12th Grade > Mathematics
VECTOR ALGEBRA MCQs
Total Questions : 60
| Page 1 of 6 pages
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 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 D. -> 1
:
D
The unit vector parallel to the sum of the vectors 2^i+4^j−5^k and b^i+2^j+3^k is
^n=(2+b)^i+6^j−2¨k√(2+b)2+62+(−2)2=(2+b)^i+6^j−2¨k√b2+4b+44
Now,(^i+^j+^k).^n=1
⇒2+b+6−2=√b2+4b+44⇒b=1
:
D
The unit vector parallel to the sum of the vectors 2^i+4^j−5^k and b^i+2^j+3^k is
^n=(2+b)^i+6^j−2¨k√(2+b)2+62+(−2)2=(2+b)^i+6^j−2¨k√b2+4b+44
Now,(^i+^j+^k).^n=1
⇒2+b+6−2=√b2+4b+44⇒b=1
Answer: Option A. -> 9
:
A
[2a+b,2b+c,2c+a][a,b,c]=∣∣
∣∣210021102∣∣
∣∣=2(4−0)−1(0−1)=8+1=9
:
A
[2a+b,2b+c,2c+a][a,b,c]=∣∣
∣∣210021102∣∣
∣∣=2(4−0)−1(0−1)=8+1=9
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 D. -> 1
:
D
The unit vector parallel to the sum of the vectors 2^i+4^j−5^k and b^i+2^j+3^k is
^n=(2+b)^i+6^j−2¨k√(2+b)2+62+(−2)2=(2+b)^i+6^j−2¨k√b2+4b+44
Now,(^i+^j+^k).^n=1
⇒2+b+6−2=√b2+4b+44⇒b=1
:
D
The unit vector parallel to the sum of the vectors 2^i+4^j−5^k and b^i+2^j+3^k is
^n=(2+b)^i+6^j−2¨k√(2+b)2+62+(−2)2=(2+b)^i+6^j−2¨k√b2+4b+44
Now,(^i+^j+^k).^n=1
⇒2+b+6−2=√b2+4b+44⇒b=1
Answer: Option A. -> all values of x
:
A
→a,→b,→care coplanar, iff [→a→b→c]=0
We have, [→a→b→c]=∣∣
∣∣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 [→a→b→c]=0
We have, [→a→b→c]=∣∣
∣∣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. -> 9
:
A
[2a+b,2b+c,2c+a][a,b,c]=∣∣
∣∣210021102∣∣
∣∣=2(4−0)−1(0−1)=8+1=9
:
A
[2a+b,2b+c,2c+a][a,b,c]=∣∣
∣∣210021102∣∣
∣∣=2(4−0)−1(0−1)=8+1=9
Answer: Option A. -> (1,1)
:
A
Given that the three vectors are linearly dependent so
→c=l→a+m→b
⇒l+2m=1
l−m=x
⇒x=3y−2
l+m=y
Also, x2+y2+1=3
10y2−12y+2=0
⇒y=1,15
x=1,−75
:
A
Given that the three vectors are linearly dependent so
→c=l→a+m→b
⇒l+2m=1
l−m=x
⇒x=3y−2
l+m=y
Also, x2+y2+1=3
10y2−12y+2=0
⇒y=1,15
x=1,−75
Answer: Option A. -> ⃗a+⃗b+⃗c=⃗0
:
A
The position vector of the centroid of the triangle is ⃗a+⃗b+⃗c3
Since, the triangle is an equilateral, therefore the orthocenter coincides with the centroid and hence ⃗a+⃗b+⃗c3=⃗0⇒⃗a+⃗b+⃗c=0
:
A
The position vector of the centroid of the triangle is ⃗a+⃗b+⃗c3
Since, the triangle is an equilateral, therefore the orthocenter coincides with the centroid and hence ⃗a+⃗b+⃗c3=⃗0⇒⃗a+⃗b+⃗c=0
Latest Videos
Latest Test Papers
Login With Google
Old way of Login
Login With Email