Free Objective Test 01 Practice Test - 11th and 12th
Question 1
If the points A,B and C have position vectors (2,1,1), (6,-1,2) and (14,-5,P) respectively and if they are collinear, then P =
2
4
6
8
SOLUTION
Solution : 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.
Question 2
If →a=(3,−2,1), →b=(−1,1,1) then the unit vector parallel to the vector →a+→b is
SOLUTION
Solution : 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)
Question 3
The points 2^i−^j−^k,^i+^j+^k,2^i+2^j+^k,2^j+5^k are
SOLUTION
Solution : 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.
Question 4
The vectors 3a-2b-4c, -a+2c, -2a+b+3c are
SOLUTION
Solution : 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.
Question 5
The ratio in which ^i+2^j+3^k divides the join of −2^i+3^j+5^k and 7^i−^k is
SOLUTION
Solution : B
Ratio =-2-1: 1-7 =-3:-6=1:2
Question 6
The vector a^i+b^j+c^k is a bisector of the angle between the vectors ^i+^j and ^j+^k if
SOLUTION
Solution : 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
Question 7
If |a+b| = |a-b| then (a,b) =
SOLUTION
Solution : 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∘
Question 8
If the angle θ between the vectors a=2x2^i+4x^j+^k and b=7^i−2^j+x^k is such that 90∘ < θ < 180∘
then x lies in the interval:
SOLUTION
Solution : 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
Question 9
A unit vector perpendicular to the plane of a=2^i−6^j−3^k,b=4^i+3^j−^k is
SOLUTION
Solution : 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
Question 10
If a is any vector then (a×^i)2+(a×^j)2+(a×^k)2 =
SOLUTION
Solution : B
(a×^i)2+(a×^j)2+(a×^k)2=2a2
Question 11
The area of the parallelogram whose diagonals are ^i−3^j+2^k,−^i+2^j is
SOLUTION
Solution : 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
Question 12
The volume of the parallellopiped whose coterminal edges are 2^i−3^j+4^k,^i+2^j−2^k,3^i−^j+^k is
SOLUTION
Solution : C
Volume =| ∣∣ ∣∣2−3412−23−11∣∣ ∣∣|=|2(2−2)+3(1+6)+4(−1−6)|=|0+21−28|=7 cubic units
Question 13
If a, b, c are the position vectors of the vertices of an equilateral triangle whose orthocenter is at the origin, then
SOLUTION
Solution : 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
Question 14
If →a=^i+^j+^k,→b=2^i−^j+^k and →c=^i+x^j+y^k, are linearly dependent and |→c|=√3 then (x,y) is
SOLUTION
Solution : 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
Question 15
If →a ′=^i+^j,→b ′=^i+^j+2^k and →c ′=2^i+^j−^k. Then altitude of the parallelopiped formed by the vectors →a,→b,→c having base formed by →b and →c is (→a,→b,→c and →a′,→b′,→c′ are reciprocal system of vectors)
SOLUTION
Solution : 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.