12th Grade > Mathematics
VECTOR ALGEBRA MCQs
Total Questions : 60
| Page 3 of 6 pages
Answer: Option C. -> (−43,0)
:
C
For the vectors ⃗a and ⃗bto be inclined at an obtuse angle, we must have
⃗a.⃗b<0 for all x∈(0,∞)
⇒c(log2x)2−12+6c(log2x)<0 for all x∈(0,∞)
⇒cy2+6cy−12<0 for all y∈R, where y=log2x
⇒c<0 and ⇒36c2+48c<0⇒c<0and c(3c+4)<0
⇒c<0 and −43<c<0
⇒c∈(−43,0)
:
C
For the vectors ⃗a and ⃗bto be inclined at an obtuse angle, we must have
⃗a.⃗b<0 for all x∈(0,∞)
⇒c(log2x)2−12+6c(log2x)<0 for all x∈(0,∞)
⇒cy2+6cy−12<0 for all y∈R, where y=log2x
⇒c<0 and ⇒36c2+48c<0⇒c<0and c(3c+4)<0
⇒c<0 and −43<c<0
⇒c∈(−43,0)
Answer: Option C. -> (−43,0)
:
A
⃗a+⃗b=<3,−2,1>+<−1,1,1>=<2,−1,2>⇒|⃗a+⃗b|=√4+1+4=√9=3
Unit vector parallel to ⃗a+⃗b is ±a+b|⃗a+⃗b|=±<2,−1,2>3
:
A
⃗a+⃗b=<3,−2,1>+<−1,1,1>=<2,−1,2>⇒|⃗a+⃗b|=√4+1+4=√9=3
Unit vector parallel to ⃗a+⃗b is ±a+b|⃗a+⃗b|=±<2,−1,2>3
Answer: Option B. -> 1 : 2
:
B
Ratio =-2-1: 1-7 =-3:-6=1:2
:
B
Ratio =-2-1: 1-7 =-3:-6=1:2
Answer: Option A. -> all values of x
:
A
⃗a,⃗b,⃗care coplanar, iff [⃗a⃗b⃗c]=0
We have, [⃗a⃗b⃗c]=∣∣
∣∣xx+1x+2x+3x+4x+5x+6x+7x+8∣∣
∣∣
=∣∣
∣∣xx+1x+2333666∣∣
∣∣[ApplyingR2→R2−R1,R3→R3−R1]
= 0 for all x[∵R1 and R2 are proportional]
:
A
⃗a,⃗b,⃗care coplanar, iff [⃗a⃗b⃗c]=0
We have, [⃗a⃗b⃗c]=∣∣
∣∣xx+1x+2x+3x+4x+5x+6x+7x+8∣∣
∣∣
=∣∣
∣∣xx+1x+2333666∣∣
∣∣[ApplyingR2→R2−R1,R3→R3−R1]
= 0 for all x[∵R1 and R2 are proportional]
Answer: Option A. -> -4
:
A
Since,⃗a=2^i+3^j+^k,⃗b=2^i+p^j+3^k and ⃗c=2^i+17^j+3^k are coplanar,
therefore [⃗a⃗b⃗c]=∣∣
∣∣2312p3217−3∣∣
∣∣=0⇒p=−4
:
A
Since,⃗a=2^i+3^j+^k,⃗b=2^i+p^j+3^k and ⃗c=2^i+17^j+3^k are coplanar,
therefore [⃗a⃗b⃗c]=∣∣
∣∣2312p3217−3∣∣
∣∣=0⇒p=−4
Answer: Option B. -> 4
:
B
⃗OA=2^i+^j+^k,⃗OB=6^i−^j+2^k,⃗OC=14^i−5^j+P^k⇒⃗AB=⃗OB−⃗OA=4^i−2^j+^k,⃗AC=⃗OC−⃗OA=12^i−6^j+(p−1)^k
A, B, C are collinear ⇒⃗AC=λ⃗AB⇒12^i−6^j+(p−1)^k=λ(4^i−2^j+^k)
⇒λ=3,p−1=3⇒p=4.
:
B
⃗OA=2^i+^j+^k,⃗OB=6^i−^j+2^k,⃗OC=14^i−5^j+P^k⇒⃗AB=⃗OB−⃗OA=4^i−2^j+^k,⃗AC=⃗OC−⃗OA=12^i−6^j+(p−1)^k
A, B, C are collinear ⇒⃗AC=λ⃗AB⇒12^i−6^j+(p−1)^k=λ(4^i−2^j+^k)
⇒λ=3,p−1=3⇒p=4.
Answer: Option A. -> linearly dependent
:
A
Box=∣∣
∣∣3−2−4−102−213∣∣
∣∣=3(0−2)+2(−3+4)−4(−1−0)=−6+2+4=0
∴Given vectors are coplanar Given vectors are linarly dependent.
:
A
Box=∣∣
∣∣3−2−4−102−213∣∣
∣∣=3(0−2)+2(−3+4)−4(−1−0)=−6+2+4=0
∴Given vectors are coplanar Given vectors are linarly dependent.
Answer: Option A. -> ±(53(^i−7^j+2^k)
:
A
The required vector →C is given by
→C=λ(^a+^b)=λ(→a|→a|+→b|→b|)=λ{19(7^i−4^j−4^k)+13(−2^i−^j+2^k)}
⇒→c=λ9(^i−7^j+2^k)⇒|→c|=±λ9√1+49+4=±λ9√54
But |→c|=5√6 (given)
∴±λ9√54=5√6⇒λ=±15
Hence, →c=±159(^i−7^j+2^k)=±53(^i−7^j+2^k)
:
A
The required vector →C is given by
→C=λ(^a+^b)=λ(→a|→a|+→b|→b|)=λ{19(7^i−4^j−4^k)+13(−2^i−^j+2^k)}
⇒→c=λ9(^i−7^j+2^k)⇒|→c|=±λ9√1+49+4=±λ9√54
But |→c|=5√6 (given)
∴±λ9√54=5√6⇒λ=±15
Hence, →c=±159(^i−7^j+2^k)=±53(^i−7^j+2^k)
Answer: Option A. -> 2^i+^j+^k√6
:
A
−−→OP=^i−^j+2^k,−−→OQ=2^i−^k,−−→OR=2^j+^k⇒−−→PQ=−−→OQ−−−→OP=^i+^j−3^k,−−→PR=−−→OR−−−→OP=−^i+3^j−^k
−−→PQ×−−→PR=∣∣
∣
∣∣^i^j^k11−3−13−1∣∣
∣
∣∣=8^i+4^j+4^k;|−−→PQ×−−→PR|=√64+16+16=√96=4√6
Required unit vectors = ±8^i+4^j+4^k4√6=±2^i+^j+^k√6
:
A
−−→OP=^i−^j+2^k,−−→OQ=2^i−^k,−−→OR=2^j+^k⇒−−→PQ=−−→OQ−−−→OP=^i+^j−3^k,−−→PR=−−→OR−−−→OP=−^i+3^j−^k
−−→PQ×−−→PR=∣∣
∣
∣∣^i^j^k11−3−13−1∣∣
∣
∣∣=8^i+4^j+4^k;|−−→PQ×−−→PR|=√64+16+16=√96=4√6
Required unit vectors = ±8^i+4^j+4^k4√6=±2^i+^j+^k√6
Answer: Option C. -> 12√3
:
C
|[abc]|=|a.(b×c)|=|a||b×c||cos(a,b×c)|=|a||b×c|=|a||b|c|sin(b,c)
=2.3.4sin(2π3)=24.√32=12√3.
:
C
|[abc]|=|a.(b×c)|=|a||b×c||cos(a,b×c)|=|a||b×c|=|a||b|c|sin(b,c)
=2.3.4sin(2π3)=24.√32=12√3.
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