12th Grade > Mathematics
VECTOR ALGEBRA MCQs
Total Questions : 60
| Page 6 of 6 pages
Answer: Option B. -> 12√21sq.units
:
B
Vector area =12(a×b)=12∣∣
∣
∣∣^i^j^k1−32120∣∣
∣
∣∣=12[(0−4)−^j(0+2)+^k(2−3)]=12(−4^i−2^j−^k)
Area =12√16+4+1=12√21sq. units
:
B
Vector area =12(a×b)=12∣∣
∣
∣∣^i^j^k1−32120∣∣
∣
∣∣=12[(0−4)−^j(0+2)+^k(2−3)]=12(−4^i−2^j−^k)
Area =12√16+4+1=12√21sq. units
Answer: Option B. -> a=c
:
B
a^i+b^j+c^k=Angle bisector of ^i+^j and ^j+^k=m [^i+^j+^j+^k√2]⇒a=m√2,b=√2.m,c=m√2
:
B
a^i+b^j+c^k=Angle bisector of ^i+^j and ^j+^k=m [^i+^j+^j+^k√2]⇒a=m√2,b=√2.m,c=m√2
Answer: Option B. -> 2a2
:
B
(a×^i)2+(a×^j)2+(a×^k)2=2a2
:
B
(a×^i)2+(a×^j)2+(a×^k)2=2a2
Answer: Option D. -> π2
:
D
|a+b|=|a−b|⇒|a+b|2=|a−b|2⇒(a+b)2=(a−b)2⇒a2+b2+2a.b=a2+b2−2a.b⇒4a.b=0⇒a.b=0⇒(a,b)=90∘
:
D
|a+b|=|a−b|⇒|a+b|2=|a−b|2⇒(a+b)2=(a−b)2⇒a2+b2+2a.b=a2+b2−2a.b⇒4a.b=0⇒a.b=0⇒(a,b)=90∘
Answer: Option C. -> 7
:
C
Volume =| ∣∣
∣∣2−3412−23−11∣∣
∣∣|=|2(2−2)+3(1+6)+4(−1−6)|=|0+21−28|=7 cubic units
:
C
Volume =| ∣∣
∣∣2−3412−23−11∣∣
∣∣|=|2(2−2)+3(1+6)+4(−1−6)|=|0+21−28|=7 cubic units
Answer: Option A. -> (23,−13,23)
:
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=(23,−13,23)
:
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=(23,−13,23)
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 B. -> coplanar but not collinear
:
B
⃗AB=−^i+2^k,⃗AC=^j+2^k,⃗AD=−2^i+^j+6^k,
A, B, C, D are not collinear.
Box=∣∣
∣∣−102012−216∣∣
∣∣=−1(6−2)+2(0+2)=−4+4=0.
∴ A, B, C, D are coplannar.
:
B
⃗AB=−^i+2^k,⃗AC=^j+2^k,⃗AD=−2^i+^j+6^k,
A, B, C, D are not collinear.
Box=∣∣
∣∣−102012−216∣∣
∣∣=−1(6−2)+2(0+2)=−4+4=0.
∴ A, B, C, D are coplannar.
Answer: Option A. -> (0,12)
:
A
90∘<θ<180∘⇒a.b<0⇒(2x2^i+4x^j+^k).(7^i−2^j+x^k)<0⇒14x2−8x+x<0⇒14x2−7x<0⇒7x(2x−1)<0⇒0<x<12
:
A
90∘<θ<180∘⇒a.b<0⇒(2x2^i+4x^j+^k).(7^i−2^j+x^k)<0⇒14x2−8x+x<0⇒14x2−7x<0⇒7x(2x−1)<0⇒0<x<12
Answer: Option C. -> 3^i−2^j+6^k7
:
C
a×b=∣∣
∣
∣∣^i^j^k2−6−343−1∣∣
∣
∣∣=^i(6+9)−^j(−2+12)+^k(6+24)=15^i−10^j+30^k
|a×b|=√225+100+900=35
Unit vector normal to the plane = 15^i−10^j+30^k35=3^i−2^j+6^k7
:
C
a×b=∣∣
∣
∣∣^i^j^k2−6−343−1∣∣
∣
∣∣=^i(6+9)−^j(−2+12)+^k(6+24)=15^i−10^j+30^k
|a×b|=√225+100+900=35
Unit vector normal to the plane = 15^i−10^j+30^k35=3^i−2^j+6^k7
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