+22−2.2 cos θ →
b.→
c
⇒2=4cos2θ+4−4 cos θ .2 cos θ
⇒−2= −4cos2θ
⇒cos2θ=1
2
⇒sec2θ=2
⇒tan2θ=1
⇒θ=π
4
1+tan θ =2.
( ) ( )
| |
Ifthevectorsareco-planar,
a+b+2 a +2b +c−b−c
b+12b −b
b+22b 1−b
=0
NowR3→R3−R2,R1→R1−R2
So
a+1 a +c−c
b+12b −b
1 0 1
=0
= (a+1)2b − (a+c)(2b +1) − c(−2b)
=2ab +2b −2ab −a−2bc −c+2bc
=2b −a−c=0
| |
| |
Question115
Letthreevectors→
a,→
band→
cbesuchthat→
a×→
b=→
c,→
b×→
c=→
aand →
a=2.
Thenwhichoneofthefollowingisnottrue?
[22Jul2021Shift2]
Options:
A.→
a×→
b+→
c×→
b−→
c=→
0
B.Projectionof→
aon →
b×→
cis2
C. →
a→
b c +→
c→
a→
b=8
D. 3→
a+→
b−2→
c
2=51
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question116
Letavector→
abecoplanarwithvectors→
b=2^
i+^
j+^
kand→
c=^
i−^
j+^
k.If→
a
isperpendicularto→
d=3^
i+2^
j+6^
k,and →
a= √10.Thenapossiblevalue
of →
a→
b→
c+→
a→
b→
d+→
a→
c→
dequalto:
[22Jul2021Shift2]
Options:
A.−42
| |
( ( ) ( ) )
( )
[ ] [ ]
| |
| |
[ ] [ ] [ ]
(1)→
a×→
b+→
c×→
b−→
c=→
0
=→
a−→
b×→
c+→
c×→
b = −2→
a×→
b×→
c
= −2→
a×→
a→
0
(2)Projectionof→
aon →
b×→
c
=
→
a.→
b×→
c
→
b×→
c
=
→
a.→
a
→
a
=→
a=2
(3) →
a→
b c +→
c→
a→
b = 2→
a→
b c =2→
a.→
b×→
c
=2→
a.→
a=2→
a|2=8
(4)→
a×→
b=→
cand→
b×→
c=→
a
⇒→
a,→
b,→
caremutuallyperpvectors.
∴→
a×→
b=→
c⇒→
a→
b=→
c⇒ →
b=
→
c
2
Also, →
b×→
c=a⇒→
b→
c=2⇒→
c=2&→
b=1
→
a+→
b−2→
c2=3→
a+→
b−2→
c. 3→
a+→
b−2→
c
=9→
a|2+→
b|2+4→
c|2
= (9×4) + 1+ (4×4)
=36 +1+16 =53
( ( ) ( ) )
( ) ( ( ) )
( )
( )
( )
| | | | | |
[ ] [ ] [ ] ( )
|
| | | | | | | | | | | || |
| | | | | | | | | | | |
| | ( ) ( )
|| |
B.−40
C.−29
D.−38
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question117
Let→
a=2^
i+^
j−2^
kand→
b=^
i+^
j.If→
cisavectorsuchthat
→
a.→
c=→
c,→
c−→
a=2√2andtheanglebetween →
a×→
band→
cisπ
6,thenthe
valueof →
a×→
b×→
cis
[20Jul2021Shift1]
Options:
A.2
3
B.4
C.3
D.3
2
Answer:D
Solution:
Solution:
| | | | ( )
| ( ) |
→
a=λ→
b+µ→
c=^
i(2λ +µ) + ^
j(λ−µ) + ^
k(λ+µ)
→
a.→
d=0=3(2λ +µ) + 2(λ−µ) + 6(λ+µ)
⇒14λ +7µ =0⇒µ= −2λ
⇒→
a= (0)^
i−3λ^
j+ (−λ)^
k
⇒→
a= √10 λ= √10⇒λ=1
⇒λ=1or−1
→
a→
b→
c=0
→
a→
b→
c+→
a→
b→
d+→
a→
c→
d = →
a→
b+→
c→
d
=
0−3λ λ
302
326
=3λ(12) + λ(6) = 42λ = −42
| | | | | |
[ ]
[ ] [ ] [ ] [ ]
| |
→
a=3=a;→
a.→
c=c
Now →
c−→
a=2√2
⇒c2+a2−2→
c.→
a=8
⇒c2+9−2(c) = 8
⇒c2−2c +1=0⇒c=1=→
c
| |
| |
| |
-------------------------------------------------------------------------------------------------
Question118
Letaandbbetwovectorssuchthat|2a+3b|=|3a+b|andtheangle
betweenaandbis60°.If1
8aisaunitvector,then|b|isequalto
[31Aug2021Shift1]
Options:
A.4
B.6
C.5
D.8
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question119
Leta,bandcbethreevectorsmutuallyperpendiculartoeachotherand
havesamemagnitude.Ifavectorrsatisfies.
a×{(r-b)×a}+b×{(r-c)×b}+c×{(r-a)×c}=0,thenrisequel
to
[31Aug2021Shift2]
Options:
A.1
3(a+b+c)
Also,→
a×→
b=2^
i−2^
j+^
k
Given →
a×→
b=→
a×→
b→
c sin π
6
= (3)(1)(1∕2)
=3∕2
( ) | | | |
|2a+3b|=|3a+b|
⇒ | 2a +3b |2= | 3a +b|2
⇒4|a|2+.9b |2+12a .b = 9|a|2+ | b|2+6a .b
⇒5|a|2−6a .b−8|b|2=0
a
8isaunitvector.
So,|a|=8
And,
5.64 −6.8|b|1
2−8|b|2=0
⇒ | b|2+3|b| −40 =0
(|b| + 8)(|b| − 5) = 0
|b|=5
As,|b|=−8Notpossible.
( )
B.1
3(2a+b-c)
C.1
2(a+b+c)
D.1
2(a+b+2c)
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question120
Ahallhasasquarefloorofdimension10m ×10m(seethefigure)and
verticalwalls.Ifthe∠GPHbetweenthediagonalsAGandBHiscos−11
5,
thentheheightofthehall(inm)is
[26Aug2021Shift2]
Options:
A.5
B.2√10
C.5√3
D.5√2
Answer:D
Solution:
a×[(r-b)×a]+b×[(r-c)×b]+c×[(r-a)×c]=0
⇒a.a(r-b)-(a.(r-b))a+b.b(r-c)-(b.(r-c))b+c.c(r-a)-(c.(r-a))c=0
⇒ | a|2(r−b) − (r.a)a+ | b|2(r−c) − (r.b)b+ | c|2(r−a) − (r.c)c=0[∵ab,caremutuallyperpendicular;∴a.b=
b.c=c.a=0]
⇒ | a|2[3r − (a+b+c)] − [(r.a)a+ (r.b)b+ (r.c)c] = 0[∵|a|=|b|=|c|]
⇒ | a|2[3r − (a+b+c) − r] = 0
∴3r-(a+b+c)-r=0
⇒r=a+b+c
2
Solution:
-------------------------------------------------------------------------------------------------
Question121
Iftheprojectionofthevector^
i+2^
j+^
konthesumofthetwovectors
2^
i+4^
j−5^
kand−λ^
i+2^
j+3^
kis1,thenλisequalto
[26Aug2021Shift2]
Answer:5
Solution:
-------------------------------------------------------------------------------------------------
Question122
Leta =^
i+5^
j+α^
k,b=^
i+3^
j+β^
kandc = −^
i+2^
j−3^
kbethreevectors
suchthat,|b×c| = 5√3andaisperpendiculartob.Then,thegreatest
amongstthevaluesof|a|2is
[27Aug2021Shift1]
LetA= (0,0,0)
B= (10,0,0)
G= (10,10,h)
H= (0,10,h)
AG =10^
i+10^
j+h^
k
BH = −10^
i+10^
j+h^
k
Since,AG .BH = | AG||BH |cos θ
(−100 +100 +h2) = h2+200 .h2+200 .1
5
⇒5h2=h2+200
⇒h2=50
⇒h=5√2
√ √ ( )
Leta = ^
i+2^
j+^
k
b = 2^
i+4^
j−5^
k
c= −λ^
i+2^
j+3^
k
Now,accordingtothequestions,
a. (b+c)
|b+c|=1
⇒a.b+a.c= | b+c|
⇒5+ (−λ+4+3) = (2−λ)^
i+6^
j−2^
k
⇒(12 −λ)2= (2−λ)2+36 +4
⇒λ2+144 −24λ =λ2−4λ +4+40
⇒λ=5
| |
Answer:90
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question123
If 50
Σ
r=1tan−11
2r2=pthenthevalueoftan pis
[26Aug2021Shift2]
Options:
A.101
102
B.50
51
C.100
D.51
50
Answer:B
Solution:
Given,a=^
i+5^
j+α^
k
b=^
i+3^
j+β^
k
andc= −^
i+2^
j−3^
k
∵a⊥b
⇒a.b=0
⇒^
i+5^
j+α^
k.^
i+3^
j+β^
k=0
⇒1+15 +αβ =0
orαβ = −16...(i)
Now,b×c=
^
i^
j^
k
1 3 β
−12−3
=^
i(−9−2β) − ^
j(−3+β) + ^
k(2+3)
b×c=^
i(−9−2β) + ^
j(3−β) + 5^
k
Given,|b×c| = 5√3
⇒ | b×c|2=75
⇒(−9−2β)2+ (3−β)2+25 =75
⇒β2+6β +8=0
⇒β= −2, −4
FromEq.(i),weget
Forβ= −2,α=8
Forβ= −4,α=4
Formaximumvalueof|a|2,α=8
∴ | α|2=1+25 +64 =90
( ) ( )
| |
-------------------------------------------------------------------------------------------------
Question124
Leta =^
i+^
j+^
kandb =^
j−^
k.Ifcisavectorsuchthata ×c=band
a.c=3,thena . (b×c)isequalto
[26Aug2021Shift1]
Options:
A.−2
B.−6
C.6
D.2
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question125
Given, 50
Σ
r=1
tan−11
2r2=p
Now,Σtan−12
1+4r2−1
=Σtan−1(2r +1) − (2r −1)
1+ (2r +1)(2r −1)
=Σ[tan−1(2r +1) − tan−1(2r −1)]
= (tan−13−tan−11) + (tan−15−tan−13) + ........ + tan−1101 −tan−199
=tan−1(101) − tan−11
=tan−1101 −1
1+101 =tan−150
51
∴tan−150
51 =p
⇒tan p =50
51
( )
[ ]
( ) ( )
Given,a×c=b
a× (a×c) = a×b
(a.c)a− (a.a)c=a×b
Wehave,a= (1,1,1), b= (0,1, −1), a.c=3
a×b=
^
i^
j^
k
1 1 1
0 1 −1
= −2^
i+^
j+^
k
So,3a −3c = −2^
i+^
j+^
k
⇒3^
i+3^
j+3^
k−3c = −2^
i+^
j+^
k
⇒3c =5^
i+2^
j+2^
k
Now,a. (b×c) = 1
3
1 1 1
01−1
5 2 2
=1
3(4−5−5) = −2
| |
( )
( ) ( )
( )
( ) | |
LetP1,P2, ........., P15be15pointsonacircle.Thenumberofdistinct
trianglesformedbypointsP^
i
,P^
j
andP^
k′suchthat^
i+^
j+^
k≠15is
[1Sep2021Shift2]
Options:
A.12
B.419
C.443
D.455
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question126
Avector→
a=α^
i+2^
j+β^
k(α,β∈R)liesintheplaneofthevectors,
→
b=^
i+^
jand→
c=^
i−^
j+4^
k.If→
abisectstheanglebetween→
band→
c,then:
[Jan.7,2020(I)]
Options:
A.→
a.^
i+3=0
B.→
a.^
i+1=0
C.→
a.^
k+2=0
D.→
a.^
k+4=0
Answer:C
^
i+^
j+^
k=15
where,^
i=1,^
j+^
k=14
⇒ ^
j=2,^
k=12 ,^
j=3,^
k=11 ,^
j=4,^
k=10,
^
j=5,^
k=9^
j=6,^
k=8...5ways
^
i=2,^
j+^
k=13
⇒ ^
j=3,^
k=10 , ..., ^
j=6,^
k=7...4ways
^
i=3,^
j+^
k=12
⇒ ^
j=4,^
k=8,^
j=5,^
k=7...2ways
^
i=4,^
j+^
k=11
⇒^
j=5,^
k=6...1way
∴Total=12ways
Then,numberofpossibletrianglesusingvertices
P^
i,P^
i,P^
ksuchthat^
i+^
j+^
k≠15is
15C3−12 =455 −12 =443
( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( )
( )
Solution:
-------------------------------------------------------------------------------------------------
Question127
Let→
a,→
band→
cbethreeunitvectorssuch
that→
a+→
b+→
c=→
0if
λ=→
a.→
b+→
b.→
c+→
c.→
aand
→
d=→
a×→
b+→
b×→
c+→
c×→
a,then
theorderedpair, λ,→
disequalto:
[Jan.7,2020(II)]
Options:
A. 3
2,3→
a×→
c
B. −3
2,3→
c×→
b
C. 3
2,3→
b×→
c
D. −3
2,3→
a×→
b
Answer:D
( )
( )
( )
( )
( )
Anglebisectorbetween→
band→
ccanbe
→
a=λ^
b+^
cor→
a=µ^
b−^
c
If→
a=λ
^
i+^
j
√2+
^
i−^
j+4^
k
3√2
=λ
3√23^
i+3^
j+^
i−^
j+4^
k
=λ
3√24^
i+2^
j+4^
k
Comparewith→
a=α^
i+2^
j+β^
k
2λ
3√2=2⇒λ=3√2
→
a=4^
i+2^
j+4^
k
Notsatisfyanyoption
Nowconsidera=µ
^
i+^
j
√2−
^
i−^
j+4^
k
3√2
→
a=µ
3√23^
i+3^
j−^
i+^
j−4^
k
=µ
3√22^
i+4^
j−4^
k
Comparewith→
a=α^
i+2^
j+β^
k
4µ
3√2=2⇒µ=3√2
2
→
a=^
i+2^
j−2^
k
∴→
a.→
k+2=^
i+2^
j−2^
k.^
k+2
= −2+2=0
( ) ( )
( )
[ ]
[ ]
( )
( )
( )
( )
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question128
Letthevolumeofaparallelopipedwhosecoterminousedgesaregiven
by→
u=^
i+^
j+λ^
k,→
v=^
i+^
j+3^
kand→
w=2^
i+^
j+^
kbe1cu.unit.Ifθbethe
anglebetweentheedges→
uand→
w,thencos θcanbe:
[Jan.8,2020(I)]
Options:
A. 7
6√6
B. 7
6√3
C.5
7
D. 5
3√3
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question129
Iet→
a=i−2j +kand→
b=i−j+khetwovectors.If→
cisavectorsuchthat
→
a+→
b+→
c2=0
3+2→
a.→
b+→
b.→
c+→
c.→
a = 0
→
a.→
b+→
b.→
c+→
c.→
a = −3
2⇒λ=−3
2
→
d=→
a×→
b+→
b× −→
a−→
b+ −→
a−→
b×a∵c= −→
a−→
b
=→
a×→
b+→
a×→
b+→
a×→
b
d=3→
a×→
b
| |
( )
( )
( ) ( ) [ ]
( )
Itisgiventhat→
u=^
i+^
j+λ^
k,→
v=^
i+^
j+3^
kand→
w=2^
i+^
j+^
k
Volumeofd=→
u.→
v.→
w
⇒±1=
1 1 λ
113
211
⇒−λ+3= ±1⇒λ=2orλ=4
Forλ=2
cos θ =2+1+2
√6√6=5
6
Forλ=4
cos θ =2+1+4
√6√18 =7
6√3
[ ]
| |
→
b×→
c=→
b×→
aand→
c.→
a=0,then→
c.→
bisequalto:
[Jan.8,2020(II)]
Options:
A.−3
2
B.1
2
C.−1
2
D.-1
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question130
Theprojectionofthelinesegmentjoiningthepoints(1,-1,3)and
(2,-4,11)onthelinejoiningthepoints(-1,2,3)and(3,-2,10)is_______.
[NAJan.9,2020(I)]
Answer:8
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question131
→
a×→
b×→
c=→
a×→
b×→
a
−→
a.→
b→
c=→
a.→
a→
b−→
a.→
b→
a
⇒−4→
c=6^
i−^
j+^
k−4^
i−2^
j−^
k
⇒−4→
c=2^
i−2^
j+2^
k
⇒→
c=1
2
^
i+^
j+^
k
⇒→
b.→
c= −1
2
( ) ( )
( ) ( ) ( )
( ) ( )
( )
LetP(1, −1,3), Q(2, −4,11), R(−1,2,3)andS(3, −2,10)
Then,PQ =^
i−3^
j+8^
k
ProjectionofPQonRS
=PQ .RS
|RS|=4+12 +56
(4)2+ (4)2+ (7)2=8
√
Let→
a,→
band→
cbethreevectorssuchthat →
a= √3,→
b=5,→
b.→
c=10and
theanglebetween→
band→
cisπ
3.If→
bisperpendiculartothevector
→
b×→
c,then →
a×→
b×→
cisequalto_______.
[NAJan.9,2020(II)]
Answer:30
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question132
Letthevectors→
a,→
b,→
cbesuchthat →
a=2,→
b=4and →
c=4.Ifthe
projectionof→
bon→
aisequaltotheprojectionof→
con→
aand→
bis
perpendicularto→
c,thenthevalueof →
a+→
b−→
cis________.
[NASep.05,2020(II)]
Answer:6
Solution:
-------------------------------------------------------------------------------------------------
| | | |
| ( ) |
| | | | | |
| |
→
b.→
c=10 ⇒→
b→
c cos π
3=10
⇒5.→
c.1
2=10 ⇒→
c=4
Since,isperpendiculartothevector→
b×→
c,then
→
a.→
b×→
c=0
Now, →
a×→
b×→
c= →
a→
b×→
c sin π
2
= √3×→
b→
c sin π
3×1
Hence, →
a×→
b×→
c=30
| | | | ( )
| | | |
( )
| ( ) | | | || ( )
| | | |
| ( ) |
∵Projectionof→
bon→
a= Projectionof→
con→
a
∴→
a.→
b=→
a.→
c
Given,→
b.→
c=0
∵→
a+→
b−→
c|2=→
a|2+→
b|2+→
c|2+2→
a.→
b−2→
b.→
c−2→
a.→
c
=4+16 +16 =36
⇒→
a+→
b−→
c|2=6
| | | |
|
Question133
Ifthevectors,→
p= (a+1)^
i+a^
j+a^
k,→
q=a^
i+ (a+1)^
j+a^
kand
→
r=a^
i+a^
j+ (a+1)^
k(a∈R)arecoplanarand3 →
p.→
q2−λ→
r×→
q|2=0,
thenthevalueofλis_______.
[NAJan.9,2020(I)]
Answer:1
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question134
Leta,b,c∈Rbesuchthata2+b2+c2=1.
Ifa cos θ −b cos θ +2π
3=c cos θ +4π
3,whereθ =π
9thentheangle
betweenthevectorsa^
i+b^
j+c^
kandb^
i+c^
j+a^
kis:
[Sep.03,2020(II)]
Options:
A.π
2
( ) |
( ) ( )
a+1 a a
a a +1 a
a a a +1
=0
⇒3a +1=0⇒a= −1
3
Thegivenvectors
→
p=2
3
^
i−1
3
^
j−1
3
^
k = 1
32^
i−^
j−^
k
→
q=1
3−^
i+2^
j−^
k
→
r=1
3−^
i−^
j+2^
k
Now,→
p.→
q=1
9(−2−2+1) = −1
3
→
r×→
q=1
9
^
i^
j^
k
−1 2 −1
−1−1 2
=1
9(i(4−1) − j(−2−1) + k(1+2))
=1
9(3i +3j +3k) = i+j+k
3
→
r×→
q=1
3√3⇒→
r×→
q|2 = 1
3
3→
p.→
q2−λ→
r×→
q|2=0
⇒3.1
9−λ.1
3=0⇒λ=1
| |
( )
( )
( )
| |
| | |
( ) |
B.2π
3
C.π
9
D.0
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question135
Letthepositionvectorsofpoints'A'and'B'be^
i+^
j+^
kand2^
i+^
j+3^
k,
respectively.Apoint'P'dividesthelinesegmentABinternallyinthe
ratiolambda :1(λ>0).IfOistheregionand →
OB .→
OP −3→
OA ×→
OP |2=6,
thenλisequalto______.
[NASep.02,2020(II)]
Answer:0.8
Solution:
|
a cos θ =b cos θ +2π
3 = c cos θ +4π
3=k
a=k
cos θ,b=k
cos θ +2π
3
,c=k
cos θ +4π
3
ab +bc +ca =k2
cos θ +4π
3+cos θ +cos θ +2π
3
cos θ +4π
3.cos θ .cos θ +2π
3
=k2cos θ +2 cos(θ+π) . cos π
3
cos θ .cos θ +2π
3.cos θ +4π
3
=k2
cos θ +2 cos θ 1
2
cos θ .cos θ +2π
3.cos θ +4π
3
=0
cos ϕ = a^
i+b^
j+c^
k.b^
i+c^
j+a^
k
a2+b2+c2.b2+c2+a2
=ab +bc +ca =0
ϕ=π
2
( ) ( )
( ) ( )
[ ( ) ( ) ]
( ) ( )
[( )
( ) ( ) ]
[( ) ( ) ]
( ) ( )
√ √
LetpositionvectorofAandBbe→
aand→
brespectively.
∴PositionvectorofPis →
OP =λ→
b+→
a
λ+1
-------------------------------------------------------------------------------------------------
Question136
If→
aand→
bareunitvectors,thenthegreatestvalueof√3→
a+→
b+→
a−→
bis
_______.
[NASep.06,2020(I)]
Answer:4
Solution:
-------------------------------------------------------------------------------------------------
Question137
If→
xand→
ybetwonon-zerovectorssuchthat →
x+→
y=→
xand2→
x+λ→
yis
perpendicularto→
y,thenthevalueofλis_______.
[NASep.06,2020(II)]
| | | |
| | | |
Given →
OB .→
OP −3→
OA ×→
OP |2=6
⇒→
b.λ→
b+→
a
λ+1−3→
a×λ→
b+→
a
λ+1|2=6
⇒
→
a.→
b+λ→
b|2
λ+1− 3λ2
(λ+1)2
→
a×→
b2=6 ∵→
a×→
b=2→
i−→
j−→
kand→
a.→
b=6
⇒6+14λ
λ+1−18λ2
(λ+1)2=6
⇒6+8λ
λ+1−18λ2
(λ+1)2=6
Let λ
λ+1=t
⇒18t2−8t =0⇒2t(9t −4) = 0
⇒t=0,4
9
∴λ
λ+1=4
9⇒λ=4
5=0.8
|
( ) |
|| | ( )
Letanglebetween→
aand→
bbeθ.
→
a+→
b= √1+1+2 cos θ = 2 cos θ
2∵ | a→
| = b=1
Similarly, →
a−→
b=2 sin θ
2
So,√3→
a+→
b+→
a−→
b = 2√3 cos θ
2+sin θ
2
∵Maximumvalueof(a cos θ +b sin θ) = a2+b2
∴Maximumvalue = 2(√3)2+ (1)2=4
| | | | [ | | ]
| | | |
| | | | [ | | | | ]
√
√
Answer:1
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question138
Let→
a,→
band→
cbethreeunitvectorssuchthat →
a−→
b
2+→
a−→
c|2=8.Then
→
a+2→
b
2+→
a+2→
c|2isequalto_______.
[NASep.02,2020(I)]
Answer:2
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question139
Ifthevolumeofaparallelopiped,whosecoterminusedgesaregivenby
thevectors→
a=^
i+^
j+n^
k,→
b=2^
i+4^
j−n^
kand→
c=^
i+n^
j+3^
k(n≥0),is
158cu.units,then:
[Sep.05,2020(I)]
Options:
A.→
a.→
c=17
B.→
b.→
c=10
| | |
| | |
∵→
x+→
y=→
x
Squaringbothsidesweget
→
x2+2→
x.→
y+→
y|2=→
x|2
⇒2→
x.→
y+→
y.→
y=0........(i)
Also2→
x+λ→
yand→
yareperpendicular
∴2→
x.→
y+λ→
y.→
y=0........(ii)
Comparing(i)and(ii),λ=1
| | | |
| | | |
→
a=→
b=→
c=1
→
a−→
b2+→
a−→
c|2=8
⇒→
a.→
b+→
a.→
c= −2
Now, →
a+2→
b2+→
a+2→
c|2
=2==→
a|2+4→
b|2+4→
c|2+4→
a.→
b+→
a.→
c=2
| | | | | |
| | |
| | |
|| | ( )
C.n =7
D.n =9
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question140
Letx0bethepointoflocalmaximaoff (x) = →
a.→
b×→
c,where
→
a=x^
i−2^
j+3^
k,→
b= −2^
i+x^
j−^
kand→
c=7^
i−2^
j+x^
kThenthevalueof
→
a.→
b+→
b.→
c+→
c.→
aatx =x0is:
[Sep.04,2020(I)]
Options:
A.-4
B.-30
C.14
D.-22
Answer:D
Solution:
Solution:
( )
Weknowthatthevolumeofparallelopiped
=→
a→
b→
c
1 1 n
24−n
1 n 3
=158
⇒(12 +n2) − 1(6+n)+n(2n −4) = 158
⇒3n2−5n −152 =0
⇒3n2−24n +19n −152 =0
⇒3n(n−8) + 19(n−8) = 0
⇒n=8orn=−19
3
⇒n=8(∵n≥0)
∴→
a=^
i+^
j+8^
k,→
b=2^
i+4^
j−8^
kand→
c=^
i+8^
j+3^
k
→
a.→
c=1+8+24 =33
→
b.→
c=2+32 −24 =10
[ ]
| |
Itisgiventhat
f(x) = →
a.→
b×→
c=
^
i^
j^
k
x−23
−2x−1
7−2x
= x3−27x +26
( ) | |
-------------------------------------------------------------------------------------------------
Question141
If→
a=2^
i+^
j+2^
k,thenthevalueof ^
i×→
a×^
i
2+^
j×→
a×^
j|2
+^
k×→
a×^
k|2isequalto________.
[NASep.04,2020(II)]
Answer:18
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question142
Thesumofthedistinctrealvaluesofmu,forwhichthevectors,
µ^
i+^
j+^
k,^
i+µ^
j+^
k,^
i+^
j+µ^
k,areco-planar,is:
[Jan.12,2019(I)]
Options:
A.-1
B.0
C.1
D.2
Answer:A
Solution:
| ( ) | | ( )
| ( )
⇒f(x) = x3−27x +26
⇒f′(x) = 3x2−27
Forcriticalpointf′(x) = 0
⇒3x2−27 =0⇒x= −3,3
Thelocalmaximaoff(x)is,x0= −3.
Thenra .b+b.c+c.a
= −2x −2x −3−14 −2x −x+7x+4+3x =3x −13
So,valueatx=x0, = a.b+b.c+c.a = 3x −13
=3× (−3) − 13 = −22
^
i×a×^
i=^
i.^
i a −^
i.a^
i = ^
j+2^
k
Similarly,^
j×a×^
j=2^
i+2^
k
^
k×^
a×^
k=2^
i+^
j
∴^
j+2^
k|2+2^
i+2^
k|2+2^
i+^
j|2 = 5+8+5=18
( ) ( ) ( )
( )
( )
| | |
Solution:
-------------------------------------------------------------------------------------------------
Question143
Let→
a=^
i+2^
j+4^
k,→
b=^
i+λ^
j+4^
kand→
c=2^
i+4^
j+ (λ2−1)^
kbecoplanar
vectors.Thenthenon-zerovector→
a×→
cis:
[Jan.11,2019(I)]
Options:
A.−10^
i−5^
j
B.−14^
i−5^
j
C.−14^
i+5^
j
D.−10^
i+5^
j
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question144
Let√3^
i+^
j,^
i+ √3^
jandβ^
i+ (1−β)^
jrespectivelybethepositionvectors
ofthepointsA,BandCwithrespecttotheoriginO.IfthedistanceofC
∵Threevectors µ^
i+^
j+^
k, ==^
i+µ^
j+^
kand ^
i+^
j+µ^
karecopalnar.
∴
µ 1 1
1 µ 1
1 1 µ
=0
⇒µ(µ2−1) + 1−µ+1−µ=0
⇒(1−µ)[2−µ(µ+1)] = 0
⇒(1−µ)[µ2+µ−2] = 0
⇒µ=1, −2
Therefore,sumofallrealvalues = 1−2= −1
( ) ( ) ( )
| |
∵→
a,→
band→
carecoplanar
∴
1 2 4
1 λ 4
24(λ2−1)
=0
⇒λ3−λ−16 +2(8−λ2+1)+4(4−2λ) = 0
⇒λ3−2λ2−9λ +18 =0
i.e.,(λ−2)(λ−3)(λ+3) = 0
Forλ=2,→
c=2^
i+4^
j+3^
k
∴→
a×→
c=
^
i^
j^
k
124
243
= −10^
i+5^
j
Forλ=3or−3,→
c=2→
a⇒→
a×c=0(Rejected)
| |
| |
fromthebisectoroftheacuteanglebetweenOAandOBis 3
√2,thenthe
sumofallpossiblevaluesofbetais:
[Jan.11,2019(II)]
Options:
A.4
B.3
C.2
D.1
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question145
Letα = (λ−2)→
a+→
band→
β= (4λ −2)→
a+3→
bbetwogivenvectorswhere
vectors→
aand→
barenon-collinear.Thevalueofλforwhichvectors→
αand
→
βarecollinear,is:
[Jan.10,2019(II)]
Options:
A.-4
B.-3
C.4
D.3
Answer:A
Since,theanglebisectorofacuteanglebetweenOAandOBwouldbey=x
Since,thedistanceofCfrombisector = 3
√2
⇒β− (1−β)
√2=3
√2 = 2β = ±3+1
β=2orβ= −1
Hence,thesumofallpossiblevalueofβ=2+ (−1) = 1
| |
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question146
Let→
a=2^
i+λ1
^
j+3^
k,→
b=4^
i+ (3−λ2)^
j+6^
kand→
c=3^
i+6^
j+ (λ3−1)^
kbe
threevectorssuchthat→
b=2→
aand→
aisperpendicularto→
cThenapossible
valueof(λ1,λ2,λ3)is:
[Jan.10,2019(I)]
Options:
A.(1,3,1)
B. −1
2,4,0
C. 1
2,4, −2
D.(1,5,1)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
( )
( )
Let→
αand→
βarecollinearforsamek
i.e.,→
α=k→
β
(λ−2)→
a+→
b=k(4λ −2)→
a+3→
b
(λ−2)→
a+→
b=k(4λ −2)→
a+3k→
b
(λ−2−k(4λ −2))→
a+→
b(1−3k) = 0
But→
aand→
barenon-collinear,then
λ−2−k(4λ −2) = 0,1−3k =0
⇒k=1
3andλ−2−1
3(4λ −2) = 0
3λ −6−4λ +2=0
λ= −4
( )
∵→
b=2→
a
∴4^
i+ (3−λ2)^
j+6^
k = 4^
i+2λ1
^
j+6^
k
∴3−λ2=2λ1......(1)
∵→
aisperpendicularto→
c
∴→
a.→
c=0
⇒6+6λ1+3(λ3−1) = 0
⇒2+2λ1+λ3−1=0
⇒λ3= −2λ1−1......(2)
Since −1
2,4,0satisfiesequation(1)and(2).Hence,oneofpossiblevalueof
λ1= −1
2,λ2=4andλ3=0
)
( )
Question147
Let→
a=^
i+^
j+ √2^
k,→
b=b1
^
i+b2
^
j+ √2^
kand→
c=5^
i+^
j+ √2^
kbethree
vectorssuchthattheprojectionvectorof→
bon→
ais→
a.
If→
a+→
bisperpendicularto→
c,then →
bisequalto:
[Jan.09,2019(II)]
Options:
A.√32
B.6
C.√22
D.4
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question148
Let→
a,→
band→
cbethreeunitvectors,outofwhichvectors→
band→
carenon-
parallel.Ifαandβaretheangleswhichvector→
amakeswithvectors→
b
and→
crespectivelyand→
a×→
b×→
c=1
2
→
b,then|α−β|isequalto
[Jan.12,2019(II)]
Options:
A.30°
B.90°
C.60°
D.45°
| |
( )
Projectionof→
bon→
a=
→
b.→
a
→
a
=b1+b2+2
4
Accordingtoquestionb1+b2+2
2= √1+1+2=2
⇒b1+b2=2........(1)
Since,→
a+→
bisperpendicularto→
c.
Hence,→
a.→
c+→
b.→
c=0
⇒8+5b1+b2+2=0........(2)
From(1)and(2),
b1= −3,b2=5
⇒→
b= −3.^
i+5^
j+ √2^
k
→
b= √9+25 +2=6
| |
| |
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question149
Let→
a=i−j,→
b=i+j+kand→
cbeavectorsuchthat→
a×→
c+→
b=→
0and
→
a.→
c=4,then →
c2isequalto:
[Jan09,2019]
Options:
A.19
2
B.9
C.8
D.17
2
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question150
Let→
a=3^
i+2^
j+2^
kand→
b=^
i+2^
j−2^
kbetwovectors.Ifavector
perpendiculartoboththevectors→
a+→
band→
a−→
bhasthemagnitude12
thenonesuchvectoris:
[April12,2019(I)]
| |
Since,→
a,→
band→
carethreeunitvectors
∴→
a=→
b=→
c=1
Then,→
a×→
b×→
c=1
2
→
b
→
a.→
c→
b−→
a.→
b→
c=1
2
→
b
∴→
a.→
c=1
2and→
a.→
b=0
→
a→
c cos β =1
2and →
a→
b cos α =0
⇒β=60°andα=90°
Hence,|α−β| = 90° − 60° | = 30°
| | | | | |
( )
( ) ( )
| | | | | | | |
∵→
a×→
c|2=→
a|2→
c|2−→
a.→
c2
⇒ −→
b|2=2→
c|2−16⇒3=2→
c|2−16
⇒→
c|2=19
2
| | | ( )
| | |
|
Options:
A.4 2^
i+2^
j+2^
k
B.4 2^
i−2^
j−2^
k
C.4 2^
i+2^
j−2^
k
D.4 −2^
i−2^
j+2^
k
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question151
Ifthevolumeofparallelopipedformedbythevectors^
i+λ^
j+^
k,^
j+λ^
k
andλ^
i+^
kisminimum,thenλisequalto:
[April12,2019(I)]
Options:
A.−1
√3
B. 1
√3
C.√3
D.−√3
Answer:B
Solution:
Solution:
( )
( )
( )
( )
Letvectorbeλ→
a+→
b×→
a−→
b
Given,a=3^
i+2^
j+2^
kand→
b=^
i+2^
j−2^
k
∴→
a+→
b=4^
i+4^
jand→
a−→
b=2^
i+4^
k
∴vector = λ 4^
i+4^
j×2^
i+4^
k
=λ 16^
i−16^
j−8^
k = 8λ 2^
i−2^
j−^
k
Giventhatmagnitudeofthevectoris12.
∴12 =8|λ|√4+4+1⇒ |λ| = 1
2
∴requiredvectoris±4 2^
i−2^
j−2^
k
[ ( ) ( ) ]
[ ( ) ( ) ]
[ ] [ ]
( )
Volumeoftheparallelepipedis,
$V=|{|\table1,\λ,1;0,1,\λ;\λ,0,1}||
={|1(1)+\λ(\λ^{2})+1(-\λ)|}$
= | λ3−λ+1|
Letf(x) = x3−x+1
Ondifferentiating,f′(x) = 3x2−1
-------------------------------------------------------------------------------------------------
Question152
Letα ∈Randthethreevectors→
a=α^
i+^
j+3^
k,→
b=2^
i+^
j−α^
kand
→
c=α^
i−2^
j+3^
k.ThenthesetS =α:→
a,→
band→
carecoplanar)
[April12,2019(II)]
Options:
A.issingleton
B.isempty
C.containsexactlytwopositivenumbers
D.containsexactlytwonumbersonlyoneofwhichispositive
Answer:B
Solution:
Solution:
(
Now,f′(x) = 0
⇒x= ± 1
√3
andf "(x) = 6x
Since,f " 1
√3>0
∴x=1
√3ispointoflocalminima.
( )
Forλ=λ1,volumeofparallelopipediszero.
∴vectorsarecoplanar.
Let,threevectors→
a,→
b,→
carecoplanar,
then →
a,→
b,→
c=0
⇒
α 1 3
2 1 −α
α−23
= 0⇒α2+6=0
∵norealvalueof'α'exist.
[ ]
| |
-------------------------------------------------------------------------------------------------
Question153
Ifaunitvector→
amakesanglesπ ∕3with^
i,π∕4with^
jandθ ∈ (0,π)
with^
k,thenavalueofis:
[April09,2019(II)]
Options:
A.5π
6
B.π
4
C.5π
12
D.2π
3
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question154
Let→
α=3^
i+^
jand→
β=2^
i−^
j+3^
k.If→
β=→
β1−→
β2,where→
β1isparallelto→
α
and→
β2isperpendicularto→
α,then→
β1×→
β2isequalto:
[April09,2019(I)]
Options:
A.−3^
i+9^
j+5^
k
B.3^
i−9^
j−5^
k
C.1
2−3^
i+9^
j+5^
k
D.1
23^
i−9^
j+5^
k
Answer:C
( )
( )
∴setSisanemptyset.
Letcos α,cos β,cos γbedirectioncosinesofa.
∴cos α =cos π
3,cos β =cos π
4andcos γ =cos θ
⇒cos2π
3+cos2π
4+cos2θ=1⇒cos2θ=1
4
⇒cos θ = ±1
2⇒θ=π
3or2π
3
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question155
Themagnitudeoftheprojectionofthevector2^
i+3^
j+^
konthevector
perpendiculartotheplanecontainingthevectors^
i+^
j+^
kand
^
i+2^
j+3^
k,is:
[April08,2019(I)]
Options:
A.√3
2
B.√6
C.3√6
D. 3
2
Answer:D
Solution:
Solution:
√
→
β=→
β1−→
β2.......(1)
Since,→
β2isperpendicularto→
α.
∴→
β2.→
α=0
Since,→
β1isparallelto→
a.
then→
β1=λ→
α(say)
→
a.→
β=→
a.→
β1−α.→
β2
⇒5=λα2⇒5=λ×10 ∵→
α= √10
⇒λ=1
2∴→
B1=α
2
∴→
β1=
→
α
2
Crossproductwith→
B1inequation(1)
⇒→
β×→
B1= −→
B2×→
B1
⇒→
β×→
B1=→
B1×→
B2⇒→
β1×→
β2=
→
β×→
α
2
⇒→
B1×→
B2=1
2
^
i^
j^
k
2−1 3
310
=1
2−3^
i−^
j(−9) + ^
k(5) = 1
2−3^
i+9^
j+5^
k
( | | )
| |
[ ] [ ]
Let→
a=^
i+^
j+^
kand→
b=^
i+2^
j+3^
k
∴vectorperpendicularto→
aand→
bis→
a×→
b
-------------------------------------------------------------------------------------------------
Question156
Let→
a=3^
i+2^
j+x^
kand→
b=^
i−^
j+^
k,forsomerealx.
Then →
a×→
b=rispossibleif:
[April08,2019(II)]
Options:
A. 3
2<r≤33
2
B.r ≥53
2
C.0 <r≤3
2
D.3 3
2<r<53
2
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question157
Let→
ubeavectorcoplanarwiththevectors→
a=2^
i+3^
j−^
kand→
b=^
j+^
k.If
→
uisperpendicularto→
aand→
u.→
b−24,then →
u2isequalto:
| |
√ √
√
√
√ √
| |
∴→
a×→
b=
^
i^
j^
k
111
123
= ^
i−2^
j+^
k
Now,projectionofvector→
c=2^
i+3= = ^
j+^
kon→
a×→
bis=
→
c.→
a×→
b
→
a×→
b
=2−6+1
√6 = 3
√6=√3
2
| |
|( )
| | | | |
Given,→
a=3^
i+2^
j+x^
kand→
b=^
i−^
j+^
k
Now,→
a×→
b=
^
i^
j^
k
3 2 x
1−11
= (2+x)^
i+ (x−3)^
j−5^
k
→
a×→
b=(2+x)2+ (x−3)2+ (−5)2 = r
⇒r=4+x2+4x +x2+9−6x +25
=2x2−2x +38 = 2 x2−x+1
4+38 −1
2
=2 x −1
2
2+75
2⇒r≥75
2⇒r≥53
2
| |
| | √
√
√√( )
√( ) √ √
[2018]
Options:
A.315
B.256
C.84
D.336
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question158
IfthepositionvectorsoftheverticesA,BandCofaΔABCare
respectively4^
i+7^
j+8^
k,2^
i+3^
j+4^
kand2^
i+5^
j+7^
k,thentheposition
vectorofthepoint,wherethebisectorof∠AmeetsBCis
[OnlineApril15,2018]
Options:
A.1
24^
i+8^
j+11^
k
B.1
36^
i+13^
j+18^
k
C.1
48^
i+14^
j+9^
k
D.1
36^
i+11^
j+15^
k
Answer:B
Solution:
Solution:
( )
( )
( )
( )
∵→
u,→
a&→
barecoplanar
∴→
u=λ→
a×→
b×→
a = λ→
a2.→
b−→
a.→
b→
a
=λ−4^
i+8^
j+16^
k = λ′ −^
i+2^
j+4^
k
Also,→
u.→
b=24 ⇒λ′ = 4
∴→
u= −4^
i+8^
j+16^
k
⇒→
u|2=336
( ) { ( ) }
{ } { }
|
SupposeangularbisectorofAmeetsBCatD(x,y,z)
Usingangularbisectortheorem,
AB
AC =BD
DC
-------------------------------------------------------------------------------------------------
Question159
Let→
a=^
i+^
j+^
k,→
c=^
j−^
kandavector→
bbesuchthat→
a×→
b=→
cand
→
a.→
b=3.Then →
bequals?
[OnlineApril16,2018]
Options:
A. 11
3
B.√11
3
C.11
√3
D.11
3
Answer:A
Solution:
Solution:
| |
√
BD
DC =(4−2)2+ (7−3)2+ (8−4)2
(4−2)2+ (7−5)2+ (8−7)2
=22+42+42
22+22+12=6
3=2
So,D(x,y,z) ≡ (2)(2) + (1)(2)
2+1,(2)(5) + (1)(3)
2+1
(2)(7) + (1)(4)
2+1
D(x,y,z) ≡ 6
3,13
3,18
3
Therefore,positionvectorofpointP=1
36^
i+13^
j+18^
k
√
√
√
√
(
)
( )
( )
∵→
a=^
i+^
j+^
k⇒→
a= √3
&→
c=^
j−^
k⇒→
c√2
Now,→
a×→
b=→
c(Given)
⇒→
a→
b sin θ =→
c
⇒→
a→
b sin θ = √2.......(i)
Also→
a.→
b=3⇒→
a→
b cos θ =3........(ii)
Dividing[i]by[ii],weget
| |
| |
| | | | | |
| | | |
| | | |
-------------------------------------------------------------------------------------------------
Question160
If→
a,→
b,and→
careunitvectorssuchthat→
a+2→
b+2→
c=→
0,then →
a×→
cis
equalto
[OnlineApril15,2018]
Options:
A.1
4
B.√15
4
C.15
16
D.√15
16
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question161
Let→
a=2^
i+^
j−2kand→
b=^
i+^
j.Let→
cbeavectorsuchthat
→
c−→
a=3,→
a×→
b×→
c=3andtheanglebetween→
cand→
a×→
bbe30°.
Then→
a.→
cisequalto:
[2017]
Options:
| |
| | | ( ) |
tan θ =√2
3∴sin θ =√2
√11
Substitutingvalueofsin θin[i]weget
√3→
b√2
√11 = √2
→
b=√11
√3
| |
| |
∵→
a+2→
b+2→
c=→
0[Given]
⇒→
a+2→
c= −2→
b⇒→
a+2→
c.→
a+2→
c = −2→
b−2→
b
⇒→
a.→
a+4→
c.→
c+4→
a.→
c=4→
b.→
b⇒1+4+4→
a.→
c=4
⇒→
a.→
c=−1
4
∵→
a.→
c|2+→
a×→
c|2=1→
aisunitvector)
⇒1
16+→
a×→
c|2=1
⇒→
a×→
c|2=15
16⇒→
a×→
c=√15
4
( ) ( ) ( ) ( )
| | (
|
| | |
A.1
8
B.25
8
C.2
D.5
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question162
Ifthevector→
b=3^
j+4^
kiswrittenasthesumofavector→
b1,parallelto
→
a=^
i+^
jandavector→
b2,perpendicularto→
a,then→
b1×→
b2isequalto:
[OnlineApril9,2017]
Options:
A.−3^
i+3^
j−9^
k
B.6^
i−6^
j+9
2
^
k
C.−6^
i+6^
j−9
2
^
k
D.3^
i−3^
j+9^
k
Answer:B
Solution:
Solution:
Given:→
a=2^
i+^
j−2k,→
b=^
i+^
j
⇒→
a=3
∴→
a×→
b=2^
i−2^
j+^
k
→
a×→
b=22+22+12=3
Wehave →
a×→
b×→
c=→
a×→
b→
c sin 30 n
⇒→
a×→
b×→
c=3→
c.1
2⇒3=3→
c.1
2
∴→
c=2
Now →
c−→
a=3
Onsquaring,weget
⇒→
c|2+→
a|2−2→
a.→
c=9⇒4+9−2→
a.→
c=9
⇒→
a.→
c=2∵→
c.→
a=→
a.→
c
| |
| | √
( ) | | | |
| ( ) | | |||
| |
| |
| |
[ ]
→
b1=
→
b1.→
a^
a
1 = 3^
j+4^
k.^
i+^
j
√2
^
i+^
j
√2
( ) {( ) ( ) } ( )
-------------------------------------------------------------------------------------------------
Question163
Thearea(insq.units)oftheparallelogramwhosediagonalsarealong
thevectors8^
i−6^
jand3^
i+4^
j−12^
k,is:
[OnlineApril8,2017]
Options:
A.26
B.65
C.20
D.52
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question164
LetABCbeatrianglewhosecircumcentreisatP.Ifthepositionvectors
A,B,CandPare→
a,→
b,→
cand
→
a+→
b+→
c
4respectively,thenthepositionvector
oftheorthocentreofthistriangle,is:
=3^
i+^
j
√2× √2=3^
i+^
j
2
→
b1+→
b2=→
b
⇒→
b2=→
b−→
b1 = 3^
j+4^
k−3
2
^
i+^
j
⇒→
b2= −3
2
^
i+3
2
^
j+4^
k
&→
b1×→
b2=
^
i^
j^
k
3
2
3
20
−3
2
3
24
⇒→
b1×→
b2=^
i(6) − ^
j(6)+^
k−9
4+9
4
⇒6^
i−6^
j+9
2
^
k
( ) ( )
( ) ( )
| |
( )
Let;d1=8^
i−6^
j+0^
k&d 2=3^
i+4^
j−12^
k∴d1×d2| =
^
i^
j^
k
8−60
3 4 −12
= 72^
i− (−96)^
j+50^
k
⇒d1×d2| = √16900 =130
∴Areaofparallelogram = 1
2d1×d2=1
2×130 =65
| | | |
| |
[OnlineApril10,2016]
Options:
A.−
→
a+→
b+→
c
2
B.→
a+→
b+c
C.
→
a+→
b+→
c
2
D.→
0
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question165
InatriangleABC,rightangledatthevertexA,ifthepositionvectorsof
A,BandCarerespectively3^
i+^
j−^
k, −^
i+3^
j+p^
kand5^
i+a^
j−4^
k,then
thepoint(pq)liesonaline:
[OnlineApril9,2016]
Options:
A.makinganobtuseanglewiththepositivedirectionofx-axis
B.paralleltox-axis
C.paralleltoy-axis
D.makinganacuteanglewiththepositivedirectionofx-axis
Answer:D
( )
( )
Positionvectorofcentriod→
G=
→
a+→
b+→
c
3
Positionvectorofcircumcentre→
C=
→
a+→
b+→
c
4
→
G=2→
C+→
r
3
3→
G=2→
C+→
r
r=3→
G−2→
C=→
a+→
b+→
c−2
→
a+→
b+→
c
4
=
→
a+→
b+→
c
2
( ) ( )
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question166
Let→
a,→
band→
cbethreeunitvectorssuchthat→
a×→
b×→
c=√3
2
→
b+→
c.If→
bis
notparallelto→
c,thentheanglebetween→
aand→
bis:
[2016]
Options:
A.2π
3
B.5π
6
C.3π
4
D.π
2
Answer:B
Solution:
Solution:
( ) ( )
AB = −4^
i+2^
j+ (p+1)^
k
AC =2^
i+ (q−1)^
j−3^
k
AB ⊥AC
⇒AB .AC =0
−8+2(q−1) − 3(p+1) = 0
3p −2q +13 =0
(p,q)lieson3x −2y +13 =0
slope = 3
2
∴Acuteanglewithx-axis
→
a×→
b×→
c=√3
2(b+c)
⇒→
a.→
c→
b−→
a.→
b→
c = √3
2
→
b+√3
2
→
c
Oncomparingbothsides
→
a.→
b= −√3
2⇒cos θ = −√3
2 ∵→
aand→
bareunitvectors]
whereθistheanglebetween→
aand→
b
θ=5π
6
( )
( ) ( )
[
-------------------------------------------------------------------------------------------------
Question167
Let→
a,→
band→
cbethreenon-zerovectorssuchthatnotwoofthemare
collinearand →
a×→
b×→
c=1
3
→
b→
c→
a.Ifθistheanglebetweenvectors→
b
and→
c,thenavalueofsin θis:
[2015]
Options:
A.2
3
B.−2√3
3
C.2√2
3
D.−√2
3
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question168
InaparallelogramABD,→
AB =a,→
AD =band →
AC =c,then →
DA .→
ABhasthe
value:
[OnlineApril11,2015]
Options:
A.1
2(a2+b2+c2)
B.1
2(a2−b2+c2)
( ) | | | |
| | | | | |
→
a×→
b×→
c=1
3
→
b→
c→
a
⇒−→
c×→
a×→
b=1
3
→
b→
c→
a
⇒− →
c.→
b→
a+→
c.→
a→
b = 1
3
→
b→
c→
a
⇒− →
b→
c cos θ →
a+→
c.→
a→
b=1
3
→
b→
c→
a
∵→
a,→
b,→
carenoncollinear,theaboveequationispossibleonlywhen
−cos θ =1
3and→
c.→
a=0
⇒cos θ = −1
3
⇒sin θ =2√2
3;θ∈IIquad
( ) | | | |
( ) | | | |
( ) ( ) | | | |
| | | |( ) | | | |
C.1
2(a2+b2−c2)
D.1
3(b2+c2−a2)
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question169
Let→
aand→
bbetwounitvectorssuchthat →
a+→
b= √3.
If→
c=→
a+2→
b+3→
a×→
b,then2 →
cisequalto:
[OnlineApril10,2015]
Options:
A.√55
B.√37
C.√51
D.√43
Answer:A
Solution:
Solution:
| |
( ) | |
Let|AB| = a, | AD | = band|AC| = c
WehaveAB +AD =AC
Onsquaringboththeside,weget
|AB|2+ | AD |2+2AB .AD = | AC|2
⇒a2+b2+2AB . (−DA) = c2
⇒2AB .DA =a2+b2−c2
⇒DA .AB =1
2(a2+b2−c2)
→
a+→
b= √3
anglebetween→
aand→
bis60°.
→
a×→
bis⊥rtoplanecontaining→
aand→
b
→
c=→
a+2→
b+3→
a×→
b
→
c=→
a2+4→
b|2+2.2 →
a|2cos 60°→
n1+3→
a→
b sin 60°→
n2+3→
a→
b sin 60° . →
n2
→
n1⊥r→
n2
→
c2= (1+4+2) + 9×3
4⇒ →
c|2=7+27 ∕4=55 ∕4
| |
( )
√| | | | | | ||| | ||
| | |
-------------------------------------------------------------------------------------------------
Question170
If^
x,^
yand^
zarethreeunitvectorsinthree-dimensionalspace,thenthe
minimumvalueof ^
x+^
y2+^
y+^
z|2+^
z+^
x|2
[OnlineApril12,2014]
Options:
A.3
2
B.3
C.3√3
D.6
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question171
If →
a=2,→
b=3and 2→
a−→
b=5,then 2→
a+→
bequals:
[OnlineApril9,2014]
Options:
A.17
B.7
C.5
D.1
Answer:C
Solution:
| | | |
| | | | | | | |
2→
c= √55
| |
^
x+^
y+^
z2≥0
⇒3+2∑^
x.^
y≥0
⇒2∑^
x.^
y≥ −3
Now, ^
x+^
y2+^
y+^
z|2+^
z+^
x|2
=6+2∑x.^
y≥6+ (−3)
⇒ ^
x+^
y2+^
y+^
z|2+^
z+^
x|2≥3
( )
| | | |
| | | |
-------------------------------------------------------------------------------------------------
Question172
If →
a×→
b→
b×→
c→
c×→
a=λ→
a,→
b,→
c
2thenλisequalto
[2014]
Options:
A.0
B.1
C.2
D.3
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question173
If→
x=3^
i−6^
j−^
k,→
y=^
i+4^
j−3^
kand→
z=3^
i−4^
j−12^
kthenthe
magnitudeoftheprojectionof→
x×→
yon→
zis:
[OnlineApril19,2014]
Options:
A.12
B.15
C.14
D.13
[ ] [ ]
Given 2→
a−→
b=5
2→
a2+→
b|2−2×2→
a→
b cos θ = 5
Puttingvaluesof →
aand →
b,weget
(2×2)2+ (3)2−24 cos θ =25
⇒cos θ =0
⇒θ=π
2
2→
a+→
b= √16 +9+24 cos θ = √25 =5
| |
√( | | ) | | |
| | | |
| |
L.H.S=→
a×→
b.→
b×→
c×→
c×→
a
=→
a×→
b.→
b×→
c.→
a→
c− →
b×→
c→
c→
a
=→
a×→
b.→
b→
c→
a→
c[∵b×c.c=0]
=→
a→
b→
c.→
a×→
b.→
c=→
a→
b→
c2
=→
a×→
b→
b×→
c→
c×→
a=→
a→
b→
c2
Soλ=1
[ ( ) [ ( ) ( ) ]
( ) [ ( ) ( ) ]
( ) ] ]
[ ] ) [ ]
[ ] ] [ ]
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question174
If →
c2=60and→
c×^
i+2^
j+5^
k=→
0,thenavalueofc . −7^
i+2^
j+3^
kis
[OnlineApril11,2014]
Options:
A.4√2
B.12
C.24
D.12√2
Answer:D
Solution:
| | ( ) ( )
Let→
x=3^
i−6^
j−^
k,→
y=^
i+4^
j−3^
kand→
z=3^
i−4^
j−12^
k
Now,→
x×→
y=
^
i^
j^
k
3−6−1
1 4 −3
= 22^
i+8^
j+18^
k
Projectionof^
x×^
yon^
z=^
x+^
y. (z)
^
z
=22(3) + 8(−4) + 18(−12)
√9+16 +144 = −182
13 = −14
Now,magnitudeofproection=14.
| |
( )
| |
Let,→
c=a^
i+b^
j+c^
k
Given,→
c×^
i+2^
j+5^
k=→
0
⇒
^
i^
j^
k
a b c
125
=→
0
⇒(5b −2c)^
i− (5a −c)^
j+ (2a −b)^
k = 0^
i+0^
j+0^
k
Comparingbothsides,weget
5b −2c =0;5a −c=0;2a −b=0
or5b =2c;5a =c;2a =b
Alsogiven →
c2=60 ⇒a2+b2+c2=60
Puttingthevalueofbandcinaboveeqn.,weget
a2+ (2a)2+ (5a)2=60
⇒a2+4a2+25a2=60$\⇒30a^{2}=60
$$a^{2}=2$
a= ±√2;b=2√2;c=5√2
Now,→
c=a^
i+b^
j+c^
k
∴→
c= √2^
i+2√2^
j+5√2^
k
( )
| |
| |
-------------------------------------------------------------------------------------------------
Question175
Ifthevectors →
AB =3^
i+4^
kand →
AC =5^
i−2^
j+4^
karethesidesofatriangle
ABC,thenthelengthofthemedianthroughAis
[2013]
Options:
A.√18
B.√72
C.√33
D.√45
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question176
If→
aand→
barenon-collinearvectors,thenthevalueofαforwhichthe
vectors→
u= (α−2)→
a+→
band→
v= (2+3α)→
a−3→
barecollinearis:
[OnlineApril23,2013]
Options:
Valueof→
c. −7^
i+2^
j+3^
kis
√2^
i+2√2^
j+5√2^
k. −7^
i+2^
j+3^
k
= −7√2+4√2+15√2=12√2
( )
( ) ( )
(c)Wehave, →
AB +→
BC +→
CA =0⇒→
BC =→
AC −→
AB
LetMbemid-pointofBC
Now, →
BM =
→
AC −→
AB
2because →
BM =
→
BC
2
Also,wehave
→
AB +→
BM +→
M A =0
⇒→
AB +
→
AC −→
AB
2=→
AM
⇒1→
AM =
→
AB +→
AC
2 = 4^
i−^
j+4^
k
⇒→
AM = √33
( )
| |
A.3
2
B.2
3
C.−3
2
D.−2
3
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question177
If^
a,^
band^
careunitvectorssatisfying^
a− √3^
b+^
c=→
0,thentheangle
betweenthevectors^
aand^
cis:
[OnlineApril22,2013]
Options:
A.π
4
B.π
3
C.π
6
D.π
2
Answer:B
Solution:
Solution:
Since,→
uand→
varecollinear,therefore →
ku +→
v=0
⇒[k(α−2) + 2+3α]→
a+(k−3)→
b=0.......(i)
Since→
aand→
barenon-collinear,thenforsomeconstantmandn,
m→
a+n→
b=0⇒m=0,n=0
Hencefromequation(i)
k−3=0⇒k=3
Andk(α−2) + 2+3α =0
⇒3(α−2) + 2+3α =0⇒α=2
3
Letanglebetween^
aand^
cbeθ.
Now,^
a− √3^
b+^
c=→
0
⇒^
a+^
c= √3^
b
⇒^
a+^
c.^
a+^
c=3^
b.^
b
⇒^
a.^
a+^
a.^
c+^
c.^
a+^
c.^
c = 3×1
⇒1+2 cos θ +1=3
( )
( ) ( ) ( )
-------------------------------------------------------------------------------------------------
Question178
Let→
a=2^
i−^
j+^
k,→
b=^
i+2^
j−^
kand→
c=^
i+^
j−2^
kbethreevectors.A
vectorofthetype→
b+λ→
cforsomescalarλ,whoseprojectionon→
aisof
magnitude 2
3is:
[OnlineApril9,2013]
Options:
A.2^
i+^
j+5^
k
B.2^
i+3^
j−3^
k
C.2^
i−^
j+5^
k
D.2^
i+3^
j+3^
k
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question179
Let→
a=2^
i+^
j−2^
k,→
b=^
i+^
j.If→
cisavectorsuchthat
→
a●→
c=→
c,→
c−→
a=2√2andtheanglebetween→
a×→
band→
cis30°,then
→
a×→
b×→
cequals:
√
| | | |
| ( ) |
⇒cos θ =1
2⇒θ=π
3
Let→
d=→
b+λ→
c
∴→
d=^
i+2^
j−^
k+λ^
i+^
j−2^
k
= (1+λ)^
i+ (2+λ)^
j− (1+2λ)^
k
Ifθbetheanglebetween→
dand→
a,thenprojectionof→
d
or →
b+λ→
con→
a
=→
d cos θ =→
d
→
d.→
a
→
d→
a
=
→
d.→
a
→
a
=2(λ+1) − (λ+2) − (2λ +1)
√4+1+1 = −λ−1
√6
Butprojectionof→
don→
a=2
3
∴−λ+1
√6=2
3⇒λ2+2λ +1
6=2
3
⇒λ2+2λ −3=0⇒λ2+3λ −λ−3=0
⇒λ(λ+3) − 1(λ+3) = 0,⇒λ=1, −3
whenλ=1,then→
b+λ→
c=2^
i+3^
j−3^
k
whenλ= −3,then→
b+λ→
c= −2^
i−^
j+5^
k
( )
( )
| | | | ( | | | | )| |
√
√
[OnlineApril25,2013]
Options:
A.1
2
B.3√3
2
C.3
D.3
2
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question180
Thevector ^
i×→
a→
b^
i+^
j×→
a.→
b^
j+^
k×→
a.→
b^
kisequalto:
[OnlineApril9,2013]
Options:
A.→
b×→
a
B.→
a
C.→
a×→
b
D.→
b
Answer:C
Solution:
( ) ( ) ( )
→
a=2^
i+^
j−2^
k,→
b=^
i+^
j
⇒→
a=3
and→
a×→
b=
^
i^
j^
k
2 1 −2
1 1 0
= 2^
i−2^
j+^
k
→
a×→
b= √4+4+1=3
Now, →
c−→
a=2√2⇒→
c−→
a|2=8
⇒→
c−→
a.→
c−→
a=8
⇒→
c|2+→
a|2−2→
c.→
a=8
⇒→
c|2+9−2→
c=8
⇒→
c−12=0⇒→
c=1
∴→
a×→
b×→
c=→
a×→
b→
c sin 30° = 3×1×1
2=3
2
| |
| |
| |
| | |
| | ( )
| |
| | |
( | | ) | |
| ( ) | | | | |
-------------------------------------------------------------------------------------------------
Question181
Statement1:Ifthepoints(1,2,2),(2,1,2)and(2,2,z)and(1,1,1)are
coplanar,thenz=2.
Statement2:Ifthe4pointsP,Q,RandSarecoplanar,thenthevolume
ofthetetrahedronPQRSis0.
[OnlineMay12,2012]
Options:
A.Statement1isfalse,,Statement2istrue.
B.Statement1istrue,Statement2isfalse.
C.Statement1istrue,Statement2istrue,Statement2isacorrectexplanationofStatement1.
D.Statement1istrue,Statement2istrue,Statement2isnotacorrectexplanationof
Statement1.
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question182
If→
a=^
i−2^
j+3^
k,→
b=2^
i+3^
j−^
kand→
c=r^
i+^
j+ (2r −1)^
karethree
vectorssuchthat→
cisparalleltotheplaneof→
aand→
b,thenrisequalto
[OnlineMay19,2012]
Options:
A.1
B.-1
C.0
D.2
^
i×→
a→
b^
i+^
j×→
a.→
b^
j+ ^
k×→
a.→
b^
k
^
i.→
a×→
b^
i+^
j.→
a×→
b^
j+ ^
k.→
a×→
b^
k ∵→
a×→
b.→
c=→
a.→
b×→
c
→
a×→
b^
i+→
a×→
b^
j+ →
a×→
b^
k = →
a×→
b
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )
Statement-1
Points(1,2,2),(2,1,2),(2,2,z)and(1,1,1)arecoplanarthenz=2whichisfalse.
∵
1−10
1 0 z−2
0−1−1
=0
⇒1(z−2) + 1(−1) = 0⇒z=3
Statement-2isthetruestatement.
| |
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question183
LetABCDbeaparallelogramsuchthat →
AB =→
q,→
AD =→
pand∠BADbean
acuteangle.If→
risthevectorthatcoincidewiththealtitudedirected
fromthevertexBtothesideAD,then→
risgivenby:
[2012]
Options:
A.→
r=3→
q−3→
p.→
q
→
p.→
p
→
p
B.→
r= −→
q+→
p.→
q
→
p.→
p
→
p
C.→
r=→
q−→
p.→
q
→
p.→
p
→
p
D.→
r= −3→
q−3→
p.→
q
→
p.→
p
→
p
Answer:B
Solution:
Solution:
( )
( )
( )
( )
( )
( )
( )
( )
Let→
a=^
i−2^
j+3^
k,and→
c=r^
i+^
j+ (2r −1)^
k
Since,→
cisparalleltotheplaneof→
aand→
btherefore,
→
a,→
band→
carecoplanar.
∴
1−23
2 3 −1
r 1 2r −1
=0
⇒1(6r −3+1) + 2(4r −2+r)+3(2−3r) = 0
⇒6r −2+10r −4+6−9r =0
⇒r=0
| |
LetABCDbeaparallelogramsuchthatAB =→
q,AD =→
pand∠BADbeanacuteangle.
Wehave
AX =
→
p.→
q
→
p
→
p
→
p
=
→
p.→
q
→
p2
→
p
Fromtrianglelaw
Let→
r=BX =BA +AX = −→
q+
→
p.→
q
→
p2
→
p
(| | )(| | )| |
| |
Question184
Let→
aand→
bbetwounitvectors.Ifthevectors→
c=^
a+2^
band→
d=5^
a−4^
b
areperpendiculartoeachother,thentheanglebetween^
aand^
bis:
[2012]
Options:
A.π
6
B.π
2
C.π
3
D.π
4
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question185
Ifa +b+c=0,→
a=3,→
b=5and →
c=7,thentheanglebetween→
aand
→
bis
[OnlineMay19,2012]
Options:
A.π
3
B.π
4
C.π
6
D.π
2
Answer:A
| | | | | |
Giventhat→
c=^
a+2^
band→
d=5^
a−4^
band ^
a=^
b=1
Since→
cand→
dareperpendiculartoeachother
∴→
c.→
d=0
⇒^
a+2^
b.5^
a−4^
b=0
⇒5+6^
a.^
b−8=0
⇒^
a.^
b=1
2⇒sin θ =1
2⇒θ=π
3
| | | |
( ) ( )
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question186
Aunitvectorwhichisperpendiculartothevector2^
i−^
j+2^
kandis
coplanarwiththevectors^
i+^
j−^
kand2^
i+2^
j−^
kis
[OnlineMay12,2012]
Options:
A.2^
j+^
k
√5
B.3^
i+2^
j−2^
k
√17
C.3^
i+2^
j+2^
k
√17
D.2^
i+2^
j−1^
k
3
Answer:D
Solution:
Leta+b+c=0⇒ (a+b) = −c
⇒(a+b)2=c2
⇒a2+b2+2a .b=c2
⇒9+25 +2.3.5 cos θ =49
∵→
a=3,→
b=5 and →
c=7
∴cos θ =1
2⇒θ=π
3
(| | | | | | )
Letx^
i+y^
j+z^
kbetherequiredunitvector.
Since^
aisperpendicularto 2^
i−^
j+2^
k.
∴2x −y+2z =0.......(i)
Sincevectorx^
i+y^
j+z^
kiscoplanarwiththevector^
i+^
j−^
kand2^
i+2^
j−^
k
∴x^
i+y^
j+z^
k=p^
i+^
j−^
k+q 2^
i+2^
j−^
k
wherepandqaresomescalars.
⇒x^
i+y^
j+z^
k= (p+2q)^
i+(p+2q)^
j− (p+q)^
k
⇒x=p+2q,y=p+2q,z= −p−q
Nowfromequation(i),
2p +4q −p−2q −2p −2q =0
⇒−p=0⇒p=0
∴x=2q,y=2q,z= −q
Sincevectorx^
i+y^
j+z^
kisaunitvector,therefore
x^
i+y^
j+z^
k=1
⇒x2+y2+z2=1
⇒x2+y2+z2=1
⇒4q2+4q2+q2=1
⇒9q2=1⇒q= ±1
3
( )
( ) ( )
| |
√
-------------------------------------------------------------------------------------------------
Question187
ABCDisparallelogram.ThepositionvectorsofAandCarerespectively,
3^
i+3^
j+5^
kand^
i−5^
j−5^
k.IfM isthemidpointofthediagonalDB,
thenthemagnitudeoftheprojectionof →
OM on →
OC,whereOistheorigin,
is
[OnlineMay7,2012]
Options:
A.7√51
B. 7
√50
C.7√50
D. 7
√51
Answer:D
Solution:
-------------------------------------------------------------------------------------------------
Whenq=1
3,thenx=2
3,y=2
3,z= −1
3
Whenq= −1
3,thenx= −2
3,y= −2
3,z=1
3
Hererequiredunitvectoris2^
i+2^
j−1^
k
3
or−2
3
^
i−2
3
^
j+1
3
^
k
Inaparallelogram,diagonalsbisecteachother.So,midpointofDBisalsothemid-pointofAC.
Mid-pointofM=2^
i−^
j
DirectionratioofOC = (1, −5, −5)
DirectionratioofOM = (2, −1,0)
AngleθbetweenOM andOCisgivenby
cos θ = (1×2) + (−5)(−1) + (−5)(0)
22+ (−1)2(1)2+ (−5)2+ (−5)2
=2+5
√5√51 =7
√5√51
Projectionof →
OM on →
OCisgivenby
|OM | . cos θ = √5×7
√5× √51 =7
√51
√ √
Question188
Statement1:Thevectors→
a,→
band→
clieinthesameplaneifandonlyif
→
a.→
b×→
c=0
Statement2:Thevectors→
uand→
vareperpendicularifandonlyif→
u.→
v=0
where→
u×→
visavectorperpendiculartotheplaneof→
uand→
v.
[OnlineMay26,2012]
Options:
A.Statement1isfalse,Statement2istrue.
B.Statement1istrue,Statement2istrue,Statement2iscorrectexplanationforStatement1.
C.Statement1istrue,Statement2isfalse.
D.Statement1istrue,Statement2istrue,,Statement2isnotacorrectexplanationfor
Statement1.
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question189
If→
u=^
j+4^
k,→
v=^
i+3^
kand→
w=cos θ ^
i+sin θ ^
jarevectorsin3-
dimensionalspace,thenthemaximumpossiblevalueof →
u×→
v.→
wis
[OnlineMay12,2012]
Options:
A.√3
B.5
C.√14
D.7
Answer:B
Solution:
( )
| |
Statement-1
Thevectors→
a,→
band→
clieinthesameplane.
⇒→
a,→
band→
carecoplanar.
Weknow,thenecessaryandsufficientconditionsforthreevectorstobecoplanaristhat →
a→
b→
c=0
i.e.→
a.→
b×→
c=0
Hence,statement-1istrue.
[ ]
( )
-------------------------------------------------------------------------------------------------
Question190
If→
a=^
i−2^
j+3^
k,→
b=2^
i+3^
j−^
kand→
c=λ^
i+^
j+ (2λ −1)^
karecoplanar
vectors,thenλisequalto
[OnlineMay7,2012]
Options:
A.0
B.-1
C.2
D.1
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question191
Let→
a,→
b,→
cbethreenon-zerovectorswhicharepairwisenon-collinear.If
→
a+3→
biscollinearwith→
cand→
b+2→
ciscollinearwith→
a,then→
a+3→
b+6→
cis
:
[2011RS]
Options:
A.→
a
Let→
u=^
j+4^
k,→
v=^
i+3^
kand→
w=cos θ ^
i+sin θ ^
j
Now,→
u×→
v=
^
i^
j^
k
0 1 4
1 0 −3
= ^
i(−3) − ^
j(−4) + ^
k(−1)
= −3^
i+4^
j−^
k
Now,
→
→
u×v.→
w = −3^
i+4^
j−^
k.cos θ ^
i+sin θ ^
j
= −3 cos θ +4 sin θ
Now,maximumpossiblevalueof|−3 cos θ +4 sin θ| = (−3)2+ (4)2= √25 =5
| |
( ) ( ) ( )
√
Since→
a=^
i−2^
j+3^
k,→
b=2^
i+3^
j−^
kand→
c=λ^
i+^
j+ (2λ −1)^
karecoplanar
therefore →
a→
b→
c=0
i.e.,
1 2 λ
−23 1
3−1 2λ −1
=0
⇒1(6λ −2) − 2(−4λ −1) + λ(−7) = 0
⇒(6λ −2) + 8λ +2+2+2λ −9λ =0
⇒7λ =0⇒λ=0
[ ]
| |
B.→
c
C.→
0
D.→
a+→
c
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question192
Ifthep^
i+^
j+^
k,^
i+q^
j+^
kand^
i+^
j+r^
k(p≠q≠r≠1)vectorare
coplanar,thenthevalueofpqr − (p+q+r)is
[2011RS]
Options:
A.2
B.0
C.-1
D.-2
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question193
Thevectors→
aand→
barenotperpendicularand→
cand→
daretwovectors
Asperquestion
→
a+3→
b=λ→
c........(i)
→
b+2→
c=µ→
a.......(ii)
Onsolvingequations(i)and(ii)
(1+3µ)→
a− (λ+6)→
c=0
As→
aand→
carenoncollinear,
∴1+3µ =0andλ+6=0
From(i)→
a+3→
b+6→
c=→
0
Thegivenvectorsarecoplanarthen
p 1 1
1 q 1
1 1 r
=0
⇒p(qr −1) + 1(1−r) + 1(1−q) = 0
⇒pqr −p+1−r+1−q=0
⇒pqr − (p+q+r) = −2
| |
satisfying→
b×→
c=→
b×→
dand→
a.→
d=0Thenthevector→
disequalto
[2011]
Options:
A.→
c+→
a.→
c
→
a.→
b
→
b
B.→
b+
→
b.→
c
→
a.→
b
→
c
C.→
c−→
a.→
c
→
a.→
b
→
b
D.→
b−
→
b.→
c
→
a.→
b
→
c
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question194
If→
a=1
√10 3^
i+^
kand→
b=1
72^
i+3^
j−6^
k,thenthevalueof
2→
a−→
b→
a×→
b×→
a+2→
bis
[2011]
Options:
A.-3
B.5
C.3
D.-5
Answer:D
Solution:
( )
( )
( )
( )
( ) ( )
( ) [ ( ) ( ) ]
Giventhat→
a.→
d≠0,→
a.→
d=0
Now,→
b×→
c=→
b×→
d
⇒→
a×→
b×→
c=→
a×→
b×→
d
⇒→
a.→
c→
b−→
a.→
b→
c = a→
.→
d→
b−→
a.→
b→
d
⇒→
a.→
b→
d= − →
a.→
c→
b+ →
a→
b→
c
→
d=→
c−
→
a.→
c
→
a.→
b
→
b
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )
( )
Solution:
-------------------------------------------------------------------------------------------------
Question195
Ifthevectors→
a=^
i−^
j+2^
k,→
b=2^
i+4^
j+^
kand→
c=λ^
i+^
j+µ^
kare
mutuallyorthogonal,then(λ,µ) =
[2010]
Options:
A.(2,-3)
B.(-2,3)
C.(3,-2)
D.(-3,2)
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question196
Let→
a=^
j−^
kand→
c=^
i−^
j−^
k.Thenthevector→
bsatisfying→
a×→
b+→
c=→
0
and→
a.→
b=3is
[2010]
Options:
A.2^
i−^
j+2^
k
B.^
i−^
j−2^
k
C.^
i+^
j−2^
k
2→
a−→
b→
a×→
b×→
a+2→
b
=2→
a−→
b.→
a×→
b×→
a+2→
a×→
b×→
b
=2→
a−→
b→
a.a→
b−→
a.→
b→
a+2→
a.→
b→
b−2→
b.b→
a
=2→
a−→
b→
b−0+0−2→
a
Fromgivenvaluesweget
→
a.→
b=0and→
b.→
b=1
= −4→
a.→
a−→
b.→
b= −5
( ) ( ( ) ( ) )
( ) ( ( ) ( ) )
( ) ( ( ) ( ) ( ) ( ) )
( ) ( )
Giventhat,→
a,→
band→
caremutuallyorthogonal
∴→
a.→
b=0,→
b.→
c=0,→
c.→
a=0
⇒2λ +4+µ=0......(i)
λ−1+2µ =0.......(ii)
Onsolving(i)and(ii),wegetλ= −3,µ=2
D.−^
i+^
j−2^
k
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question197
If→
u,→
v,→
warenon-coplanarvectorsandp,qarerealnumbers,thenthe
equality 3→
up→
vp→
ω−p→
v→
ωq→
u−2→
ωq→
vq→
u=0holdsfor:
[2009]
Options:
A.exactlytwovaluesof(p,q)
B.morethantwobutnotallvaluesof(p,q)
C.allvaluesof(p,q)
D.exactlyonevalueof(p,q)
Answer:D
Solution:
Solution:
[ ] [ ] [ ]
Giventhat
→
c=→
b×→
a⇒→
b.→
c=→
b.→
b×→
a ⇒→
b.→
c=0
⇒b1
^
i+b2
^
j+b3
^
k.^
i−^
j−^
k=0
where→
b=b1
^
i+b2
^
j+b3
^
k
b1−b2−b3=0......(i)
and→
a.→
b=3⇒^
j−^
k. b1
^
i+b2
^
j+b3
^
k=3
⇒b2−b3=3
Fromequation(i)
b1=b2+b3= (3+b3) + b3 = 3+2b3
→
b= (3+2b3)^
i+ (3+b3)^
j+b3
^
k
Fromtheoptiongiven,itisclearthatb3equaltoeither2or-2
Ifb3=2then→
b=7^
i+5^
j+2^
kwhichisnotpossibleIf
b3= −2,then→
b= −^
i+^
j−2^
k
( )
( ) ( )
( ) ( )
→
u,→
v,→
warenon-coplanarvectors
∴ →
u,→
v,→
w≠0
Now, 3→
u,p→
v,p→
w−p→
v,p→
w,q→
u− 2→
w,q→
v,q→
u=0
⇒3p2→
u,→
v,→
w−pq →
v,→
w,→
u−2q2→
w,→
v,→
u=0
⇒3p2→
u,→
v,→
w−pq →
u,→
v,→
w−2q2→
u,→
v,→
w=0
⇒(3p2−pq +2q2)→
u,→
v,→
w=0
⇒3p2−pq +2q2=0
∵→
u,→
v,→
w=0
⇒2p2+p2−pq +q2
4+7q2
4=0
[ ]
[ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ] [ ]
[ ]
( [ ] )
-------------------------------------------------------------------------------------------------
Question198
Thevector→
a=α^
i+2^
j+β^
kliesintheplaneofthevectors→
b=^
i+^
jand
→
c=^
j+^
kandbisectstheanglebetween→
band→
c.Thenwhichoneofthe
followinggivespossiblevaluesofαandβ?
[2008]
Options:
A.α =2,β=2
B.α =1,β=2
C.α =2,β=1
D.α =1,β=1
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question199
Thenon-zerovectorsare→
a,→
band→
carerelatedby→
a=8→
band→
c= −7→
b.
Thentheanglebetween→
aand→
cis
[2008]
Options:
A.0
B.π
4
C.π
2
D.pi
Answer:D
⇒2p2+p−q
2
2+7
4q2=0
⇒p=0,q=0,p=q∕2
Thisispossibleonlywhenp=0,q=0
∴Thereisexactlyonevalueof(p,q)
( )
∵→
aliesintheplaneof→
band→
c
∴→
a=→
b+λ→
c
⇒α^
i+2^
j+β^
k=^
i+^
j+λ^
j+^
k
⇒α=1,2=1+λ,β=λ
⇒α=1,β=1
( )
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question200
Let→
a=^
i+^
j+^
k,→
b=^
i−^
j+2^
kand→
c=x^
i+ (x−2)^
j−^
k.Ifthevector→
clies
intheplaneof→
aand→
b,thenxequals
[2007]
Options:
A.-4
B.-2
C.0
D.1.
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question201
If^
uand^
vareunitvectorsandθistheacuteanglebetweenthem,then
2^
u×3^
visaunitvectorfor
[2007]
Options:
A.novalueofθ
B.exactlyonevalueofθ
Clearly→
a= −8
7
→
c
⇒→
a∥→
candareoppositeindirection
∴Anglebetween→
aand→
cisπ
Given→
a=^
i+^
j+^
k,→
b=^
i−^
j+2^
kand→
c=x^
i+ (x−2)^
j−^
k.
Giventhat→
cliesintheplaneof→
aand→
b,then→
a,→
band→
carecoplanar
∴→
a→
b→
c=0
i.e.
1 1 1
1−12
x(x−2) −1
=0
⇒1[1−2(x−2)] − 1[−1−2x]+1[x−2+x] = 0
⇒1−2x +4+1+2x +2x −2=0
⇒2x = −4⇒x= −2
[ ]
| |
C.exactlytwovaluesofθ
D.morethantwovaluesofθ
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question202
Aparticlejustclearsawallofheightbatadistanceaandstrikesthe
groundatadistancecfromthepointofprojection.Theangleof
projectionis
[2007]
Options:
A.tan−1bc
a(c−a)
B.tan−1bc
a
C.tan−1b
ac
D.45°.
Answer:A
Solution:
Giventhat 2^
u×3^
v=1andθisacuteanglebetween^
uand^
v,^
u=1,^
v=1
⇒2^
u×3^
v=6^
u^
v sin θ =1
⇒6|sin θ =1⇒sin θ =1
6
Hence,thereisexactlyonevalueofθforwhich2^
u×3^
visaunitvector.
| | | | | |
| | | | | | | |
|
LetBbethetopofthewallwhosecoordinateswillbe(a,b).Range(R)=c
Bliesonthetrajectory
∴y=x tan α −1
2gx2
u2cos2α
⇒b=a tan α −1
2ga2
u2cos2α
⇒b=a tan α 1 −ga
2u2cos2α tan α
=a tan α 1 −a
2u2
gcos2α.sin α
cos α
[ ]
[ ]
-------------------------------------------------------------------------------------------------
Question203
Abodyweighing13kgissuspendedbytwostrings5mand12mlong,
theirotherendsbeingfastenedtotheextremitiesofarod13mlong.If
therodbesoheldthatthebodyhangsimmediatelybelowthemiddle
point,thentensionsinthestringsare
[2007]
Options:
A.5kgand12kg
B.5kgand13kg
C.12kgand13kg
D.5kgand5k
Answer:A
Solution:
=a tan α 1 −a
u2.2 sin α cos α
g
=a tan α 1 −a
u2sin 2 α
g
=a tan α 1 −a
R∵R=u2sin2α
g
⇒b=a tan α 1 −a
c
⇒b=a tan α .c−a
c
⇒tan α =bc
a(c−a)
Theangleofprojection,
α=tan−1bc
a(c−a)
[ ]
[ ]
[ ] ( )
[ ]
( )
InΔABC
∵132=52+122⇒AB2=AC2+BC2
⇒∠ACB =90°
MismidpointofthehypotenuseAB,thereforeM A =M B = M C
⇒∠A= ∠ACM =θ
ApplyingLami'stheorematC,weget
T1
sin(180 −θ)=T2
sin(90 +θ) = 13kg
sin 90°
⇒T1=13 sin θandT2=13 cos θ
-------------------------------------------------------------------------------------------------
Question204
TheresultantoftwoforcesPnand3nisaforceof7n.Ifthedirectionof
3nforcewerereversed,theresultantwouldbe√19n.ThevalueofPis
[2007]
Options:
A.3n
B.4n
C.5n
D.6n.
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question205
ABCisatriangle,rightangledatA.Theresultantoftheforcesacting
alongAB,BCwithmagnitudes 1
ABand 1
ACrespectivelyistheforcealong
AD,whereDisthefootoftheperpendicularfromAontoBC.The
magnitudeoftheresultantis
[2006]
Options:
A. AB2+AC2
(AB)2(AC)2
⇒T1=13 ×5
13andT2=13 ×12
13
⇒T1=5kgandT2=12kg
Giventhat:ForceP=Pn,Q=3n,resultantR=7n&P′ = Pn,Q′ = (−3)n,R′ = √19n
WeknowthatR2=P2+Q2+2PQ cos α
⇒(7)2=P2+ (3)2+2×P×3 cos α
⇒49 =P2+9+6P cos α
⇒40 =P2+6P cos α.....(i)
and(√19)2=P2+ (−3)2+2P×−3 cos α
⇒19 =P2+9−6P cos α
⇒10 =P2−6P cos α.....(ii)
Adding(i)and(ii)50 =2P2
⇒P2=25 ⇒P=5n
B.(AB)(AC)
AB +AC
C. 1
AB +1
AC
D. 1
AD
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question206
Thevaluesofa,forwhichpointsA,B,Cwithpositionvectors
2^
i−^
j+^
k,^
i−3^
j−5^
kanda^
i−3^
j+^
krespectivelyaretheverticesofa
rightangledtrianglewithC =π
2are
[2006]
Options:
A.2and1
B.-2and-1
C.-2and1
D.2and-1
Answer:A
Ifweconsiderunitvectors^
iand^
jinthedirectionABandACrespectivelyanditsmagnitude 1
ABand 1
ACrespectively,
thenasperquestion,forcesalongABandACrespectivelyare
1
AB
^
iand 1
AC
^
j
∴TheirresultantalongAD =1
AB
^
i+1
AC
^
j
∴Magnitudeofresultantis
=1
AB
2+1
AC
2 = AC2+AB2
AB2+AC2[∵AC2+AB2=BC2]
=BC
AB .AC
( ) ( )
( ) ( )
√( ) ( ) √
∵ΔABC ∼ΔDBA
⇒BC
AB =AC
AD⇒ BC
AB ×AC =1
AD
∴Therequiredmagnitudeofresultantbecomes 1
AD
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question207
If →
a×→
b×→
c=→
a×→
b×→
cwhere→
a,→
band→
careanythreevectorssuch
that→
a→
b≠0,→
b.→
c≠0then→
aand→
care
[2006]
Options:
A.inclinedatanangleofπ
3betweenthem
B.inclinedatanangleofπ
6betweenthem
C.perpendicular
D.parallel
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question208
AbodyfallingfromrestundergravitypassesacertainpointP.Itwasat
adistanceof400mfromP,4spriortopassingthroughP.If
g=10m ∕s2,thentheheightabovethepointPfromwherethebody
begantofallis
[2006]
Options:
A.720m
B.900m
C.320m
D.680m
( ) ( )
→
CA = (2−a)^
i+2^
j; →
CB = (1−a)^
i−6^
k ∵ →
CA ⊥→
CB
∴→
CA .→
CB =0⇒(2−a)(1−a) = 0
⇒a=2,1
[ ]
(a×b) × c=a× (b×c),ab ≠0,b.c≠0
⇒(a.c) . b− (b.c)a = (a.c) . b− (a.b) . c
⇒(a.b) . c= (b.c)a⇒a∥c
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question209
Aparticlehastwovelocitiesofequalmagnitudeinclinedtoeachother
atanangleθ.Ifoneofthemishalved,theanglebetweentheotherand
theoriginalresultantvelocityisbisectedbythenewresultant.Thenθis
[2006]
Options:
A.90°
B.120°
C.45°
D.60°
Answer:B
Solution:
Weknowthath=1
2gt2andh+400 =1
2g(t+4)2
Subtracting,weget400 =8g +4gt
⇒t=8sec
∴h=1
2×10 ×64 =320m
∴Requiredheight = 320 +400 =720m
Lettwovelocitiesuanduatanangleθtoeachothertheresultantisgivenby
R2=u2+u2+2u2cos θ = 2u2(1+cos θ)
⇒R2=4u2cos2θ∕2orR=2u cos θ
2
Nowinsecondcase,thenewresultantAE (i.e.,R′ )bisects∠CAB,thereforeusinganglebisectortheoreminΔABC,we
get
AB
AC =BE
E C ⇒u
R=u∕2
u∕2⇒R=u
-------------------------------------------------------------------------------------------------
Question210
IfCisthemidpointofABandPisanypointoutsideAB,then
[2005]
Options:
A.PA +PB =2PC
B.PA +PB =PC
C.PA +PB +2PC =0
D.PA +PB +PC =0
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question211
Leta,bandcbedistinctnon-negativenumbers.Ifthevectors
a^
i+a^
j+c^
k,^
i+^
kandc^
i+c^
j+b^
klieinaplane,thencis
[2005]
Options:
A.theGeometricMeanofaandb
B.theArithmeticMeanofaandb
C.equaltozero
D.theHarmonicMeanofaandb
⇒2u cos θ
2=u
⇒cos θ
2=1
2=cos 60°⇒ θ
2=60°
orθ=120°
PA +AP =0andPC +CP =0
⇒PA +AC +CP =0........(i)
Similarly,PB +BC +CP =0......(ii)
Addingeqn.(i)and(ii),weget
PA +PB +AC +BC +2CP =0
SinceAC = −BC&CP = −PC
⇒PA +PB −2PC =0
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question212
Let→
a=^
i−^
k,→
b=x^
i+^
j+ (1−x)^
kand→
c=y^
i+x^
j+ (1+x−y)^
k.Then
→
a,→
b,→
cdependson
[2005]
Options:
A.onlyy
B.onlyx
C.bothxandy
D.neitherxnory
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question213
If→
a,→
b,→
carenoncoplanarvectorsandλisarealnumberthen
λ→
a+→
b λ2→
b λ→
c=→
a→
b+→
c→
bfor
[2005]
Options:
[ ]
[( ) ] [ ]
Vectora^
i+a^
j+c^
k,^
i+^
kandc^
i+c^
j+b^
karecoplanar
a a c
1 0 1
c c b
=0⇒c2=ab ⇒c= √ab
∴cisG.M.ofaandb.
| |
Giventhat→
a=^
i−^
k,→
b=x^
i+^
j+ (1−x)^
kand→
c=y^
i+x^
j+ (1+x−y)^
k
∴→
a→
b→
c=→
a.→
b×→
c=
1 0 −1
x 1 1−x
y x 1+x−y
=1[1+x−y−x+x2]−[x2−y]
=1−y+x2−x2+y=1
Hence →
a→
b→
cisindependentofxandyboth.
[ ] ( ) | |
[ ]
A.exactlyonevalueofλ
B.novalueofλ
C.exactlythreevaluesofλ
D.exactlytwovaluesofλ
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question214
Foranyvector→
a,thevalueof →
a×^
i
2+→
a×^
j
2+→
a×^
k
2isequalto
[2005]
Options:
A.3→
a2
B.→
a2
C.2→
a2
D.4→
a2
Answer:C
Solution:
( ) ( ) ( )
Let→
a=a1
^
i+a2
^
j+a3
^
k
→
b=b1
^
i+b2
^
j+b3
^
k→
c=c1
^
i+c2
^
j+c3
^
kGiventhat λ→
a+→
b λ2→
b λ→
c = →
a→
b+→
c→
b
λ(a1+b1)λ(a2+b2)λ(a3+b3)
λ2b1λ2b2λ2b3
λc1λc2λc3
=
a1a2a3
b1+c1b2+c2b3+c3
b1b2b3
⇒λ4
a1+b1a2+b2a3+b3
b1b2b3
c1c2c3
=
a1a2a3
b1+c1b2+c2b3+c3
b1b2b3
R1→R1−R2inlstdet.
andR2→R2−R3in2nddet.
⇒λ4
a1a2a3
b1b2b3
c1c2c3
=
a1a2a3
c1c2c3
b1b2b3
⇒λ4= −1
Henceλhasnorealvalues.
[( ) ] [ ]
| | | |
| | | |
| | | |
-------------------------------------------------------------------------------------------------
Question215
TheresultantRoftwoforcesactingonaparticleisatrightanglesto
oneofthemanditsmagnitudeisonethirdoftheotherforce.Theratio
oflargerforcetosmalleroneis:
[2005]
Options:
A.2:1
B.3 : √2
C.3:2
D.3 :2√2
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question216
AandBaretwolikeparallelforces.AcoupleofmomentH liesinthe
planeofAandBandiscontainedwiththem.TheresultantofAandB
aftercombiningisdisplacedthroughadistance
[2005]
Options:
A. 2H
A−B
B. H
A+B
Let→
a=x→
i+y→
j+z→
k
→
a×→
i=z→
j−y→
k⇒ →
a×i|2=y2+z2
Similarly, →
a×→
j2=x2+z2and →
a×→
k2=x2+y2
Addingallaboveequation
⇒→
a×→
i2+→
a×→
j2+→
a×→
k2
=2(x2+y2+z2) = 2→
a|2
|
| | | |
( ) ( ) ( )
|
AccordingtoquestionF′ = 3F cos θandF=3F sin θ
⇒F′ = 2√2F
⇒F:F′::3:2√2
C. H
2(A+B)
D. H
A−B
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question217
AparticleisprojectedfromapointOwithvelocityuatanangleof60°
withthehorizontal.Whenitismovinginadirectionatrightanglesto
itsdirectionatO,itsvelocitythenisgivenby
[2005]
Options:
A.u
3
B.u
2
C.2u
3
D. u
√3
Answer:D
Solution:
-------------------------------------------------------------------------------------------------
Question218
If→
a,→
b,→
carenon-coplanarvectorsandlambdaisarealnumber,thenthe
LetAandBbedisplacedbyadistancexthenChangeinmomentof(A+B) = appliedmoments
⇒(A+B) × x=H ⇒x=H
A+B
Asperquestionu cos 60° = v cos 30°
(ashorizontalcomponentofvelocityremainsthesame)
⇒u.1
2=v.√3
2orv=1
√3u
vectors→
a+2→
b+3→
c,λ→
b+4→
cand(2λ −1)→
carenoncoplanarfor
[2004]
Options:
A.novalueofλ
B.allexceptonevalueofλ
C.allexcepttwovaluesofλ
D.allvaluesofλ
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question219
Let→
a,→
band→
cbethreenon-zerovectorssuchthatnotwooftheseare
collinear.Ifthevector→
a+2→
biscollinearwith→
cand→
b+3→
ciscollinear
with→
a(λbeingsomenon-zeroscalar)then→
a+2→
b+6→
cequals
[2004]
Options:
A.0
B.λ→
b
C.λ→
c
D.λ→
a
Answer:C
Solution:
Ifvectors→
a+2→
b+3→
c,λ→
b+4→
cand(2λ −1)→
carecoplanarthen
1 2 3
0 λ 4
002λ −1
=0
⇒λ(2λ −1) = 0⇒λ=0or1
2
∴Forcesarenoncoplanarforallλ,exceptλ=0,1
2
| |
Giventhat→
a+2→
biscollinearwith→
cand→
b+3→
ciscollinearwith→
a
Let→
a+2→
b=t→
cand→
b+3→
c=s→
a,wheretandsarescalars
∴→
a+2→
b+6→
c=t→
c+6→
c
= (t+6)→
cusing→
a+2→
b=t→
c
=λ→
c,whereλ=t+6
[]
Question220
Let→
u,→
v,→
wbesuchthat →
u=1,→
v=2,→
w=3.Iftheprojection→
valong→
u
isequaltothatof→
walong→
uand→
v,→
wareperpendiculartoeachother
then →
u−→
v+→
wequals
[2004]
Options:
A.14
B.√7
C.√14
D.2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question221
Let→
a,→
band→
cbenon-zerovectorssuchthat →
a×→
b×→
c=1
3
→
b→
c→
a.Ifθ
istheacuteanglebetweenthevectors→
band→
c,thensin θequals
[2004]
Options:
A.2√2
3
B.√2
3
C.2
3
D.1
3
| | | | | |
| |
( ) | | | |
Projectionof→
valong→
u=
→
v.→
u
→
u
=→
v.→
u
projectionof→
walong→
u=
→
w.→
u
→
u
=→
w→
u
Given→
v.→
u=→
w.→
u.......(1)
Also,→
v.→
w=0∵→
v⊥→
w........(2)
Now →
u−→
v+→
w2
=→
u|2+→
v|2+→
w|2−2→
u.→
v−2→
v.→
w+2→
u.→
w
=1+4+9+0[From(1)and(2)] = 14
∴→
u−→
v+→
w= √14
| |
| |
[ ]
| |
| | |
| |
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question222
Ift1andt2arethetimesofflightoftwoparticleshavingthesameinitial
velocityuandrangeRonthehorizontal,thent1
2+t2
2isequalto
[2004]
Options:
A.1
B.4u2∕g2
C.u2∕2g
D.u2∕g
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question223
Giventhat →
a×→
b×→
c=1
3
→
b→
c→
a
Clearly→
aand→
barenoncollinear
⇒→
a.→
c→
b−→
b.→
c→
a = 1
3
→
b→
c→
a
Comparingbothside.
∴→
a.→
c=0and−→
b.→
c=1
3
→
b→
c
⇒cos θ =−1
3
∴sin θ =1−1
9=2√2
3
[θisacuteanglebetween→
band→
c
( ) | | | |
( ) ( ) | | | |
| | | |
√
]
Forsamehorizontalrangetheanglesofprojectionmustbeαandπ
2−α
∴t1=2u sin α
g....(i)
t2=
2u sin π
2−α
g = 2u cos α
g....(ii)
Squaringandaddingeqn.(i)and(ii),
∴t1
2+t2
2=4u2
g2
( )
Avelocity1
4m∕sisresolvedintotwocomponentsalongOAandOB
makingangles30°and45°respectivelywiththegivenvelocity.Thenthe
componentalongOBis
[2004]
Options:
A.1
8(√6− √2)m∕s
B.1
4(√3−1)m∕s
C.1
4m∕s
D.1
8m∕s
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question224
ApaticlemovestowardseastfromapointAtoapointBattherateof4
km/handthentowardsnorthfromBtoCattherateof5km/hr.IfAB=
12kmandBC=5km,thenitsaveragespeedforitsjourneyfromAtoC
andresultantaveragevelocitydirectfromAtoCarerespectively
[2004]
Options:
A.13
9km ∕hand17
9km ∕h
B.13
4km ∕hand17
4km ∕h
C.17
9km ∕hand13
9km ∕h
D.17
4km ∕hand13
4km ∕h
Answer:D
Solution:
Givenv=1
4m∕s,componentalongOB
=v sin 30°
sin(45° + 30°) =
14 ×1
2
√3+1
2√2
= √6− √2
8
-------------------------------------------------------------------------------------------------
Question225
Threeforces→
P,→
Qand→
RactingalongI A,I BandI C,whereI isthe
incentreofaΔABCareinequilibrium.Then→
P:→
Q:→
Ris
[2004]
Options:
A.cosec A
2:cosec B
2:cosec C
2
B.sin A
2:sin B
2:sin C
2
C.sec A
2:sec B
2:sec C
2
D.cos A
2:cos B
2:cos C
2
Answer:D
Solution:
Solution:
TimetakenbytheparticleincompletejourneyT=12
4+5
5=4hr
∴Averagespeed = 12 +5
4=17
4
Averagevelocity = 122+52
4=13
4
√
LetIisincentreofΔABC.
∴I A,I B,I CarebisectorsoftheanglesA,BandC.
Now∠BI C =180 −B
2−C
2=90° + A
2etc
ApplyingLami'stheorematI
P
sin 90° + A
2
=Q
sin 90° + B
2
= R
sin 90° + C
2
⇒P:Q:R=cos A
2:cos B
2:cos C
2
( ) ( ) ( )
Question226
InarightangleΔABC, ∠A=90°andsidesa,b,carerespectively,
5cm,4cmand3cm.Ifaforce→
Fhasmoments0,9and16inN cm.units
respectivelyaboutverticesA,BandC,thenmagnitudeof→
Fis
[2004]
Options:
A.9
B.4
C.5
D.3
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question227
Withtwoforcesactingatpoint,themaximumaffectisobtainedwhen
theirresultantis4N .Iftheyactatrightangles,thentheirresultantis
3N .Thentheforcesare
[2004]
Options:
A. 2 +1
2√3N and 2 −1
2√3N
( ) ( )
Since,themomentaboutAiszero,hence→
FpassesthroughA.TakingAasorigin.Letthelineofactionofforce
→
Fbey=mx.(seefigure)
MomentaboutB=3m
1+m2
→
F=9.......(1)
√| |
MomentaboutC=4
1+m2
→
F=16.
Dividing(1)by(2),weget
m=3
4⇒→
F=5N
√| |
| |
B.(2+ √3)N and(2− √3)N
C. 2 +1
2√2N and 2 −1
2√2N
D.(2+ √2)N and(2− √2)N
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question228
Aparticleisacteduponbyconstantforces4^
i+^
j−3^
kand3^
i+^
j−^
k
whichdisplaceitfromapoint^
i+2^
j+3^
ktothepoint5^
i+4^
j+^
k.The
workdoneinstandardunitsbytheforcesisgivenby
[2004]
Options:
A.15
B.30
C.25
D.40
Answer:D
Solution:
-------------------------------------------------------------------------------------------------
Question229
ConsiderpointsA,B,CandDwithpositionvectors
7^
i−4^
j+7^
k,^
i−6^
j+10^
k,−^
i−3^
j+4^
kand5^
i−^
j+5^
krespectively.Then
ABCDisa
( ) ( )
LetforcesbePandQ.thenP+Q=4....(1)
andP2+Q2=32....(2)
Solvingeqns.(1)and(2),wegettheforces
2+1
2√2Nand 2−1
2√2N
( ) ( )
Resultantofforces
→
F=4^
i+^
j−3^
k+3^
i+^
j−^
k
Displacement
→
d=5^
i+4^
j+^
k−^
i+2^
j+3^
k = 4^
i+2^
j−2^
k
∴ Workdone = →
F.→
d=28 +4+8=40
( )
[2003]
Options:
A.parallelogrambutnotarhombus
B.square
C.rhombus
D.None
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question230
If
aa21+a3
bb21+b3
cc21+c3
=0andvectors(1,a,a2),(1,b,b2)and(1,c,c2)are
non-coplanar,thentheproductabcequals
[2003]
Options:
A.0
B.2
C.-1
D.1
Answer:C
Solution:
Solution:
| |
GiventhatA= (7, −4,7), B= (1, −6,10), C= (−1, −3,4)
andD= (5, −1,5)
∴AB =(7−1)2+ (−4+6)2+ (7−10)2
= √36 +4+9=7
Similarly,BC =7,CD = √41,DA = √17
∴Noneoftheoptionsissatisfied.
√
Given
aa21+a3
bb21+b3
cc21+c3
=0
| |
-------------------------------------------------------------------------------------------------
Question231
Thevectors →
AB =3^
i+4^
k&→
AC =5^
i−2^
j+4^
karethesidesofatriangle
ABC.ThelengthofthemedianthroughAis
[2003]
Options:
A.√288
B.√18
C.√72
D.√33
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question232
→
a,→
b,→
care3vectors,suchthat→
a+→
b+→
c=0,→
a=1,→
b=2,→
c=3,then
→
a.→
b+→
b.→
c+→
c.→
aisequalto
| | | | | |
⇒
a a21
b b21
c c21
+
a a2a3
b b2b3
c c2c3
= 0
⇒(1+abc)
1 a a2
1 b b2
1 c c2
=0
Giventhat(1,a,a2), (1,b,b2)and(1,c,c2)arenon-coplanar
∴
1 a a2
1 b b2
1 c c2
≠0(givencondition)
∴1+abc =0⇒abc = −1
| | | |
| |
| |
GiventhatADismedianofΔABC.
∴AD =(3+5)^
i+ (0−2)^
j+ (4+4)^
k
2 = 4i −j+4k
|AD| = √16 +16 +1= √33
[2003]
Options:
A.1
B.0
C.-7
D.7
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question233
If→
u,→
vand→
warethreenon-coplanarvectors,then
→
u+→
v−→
w.→
u−→
v×→
v−→
wequals
[2003]
Options:
A.3→
u.→
v×→
w
B.0
C.→
u.→
v×→
w
D.→
u.→
w×→
v
Answer:C
Solution:
( ) ( ) ( )
( )
Giventhat→
a+→
b+→
c=0
⇒→
a+→
b+→
c.→
a+→
b+→
c=0
⇒→
a|2+→
b|2+→
c|2+2→
a.→
b+→
b.→
c+→
c.→
a=0
1+4+9+2→
a.→
b+→
b.→
c+→
c.→
a=0
→
a.→
b+→
b.→
c+→
c.→
a = −1−4−9
2= −7
( ) ( )
|| | ( )
( )
→
u+→
v−→
w.→
u−→
v×→
v−→
w
=→
u+→
v−→
w.→
u×→
v−→
u×→
w−→
v×→
v+→
v×→
w
=→
u+→
v−→
w.→
u×→
v−→
u×→
w+→
v×→
w∵→
v×→
v=0
=→
u.→
u×→
v−→
u.→
u×→
w+→
u.→
v×→
w+→
v.→
u×→
v−→
v.→
u×→
w+→
v.→
v×→
w−→
w.→
u×→
v
+→
w.→
u×→
w−→
w.→
v×→
w
Weknowthat →
a,→
a,→
b=0
=→
u.→
v×→
w−→
v.→
u×→
w−→
w.→
u×→
v
=→
u→
v→
w+→
v→
w→
u−→
w→
u→
v = →
u.→
v×→
w
( ) ( ) ( )
( ) ( )
( ) ( ) [ ]
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
[ ]
( ) ( ) ( )
[ ) ] [ ) ] [ ] ( )
Question234
AtetrahedronhasverticesatO(0,0,0), A(1,2,1)B(2,1,3)and
C(−1,1,2).ThentheanglebetweenthefacesOABandABCwillbe
[2003]
Options:
A.90°
B.cos−119
35
C.cos−117
31
D.30°
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question235
Let→
u=^
i+^
j,→
v=^
i−^
jand→
w=^
i+2^
j+3^
k.If^
nisaunitvectorsuchthat
→
u.^
n=0and→
v.^
n=0,then →
w.^
nisequalto
[2003]
Options:
A.3
B.0
C.1
D.2
Answer:A
( )
( )
| |
NormalvectorofthefaceOAB
=→
OA ×→
OB =
^
i^
j^
k
121
213
= 5^
i−^
j−3^
k
NormalvectorofthefaceABC
=→
AB ×→
AC =
^
i^
j^
k
1−12
−2−11
= ^
i−5^
j−3k
Anglebetweenthefaces=anglebetweentheirnormals
cos θ =5+5+9
√35√35 =19
35orθ=cos−119
35
| |
| |
| | ( )
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question236
LetR1andR2respectivelybethemaximumrangesupanddownan
inclinedplaneandRbethemaximumrangeonthehorizontalplane.
ThenR1,R,R2arein
[2003]
Options:
A.H.P
B.A.G..P
C.A.P
D.G..P.
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question237
Twoparticlesstartsimultaneouslyfromthesamepointandmovealong
twostraightlines,onewithuniformvelocity→
uandtheotherfromrest
withuniformacceleration→
f.Letαbetheanglebetweentheirdirections
ofmotion.Therelativevelocityofthesecondparticlew.r.t.thefirstis
Giventhat→
u.^
n=0and→
v.→
n=0
⇒→
nisperpendicularboth→
uand→
v,
∴^
n=
→
u×→
v
→
u→
v
^
n=
^
i^
j^
k
110
1−1 0
√2× √2
=−2^
k
2= −k
→
ω.^
n= | (i+2j +3k) . (−k) | = | −3| = 3
| | | |
| |
| |
Letβbetheinclinationoftheplanetothehorizontalandubethevelocityofprojectionoftheprojectile
WehaveR1=u2
g(1+sin β)andR2=u2
g(1−sin β)
Addingaboveequations
1
R1
+1
R2
=2g
u2or 1
R1
+1
R2
=2
R ∵R=u2
g
∴R1,R,R2areinH.P.
[ ]
leastafteratime
[2003]
Options:
A.u cos α
f
B.u sin α
f
C.f cos α
u
D.u sin α
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question238
Twostonesareprojectedfromthetopofacliffhmetreshigh,withthe
samespeedu,soastohitthegroundatthesamespot.Ifoneofthe
stonesisprojectedhorizontallyandtheotherisprojectedhorizontally
andtheotherisprojectedatanangleθtothehorizontalthentan θ
equals
[2003]
Options:
A.u 2
gh
√
Letthetwovelocitiesbe→
v1=u^
iand→
v2= (f t cos α)^
i+ (f t sin α)^
j
∴Relativevelocityofsecondwithrespecttofirst
→
v=→
v2−→
v1 = (f t cos α −u)^
i+f t sin α ^
j
⇒ →
v2= (f t cos α −u)2+ (f t sin α)2
=f2t2+u2−2uf t cos α
For →
vtobeminandmax.weshouldhave
d|v|2
d t =0⇒2f 2t−2uf cos α =0
⇒t=u cos α
f
Alsod2|v|2
d t2=2f 2= +ve
∴ | v|2andhence|v|isleastatthetimeu cos α
f
| |
| |
B. 2u
gh
C.2g u
h
D.2h u
g
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
√
√
√
Giventhatthestoneprojectedhorizontally.Forhorizontalmotion,
Distance=speed×time⇒R=ut
andforverticalmotion
h=0×t+1
2gt2
⇒t=2h
g
∴WegetR=u2h
g......(1)
Whenthestoneprojectedatanangleθ,forhorizontalandverticalmotions,wehave
√
√
R=u cos θ ×t....(2)
andh= −u sin θ ×t+1
2gt2....(3)
Fromeqns.(1)and(2)weget
u2h
g=u cos θ ×t
⇒t=1
cos θ
2h
g
Puttingthevalueoftineq(3)weget
h= −u sin θ
cos θ
2h
g+1
2g2h
gcos2θ
h= −u2h
gtan θ +hsec2θ
h= −u2h
gtan θ +htan2θ+h
tan2θ−u2
hg tan θ =0;∴tan θ =u2
hg
√
√
√[ ]
√
√
√ √
Question239
Abodytravelsadistancesintseconds.Itstartsfromrestandendsat
rest.Inthefirstpartofthejourney,itmoveswithconstantacceleration
f andinthesecondpartwithconstantretardationr.Thevalueoftis
givenby
[2003]
Options:
A. 2s 1
f+1
r
B.2s 1
f+1
r
C. √2s
1
f+1
r
D.√2s(f+r)
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question240
√( )
( )
LetthebodytravelsfromAtoBwithconstantaccelerationtandfromBtoCwithconstantretardationr.
IfAB =x,BC =y,timetakenfromAtoB=t1andtimetakenfromBtoC=t2,thens=x+yandt=t1+t2
ForthemotionfromAtoB
v2=u2+2f s⇒v2=2f x(∵u=0)
⇒x=v2
2f .......(1)
andv=u+f t ⇒v=f t1
⇒t1=v
f........(2)
ForthemotionfromBtoC
v2=u2+2f s
⇒0=v2−2ry⇒y=v2
2r.......(3)
andv=u+f t ⇒0=v−rt2
⇒t2=v
r
Addingequations(1)and(3),weget
x+y=v2
2
1
f+1
r=s
Addingequations(2)and(4),weget
t1+t2=v1
f+1
r=t
t2
2s =
v21
f+1
r
2
2×v2
2
1
f+1
r
= 1
f+1
r
t=2s 1
f+1
r
[ ]
[ ]
[ ]
( )
√( )
Theresultantofforces→
Pand→
Qis→
R.If→
Qisdoubledthen→
Risdoubled.If
thedirectionof→
Qisreversed,then→
Risagaindoubled.ThenP2:Q2:R2is
[2003]
Options:
A.2:3:1
B.3:1:1
C.2:3:2
D.1:2:3.
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question241
Acoupleisofmoment→
Gandtheforceformingthecoupleis→
P.If→
Pis
turnedthrougharightanglethemomentofthecouplethusformedis→
H
.Ifinstead,theforce→
Pareturnedthroughanangleα,thenthemoment
ofcouplebecomes
[2003]
Options:
A.→
Hsin α −→
Gcos α
B.→
Gsin α −→
Hcos α
C.→
Hsin α +→
Gcos α
D.→
Gsin α +→
Hcos α
Answer:C
Solution:
R2=P2+Q2+2PQ cos θ....(1)
When→
Qand→
Raredoubled
4R2=P2+4Q2+4PQ cos θ....(2)
When→
Qisreversedand→
Risdoubled
4R2=P2+Q2−2PQ cos θ....(3)
Adding(1)and(3), 5R2=2P2+2Q2
⇒2P2+2Q2−5R2=0....(4)
Applying(3) × 2+ (2), 12R2 = 3P2+6Q2
⇒3P2+6Q2−12R2=0....(5)
From(4)and(5) P2
−24 +30 =Q2
24 −15 =R2
12 −6
P2
6=Q2
9=R2
6orP2:Q2:R2=2:3:2
Solution:
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Question242
Aparticleactedonbyconstantforces4^
i+^
j−3^
kand3^
i+^
j−^
kis
displacedfromthepoint^
i+2^
j−3^
ktothepoint5^
i+4^
j+^
k.Thetotal
workdonebytheforcesis
[2003]
Options:
A.50units
B.20units
C.30units
D.40units.
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question243
If →
a=5,→
b=4,→
c=3thuswhatwillbethevalueof →
a.→
b+→
b.→
c+→
c.→
a,
giventhat→
a+→
b+→
c=0
[2002]
Options:
A.25
B.50
C.-25
D.-50
| | | | | | | |
Weknowthat→
G=→
r×→
p;→
G=→
r→
p sin θ
→
H=→
r→
p cos θ[∵sin(90° + θ) = cos θ]
G=→
r→
p sin θ.....(1)
H=→
r→
p cos θ.....(2)
x=→
r→
p sin(θ+α).....(3)
From(1), (2)&(3), x=→
G cos α +→
H sin α
| | | | | |
| | | | | |
| | | |
| | | |
| | | |
→
F=→
F1+→
F2 = 7i +2j −4k
→
d= PositonVectorof→
B−PositionVectorof→
A
=4i +2j −2k
W=→
F.→
d = 28 +4+8=40unit
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question244
Ifsdaa→
a,→
b,→
carevectorssuchthat→
a+→
b+→
c=0and
→
a=7,→
b=5,→
c=3thenanglebetweenvector→
band→
cis
[2002]
Options:
A.60°
B.30°
C.45°
D.90°
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question245
If→
a×→
b=→
b×→
c=→
c×→
athen→
a+→
b+→
c=
[2002]
Options:
A.abc
B.-1
C.0
| | | | | |
Giventhat→
a+→
b+→
c=0
⇒ →
a+→
b+→
c2=0
⇒→
a|2+→
b|2+→
c|2+2→
a.→
b+→
b.→
c+→
c.→
a=0⇒25 +16 +9+2→
a.→
b+→
b.→
c+→
c.→
a=0
⇒→
a.→
b+→
b.→
c+→
c.→
a= −25
∴→
a.→
b+→
b.→
c+→
c.→
a=25
| |
|| | ( ) ( )
( )
| |
Giventhat→
a+→
b+→
c=0⇒→
b+→
c= −→
a
⇒→
b+→
c|2=→
a|2=52+32+2→
b.→
c=72
⇒2→
b→
c cos θ =49 −34 =15;
⇒2×5×3 cos θ =15;
⇒cos θ =1∕2; ⇒θ=π
3=60°
| |
| | | |
D.2
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question246
→
a=3^
i−5^
jand→
b=6^
i+3^
jaretwovectorsand→
cisavectorsuchthat
→
c=→
a×→
bthen →
a:→
b:→
c
[2002]
Options:
A.√34 : √45 : √39
B.√34 : √45 :39
C.34:39:45
D.39:35:34
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question247
Ifthevectors→
c,→
a=x^
i+y^
j+z^
kand→
b=^
jaresuchthat→
a,→
cand→
bforma
righthandedsystemthen→
cis:
[2002]
Options:
| | | | | |
Let→
a+→
b+→
c=→
rThen→
a×→
a+→
b+→
c=→
a×→
r
⇒0+→
a×→
b+→
a×→
c=→
a×→
r
⇒→
a×→
b−→
c×→
a=→
a×→
r⇒→
a×→
r=→
0 ∵→
a×→
b=→
c×→
a
Similarly→
b×→
r=→
0&→
c×→
r=→
0
Abovethreeconditionscanbeholdifandonlyif→
r=→
0
( )
[ ]
Wehave→
a×→
b=
^
i^
j^
k
3−50
630
= 39^
k=→
c
Also →
a= √34,→
b= √45,→
c=39;
∴→
a:→
b:→
c = √34 : √45 :39
| |
| | | | | |
| | | | | |
A.z^
i−x^
k
B.→
0
C.y^
j
D.−z^
i+x^
k
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question248
If→
a,→
b,→
carevectorssuchthat →
a→
b→
c=4then →
a×→
b→
b×→
c→
c×→
a=
[2002]
Options:
A.16
B.64
C.4
D.8
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question249
If →
a=4,→
b=2andtheanglebetween→
aand→
bisπ ∕6then →
a×→
b2is
equalto
[ ] [ ]
| | | | ( )
Giventhat→
a,→
c,→
bformarighthandedsystem,
∴→
c=→
b×→
a=
^
i^
jk
0 1 0
x y z
=z^
i−x^
k
| |
→
a×→
b→
b×→
c→
c×→
a
→
a×→
b.→
b×→
c×→
c×→
a∵→
a×→
b×→
c=→
a.→
c→
b− →
a.→
b→
c
=→
a×→
b.→
m.→
a→
c−→
m.→
c→
awhere→
m=→
b×→
c
=
→
→
a×b.→
c.→
a.→
b×→
c = →
a→
b→
c2=42=16
[ ]
( ) { ( ) ( ) } ( ) ( ) ( )
( ) { ( ) ( ) ) } ( )
{( ) }{ ( ) } [ ]
[2002]
Options:
A.48
B.16
C.→
a
D.Noneofthese
Answer:B
Solution:
-------------------------------------------------------------------------------------------------
Question250
Abeadofweightwcanslideonsmoothcircularwireinaverticalplane.
Thebeadisattachedbyalightthreadtothehighestpointofthewire
andinequilibrium,thethreadistautandmakeanangleθwiththe
verticalthentensionofthethreadandreactionofthewireonthebead
are
[2002]
Options:
A.T =w cos θ R =w tan θ
B.T =2w cos θ R =w
C.T =w R =w sin θ
D.T =w sin θ R =w cot θ
Answer:B
Solution:
Since,→
a.→
b=→
a→
b cos π
6 = 4×2×√3
2=4√3.
Weknowthat,
→
a×→
b2+→
a.→
b2=→
a|2→
b|2
⇒→
a×→
b2+48 =16 ×4
⇒→
a×→
b2=16
| | | |
( ) ( ) | |
( )
( )
Fromfigureangle T Q W =180 −θ; ∠RQW =2θ;
∠RQT =180 −θ
-------------------------------------------------------------------------------------------------
Question251
Thesumoftwoforcesis18Nandresultantwhosedirectionisatright
anglestothesmallerforceis12N.Themagnitudeofthetwoforcesare
[2002]
Options:
A.13,5
B.12,6
C.14,4
D.11,7
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
ApplyingLami'stheorematQ.
T
sin 2 θ =R
sin(180 −θ) = W
sin(180 −θ)
⇒R=WandT=2W cos θ
GiventhatP+Q=18.......(1)
Weknowthat
P2+Q2+2PQ cos α =144......(2)
tan 90° = Q sin α
P+Q cos α
⇒P+Q cos α =0......(3)
From(2)and(3),
Q2−P2=144⇒(Q−P)(Q+P) = 144
Q−P=144
18 =8
From(1),Onsolving,wegetQ=13,P=5