# Free Objective Test 01 Practice Test - 11th and 12th

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 =

A.

2

B.

4

C.

6

D.

8

#### SOLUTION

Solution : B

OA=2^i+^j+^k,OB=6^i^j+2^k,OC=14^i5^j+P^kAB=OBOA=4^i2^j+^k,AC=OCOA=12^i6^j+(p1)^k
A, B, C are collinear AC=λAB12^i6^j+(p1)^k=λ(4^i2^j+^k)
λ=3,p1=3p=4.

If a=(3,2,1), b=(1,1,1) then the unit vector parallel to the vector a+b is

A. (23,13,23)
B. (25,15,25)
C. (23,13,23)
D. (23,13,23)

#### 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+ba+b=±(2,1,2)3=(23,13,23)

The points 2^i^j^k,^i+^j+^k,2^i+2^j+^k,2^j+5^k are

A. collinear
B. coplanar but not collinear
C. noncoplanar
D. none

#### SOLUTION

Solution : B

A, B, C, D are not collinear.
Box=∣ ∣102012216∣ ∣=1(62)+2(0+2)=4+4=0.
A, B, C, D are coplannar.

The vectors 3a-2b-4c, -a+2c, -2a+b+3c are

A. linearly dependent
B. linearly independent
C. collinear
D. none

#### SOLUTION

Solution : A

Box=∣ ∣324102213∣ ∣=3(02)+2(3+4)4(10)=6+2+4=0
Given vectors are coplanar Given vectors are linarly dependent.

The ratio in which ^i+2^j+3^k divides the join of 2^i+3^j+5^k and 7^i^k is

A. -3 : 2
B. 1 : 2
C. 2 : 3
D. -4 : 3

#### SOLUTION

Solution : B

Ratio =-2-1: 1-7 =-3:-6=1:2

The vector a^i+b^j+c^k is a bisector of the angle between the vectors ^i+^j and ^j+^k if

A. a=b
B. a=c
C. c=a+b
D. a =b=c

#### SOLUTION

Solution : B

a^i+b^j+c^k=Angle bisector of ^i+^j and ^j+^k=m [^i+^j+^j+^k2]a=m2,b=2.m,c=m2

If |a+b| = |a-b| then (a,b) =

A. π6
B. π4
C. π3
D. π2

#### SOLUTION

Solution : D

|a+b|=|ab||a+b|2=|ab|2(a+b)2=(ab)2a2+b2+2a.b=a2+b22a.b4a.b=0a.b=0(a,b)=90

If the angle θ between the vectors a=2x2^i+4x^j+^k and b=7^i2^j+x^k is such that 90 < θ < 180
then x lies in the interval:

A. (0,12)
B. (12,1)
C. (1,32)
D. (12,32)

#### SOLUTION

Solution : A

90<θ<180a.b<0(2x2^i+4x^j+^k).(7^i2^j+x^k)<014x28x+x<014x27x<07x(2x1)<00<x<12

A unit vector perpendicular to the plane of a=2^i6^j3^k,b=4^i+3^j^k is

A. 4^i+3^j^k26
B. 2^i6^j3^k7
C. 3^i2^j+6^k7
D. 2^i3^j6^k7

#### SOLUTION

Solution : C

a×b=∣ ∣ ∣^i^j^k263431∣ ∣ ∣=^i(6+9)^j(2+12)+^k(6+24)=15^i10^j+30^k
|a×b|=225+100+900=35
Unit vector normal to the plane = 15^i10^j+30^k35=3^i2^j+6^k7

If a is any vector then (a×^i)2+(a×^j)2+(a×^k)2 =

A. a2
B. 2a2
C. 3a2
D. 4a2

#### SOLUTION

Solution : B

(a×^i)2+(a×^j)2+(a×^k)2=2a2

The area of the parallelogram whose diagonals are ^i3^j+2^k,^i+2^j is

A. 429sq.units
B. 1221sq.units
C. 103sq.units
D. 12270sq.units

#### SOLUTION

Solution : B

Vector area =12(a×b)=12∣ ∣ ∣^i^j^k132120∣ ∣ ∣=12[(04)^j(0+2)+^k(23)]=12(4^i2^j^k)
Area =1216+4+1=1221sq. units

The volume of the parallellopiped whose coterminal edges are 2^i3^j+4^k,^i+2^j2^k,3^i^j+^k is

A. 5
B. 6
C. 7
D. 8

#### SOLUTION

Solution : C

Volume =| ∣ ∣234122311∣ ∣|=|2(22)+3(1+6)+4(16)|=|0+2128|=7 cubic units

If a, b, c are the position vectors of the vertices of an equilateral triangle whose orthocenter is at the origin, then

A. a+b+c=0

B. a2=b2+c2

C. a+b=c
D. None of these

#### 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=0a+b+c=0

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

A. (1,1)
B. (2,0)
C. (15,75)
D. (75,35)

#### SOLUTION

Solution : A

Given that the three vectors are linearly dependent so
c=la+mb
l+2m=1
lm=x
x=3y2
l+m=y
Also, x2+y2+1=3
10y212y+2=0
y=1,15
x=1,75

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)

A. 1
B. 322
C. 16
D. 12

#### 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=14a b×c=24=122
length of altitude = 14×22=12.