12th Grade > Mathematics
VECTOR ALGEBRA MCQs
Total Questions : 60
| Page 5 of 6 pages
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 D. -> 163
:
D
[−−→AB−−→AC−−→AD]=∣∣
∣∣31−1404525∣∣
∣∣=3(0−8)−1(20−20)−1(8−0)=−24−0−8=−32
Volume of the tetrahedron =16(32)=163 cubic unit.
:
D
[−−→AB−−→AC−−→AD]=∣∣
∣∣31−1404525∣∣
∣∣=3(0−8)−1(20−20)−1(8−0)=−24−0−8=−32
Volume of the tetrahedron =16(32)=163 cubic unit.
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. -> 73(^i+2^j+2^k)
:
C
The required vectors
=|→b|^a=|→b||→a|→a=73(^i+2^j+2^k)
:
C
The required vectors
=|→b|^a=|→b||→a|→a=73(^i+2^j+2^k)
Answer: Option A. -> r.(^i+^j+^k)=2
:
A
−−→AB=−−→OB−−−→OA=(3^i+5^j−3^k)−(^i+3^j−5^k)=2^i+2^j+2^k
Midpoint of AB is (2, 4, -4)
Vector equation of the plane is [r−(2^i+4^j−4^k)].(2^i+2^j+2^k)=0
⇒r.(^i+^j+^k)=2+4−4⇒r.(^i+^j+^k)=2
:
A
−−→AB=−−→OB−−−→OA=(3^i+5^j−3^k)−(^i+3^j−5^k)=2^i+2^j+2^k
Midpoint of AB is (2, 4, -4)
Vector equation of the plane is [r−(2^i+4^j−4^k)].(2^i+2^j+2^k)=0
⇒r.(^i+^j+^k)=2+4−4⇒r.(^i+^j+^k)=2
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. -> are collinear
:
C
If A, B, C are the given points respectively, then
→OA=^i+2^j+3^k,→OB=3^i+4^j+7^k,→OC=−3^i−2^j−5^k,→AB=→OB−→OA=2^i+2^j+4^k,→AC=→OC−→OA=−4^i−4^j−8^k=−2→AB
∴→AB,→AC are collinear ⇒A,B,C are collinear.
:
C
If A, B, C are the given points respectively, then
→OA=^i+2^j+3^k,→OB=3^i+4^j+7^k,→OC=−3^i−2^j−5^k,→AB=→OB−→OA=2^i+2^j+4^k,→AC=→OC−→OA=−4^i−4^j−8^k=−2→AB
∴→AB,→AC are collinear ⇒A,B,C are collinear.
Answer: Option D. -> 1√2
:
D
Volume of the parallelepiped formed by →a′,→b′,→c′ is 4
∴ Volume of the parallelepiped formed by→a,→b,→c is 14
→b×→c=14→a′∴∣∣→b×→c∣∣=√24=12√2
∴ length of altitude = 14×2√2=1√2.
:
D
Volume of the parallelepiped formed by →a′,→b′,→c′ is 4
∴ Volume of the parallelepiped formed by→a,→b,→c is 14
→b×→c=14→a′∴∣∣→b×→c∣∣=√24=12√2
∴ length of altitude = 14×2√2=1√2.
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
Latest Videos
Latest Test Papers
Login With Google
Old way of Login
Login With Email