# Free Objective Test 02 Practice Test - 11th and 12th

The three points whose position vectors are ^i+2^j+3^k,3^i+4^j+7^k and3^i2^j5^k

A. form the vertices of an equilateral triangle
B. form the vertices of a right angled triangle
C. are collinear
D.  form the vertices of an isosceles triangle.

#### SOLUTION

Solution : 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^i2^j5^k,AB=OBOA=2^i+2^j+4^k,AC=OCOA=4^i4^j8^k=2AB
AB,AC are collinear A,B,C are collinear.

If a,b represent AB,BC respectively of a regular hexagon ABCDEF then CD,DE,EF,FA are

A. b-a, -a, -b, a-b
B. A-b, a, b, b-a
C. b-a, a, b, a-b
D. A-b,-a,-b, b-a

#### SOLUTION

Solution : A

ABCDEF is a regular hexagon
AD=2BC,ED=AB,FE=BC,FA=DC
Given AB=a,BC=b
Now AB+BC+CD=ADa+b+CD=2BC
CD=2b(a+b)=baDE=BA=AB=a,EF=CB=BC=bFA=DC=CD=(ba)=ab If A=(1,3,-5) and B=(3,5,-3), then the vector equation of the plane passing through the midpoint of AB and perpendicular to AB is

A. r.(^i+^j+^k)=2
B. r.(^i+^j^k)=2
C. r.(^i^j+^k)=2

D. None

#### SOLUTION

Solution : A

AB=OBOA=(3^i+5^j3^k)(^i+3^j5^k)=2^i+2^j+2^k
Midpoint of AB is (2, 4, -4)
Vector equation of the plane is [r(2^i+4^j4^k)].(2^i+2^j+2^k)=0
r.(^i+^j+^k)=2+44r.(^i+^j+^k)=2

A unit vector perpendicular to the plane determined by the points P(1,-1,2), Q(2,0,-1) and R(0,2,1) is

A. 2^i+^j+^k6
B. 2^i+^j+^k3
C. 2^i^j^k3
D. 2^i^j^k3

#### SOLUTION

Solution : A

OP=^i^j+2^k,OQ=2^i^k,OR=2^j+^kPQ=OQOP=^i+^j3^k,PR=OROP=^i+3^j^k
PQ×PR=∣ ∣ ∣^i^j^k113131∣ ∣ ∣=8^i+4^j+4^k;|PQ×PR|=64+16+16=96=46
Required unit vectors = ±8^i+4^j+4^k46=±2^i+^j+^k6

If x.a=0,x×b=c×b then x =

A. cc.ab.ab
B. cc.ab.aa
C. ac.ab.ab
D. bc.ab.ab

#### SOLUTION

Solution : A

x×b=c×b(xc)×b=0xc is parallel to b
xc=λb for some scalar λx=c+λb
x.a=0(c+λb).a=0c.a+λb.a=0λ=c.ab.ax=cc.ab.ab

If the vector a is perpendicular to b and c, |a|=2, |b|=3, |c|=4 and the angle between b and c is 2π3 then |[a b c ]| =

A. 24
B. 12
C. 123
D. 243

#### SOLUTION

Solution : C

|[abc]|=|a.(b×c)|=|a||b×c||cos(a,b×c)|=|a||b×c|=|a||b|c|sin(b,c)
=2.3.4sin(2π3)=24.32=123.

The volume of the tetrahedron with vertices at (1,2,3), (4,3,2), (5,2,7), (6,4,8) is

A. 223
B. 113
C. 13
D. 163

#### SOLUTION

Solution : D

[ABACAD]=∣ ∣311404525∣ ∣=3(08)1(2020)1(80)=2408=32
Volume of the tetrahedron =16(32)=163 cubic unit.

If a,b,c are linearly independent, then [2a+b, 2b+c, 2c+a][a, b, c]=

A. 9
B. 8
C. 7
D. None

#### SOLUTION

Solution : A

[2a+b,2b+c,2c+a][a,b,c]=∣ ∣210021102∣ ∣=2(40)1(01)=8+1=9

Let ABCD be a parallelogram whose diagonals intersect at P and let O be the origin, then OA+OB+OC+OD equals

A. OA
B. 2OP
C. 3OP
D. 4OP

#### SOLUTION

Solution : D

Since, the diagonals of a parallelogram bisect each other.  Therefore, P is the middle point of AC and BD both.
OA+OC=2OPandOB+OD=2OPOA+OB+OC+OD=4OP

If a=^i+2^j+2^k and b=3^i+6^j+2^k, then the vector in the direction of a and having magnitude as |b|, is

A. 7(^i+2^j+2^k)
B. 79(^i+2^j+2^k)
C. 73(^i+2^j+2^k)
D. None of these

#### SOLUTION

Solution : C

The required vectors
=|b|^a=|b||a|a=73(^i+2^j+2^k)

The value of b such that the scalar product of the vector ^i+^j+^k with the unit vector parallel to the sum of the vectors 2^i+4^j+5^k and b^i+2^j+3^k is one, is

A. -2
B. -1
C. 0
D. 1

#### SOLUTION

Solution : D

The unit vector parallel to the sum of the vectors 2^i+4^j5^k and b^i+2^j+3^k is
^n=(2+b)^i+6^j2¨k(2+b)2+62+(2)2=(2+b)^i+6^j2¨kb2+4b+44
Now,(^i+^j+^k).^n=1
2+b+62=b2+4b+44b=1

The vector C, directed along the internal bisector of the angle between the vectors a=7^i4^j4^k and b=2^i^j+2^k with |c|=56, is

A.  ±(53(^i7^j+2^k)
B. 53(5^i5^j+2^k)

C. 53(^i7^j+2^k)

D. 53(5^i5^j+2^k)

#### SOLUTION

Solution : A

The required vector C is given by
C=λ(^a+^b)=λ(a|a|+b|b|)=λ{19(7^i4^j4^k)+13(2^i^j+2^k)}
c=λ9(^i7^j+2^k)|c|=±λ91+49+4=±λ954
But |c|=56 (given)
±λ954=56λ=±15
Hence, c=±159(^i7^j+2^k)=±53(^i7^j+2^k)

If the vectors a=(clog2x)^i6^j+2^k and b=(log2x)^i+2^j+3(clog2x)^k make an obtuse angle for any x(0,) then c belongs to

A. (,0)
B. (,43)
C. (43,0)
D. (43,)

#### SOLUTION

Solution : C

For the vectors a and b to be inclined at an obtuse angle, we must have
a.b<0 for all x(0,)
c(log2x)212+6c(log2x)<0 for all x(0,)
cy2+6cy12<0 for all yR, where y=log2x
c<0 and 36c2+48c<0c<0and c(3c+4)<0
c<0 and 43<c<0
c(43,0)

The vectors a=x^i+(x+1)^j+(x+2)^k,b=(x+3)^i+(x+4)^j+(x+5)^k and c=(x+6)^i+(x+7)^j+(x+8)^k are coplanar for

A. all values of x
B. x < 0
C. x > 0
D. None of these

#### SOLUTION

Solution : A

a,b,c are coplanar, iff [a b c]=0
We have, [a b c]=∣ ∣xx+1x+2x+3x+4x+5x+6x+7x+8∣ ∣
=∣ ∣xx+1x+2333666∣ ∣[Applying R2 R2 R1,R3 R3 R1]
= 0 for all x [R1 and R2 are proportional]

If a=2^i+3^j+^k,b=2^i+p^j+3^k and c=2^i+17^j+3^k are coplanar vectors, then the value of p is

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

#### SOLUTION

Solution : A

Since,a=2^i+3^j+^k,b=2^i+p^j+3^k and c=2^i+17^j+3^k are coplanar,
therefore [abc]=∣ ∣2312p32173∣ ∣=0p=4