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

If the perpendicular distance of a point other than the origin from the plane x+y+z=p  is equal to the distance of the plane from the origin, then the coordinates of the point  are

A. (p, 2 p, 0)
B. (0, 2p, -p)
C. (2p, p, -p)
D. (2p, -p, 2p)

#### SOLUTION

Solution : C

The perpendicular distance of the origin (0, 0, 0) from the plane x + y + z = p is p1+1+1=|p|3
If the coordinates of P are (x, y, z), then we must have
x+y+zp3=|p|3|x+y+zp|=|p|
Which is satisfied by (c)

The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x2+y2+z2+4x2y6z=155 is

A. 1134
B. 13
C. 39
D. 26

#### SOLUTION

Solution : B

The centre of the sphere is (-2, 1, 3) and its radius is 4+1+9+155=13.
Length of the perpendicular from the centre of the sphere on the plane is 24+4+9327144+16+9=33813=26
So the plane is outside the sphere and the required distance is equal to 26 - 13 = 13.

The ratio in which yz-  plane divides the segment joining the points (-2, 4, 7) and (3, -5, 8) is

A. 2 : 3
B. 3 : 2
C. 4 : 5
D. -7 : 8

#### SOLUTION

Solution : A

let YZ-plane divide the segment joining (-2, 4, 7) and (3, -5, 8) in the ratio λ:1. Then 3λ2λ+1=0λ=23 and the required ratio is 2 : 3

If (l1,m1,n1) and (l2,m2,n2,) are d.c.'s of ¯¯¯¯¯¯¯¯¯¯OA, ¯¯¯¯¯¯¯¯OB such that AOB=θ where ‘O’ is the origin,  then the d.c.’s of the internal bisector of the angle AOB  are

A. l1+l22sinθ2,m1+m22sinθ2,n1+n22sinθ2
B. l1+l22cosθ2,m1+m22cosθ2,n1+n22cosθ2
C. l1l22sinθ2,m1m22sinθ2,n1n22sinθ2
D. l1l22cosθ2,m1m22cosθ2,n1n22cosθ2

#### SOLUTION

Solution : B

Let OA and OB be two lines with d.c’s l1,m1,n1 and l2,m2,n2,. Let OA = OB = 1. Then, the coordinates of A and B are (l1,m1,n1) and (l2,m2,n2), respectively. Let OC be the bisector of AOB .Then, C is the mid point of AB and so its coordinates are (l1+l22,m1+m22,n1+n22).
d.r's of OC are l1+l22,m1+m22,n1+n22
We have, OC=(l1+l22)2+(m1+m22)2+(n1+n22)2=12(l21+m21+n21)+(l22+m22+n22)+(l1l2+m1m2+n1n2)=122+2cosθ[Q cos θ=l1l2+m1m2+n1n2]=122(1+cosθ)=cos(θ2)

d.c's of OC are
l1+l22(OC),m1+m22(OC),n1+n22(OC)

The projection of any line on co-ordinate axes be respectively 3, 4, 5 then its length is

A. 12
B. 50
C. 52
D. 25

#### SOLUTION

Solution : C

Let the line segment be AB, then as given
ABcosα=3,ABcosβ=4,ABcosγ=5AB2(cos2α+cos2β+cos2γ)=32+42+52AB=9+16+25=52
where α,β and γ are the angles made by the line with the axes.

A variable plane which remains at a constant distance p from the origin cuts the coordinate axes in A, B, C. The locus of the centroid of the tetrahedron OABC is y2z2+z2x2+x2y2=kx2y2z2, where k is equal to

A. 9p2
B. 9p2
C. 7p2
D. 16p2

#### SOLUTION

Solution : D

Let equation of plane is
lx+my+nz=p
or x(pl)+y(pm)+z(pn)=1
Coordinates of A, B, C are (pl,0,0)(0,pm,0) and
(0,0,pn) respectively.
Centroid of OABC is (p4l,p4m,p4n)
x1=p4l,y1=p4m,z1=p4nl2+m2+n2=1p216x21+p216y21+p216z21=1
or x21y21+y21z21+z21x21=16/p2x21y21z21
Locus is x2y2+y2z2+z2x2=16p2x2y2z2
k=16p2

The line joining the points (1, 1, 2) and (3, -2, 1) meets the plane 3x + 2y + z = 6 at the point

A. (1, 1, 2)
B. (3, -2, 1)
C. (2, -3, 1)
D. (3, 2, 1)

#### SOLUTION

Solution : B

The straight line joining the points (1, 1, 2) and (3, -2, 1) is x12=y13=z21=r (say)
Point is (2r + 1, 1 -3r, 2 –r )
Which lies on 3x + 2y + z = 6
3(2r+1)+2(13r)+2r=6
r=1
Required point is (3, -2, 1).

The equation of the plane passing through the line of intersection of the planes 3x-y-4z=0 and x+3y+6=0 whose distance from the origin is 1, is

A. x -2y -2z-3 =0, 2x+y -2z+3 =0
B. x -2y+2z-3 =0, 2x+y+2z+3 =0
C. x+2y-2z-3=0, 2x-y-2z+3 =0
D. None of these

#### SOLUTION

Solution : A

Equation of planes passing through intersection of the planes 3x-y-4z=0 and x+3y+6=0 is, (3xy4z)+λ(x+3y+6)=0.....(i)
Given, distance of plane (i) from origin is 1.
6λ(3+λ)2+(3λ1)2+42=1
or 36λ2=10λ2+26 or λ=±1
Put the value of λ in (i),
(3xy4z)±(x+3y+6)=0
or 4x+2y -4z+6=0 i.e 2x+y-2z+3 =0
and 2x-4y-4z-6=0
Thus the required planes are x-2y-2z-3=0 and 2x+y-2z+3=0.

The equation of the plane passing through the points (1,-3,-2) and perpendicular to planes x +2y+2z=5 and 3x+3y+2z=8, is

A. 2x -4y +3z-8 =0
B. 2x-4y-3z+8 =0
C. 2x+4y+3z+8 =0
D. x-3y+z-5=0

#### SOLUTION

Solution : A

l+2m+2n=0, 3l+3m+2n=0, l2+m2+n2=1, we get l, m, n from these equations and then putting the values in l(x-1)+m(y+3) +n(z+2)=0, we get the required result.
Trick: Checking conversely,
2(x)-4(y)+3(z)-8 =0,
So, it passes through given point.
1(2)+2(-4)+2(3) =0,
So, it is perpendicular to x+2y+2z=5.
3(2)+3(-4)+2(3)=0,
So, it is perpendicular to 3x+3y +2z=8.

The distance between two points P and Q is d and the length of their projections of PQ on the coordinate planes are d1,d2,d3. Then ​d21+d22+d23=kd2, where k is

A. 1
B. 5
C. 3
D. 2

#### SOLUTION

Solution : D

Here, d1=d cos(90α)
d2=d cos(90β)
and  d3=d cos(90γ)
d1=dsinα
d2=dsinβ
and   d3=dsinγ
d21+d22+d23=kd2
d2(sin2α+sin2β+sin2γ)=k(d)2
k=2

If a line makes the angle α,β,γ with three dimensional co-ordinate axes respectively, then cos2α+cos2β+cos2γ

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

#### SOLUTION

Solution : B

cos2α+cos2β+cos2γ=2cos2α1+2cos2β1+2cos2γ1=2(cos2α+cos2β+cos2γ)3=23=1

Perpendicular distance of the point (3, 4, 5) from the y-axis, is

A. 34
B. 41
C. 4
D. 5

#### SOLUTION

Solution : A

Perpendicular distance of (x,y,z) from y axis is given by =x2+z2.
=> Required distance =32+52=34.

The perpendicular distance of the origin form the plane which makes intercepts 12, 3 and 4 on x, y, z axes respectively, is

A. 13
B. 11
C. 17
D. 6213

#### SOLUTION

Solution : D

Let equation of plane be lx + my + nz = p, where p is the perpendicular distance of the plane from origin and (l,m,n) are the direction cosines of the normal to the plane
Or  x(pl)+y(pm)+z(pn)=1
According to question,
pl=12,pm=3,pn=4
or  p12=l,p3=m,p4=n
or  p2144+p29+p216=l2+m2+n2=1
p2(1144+19+116)=1p2(1+16+9144)=1
Or p2=14426=7213
p=6213

The equation of the plane passing through the points (2, –4, 0), (3, –4, 5) and parallel to the line 2x = 3y = 4z is :

A. 125x – 90y – 79z = 340
B. 32x – 21y – 36z = 85
C. 73x + 61y – 22z = 85
D. 29x – 27y – 22z = 85

#### SOLUTION

Solution : D

Equation of the plane through the point (2, –1, 0) is
a(x – 2) + b(y + 1) + c(z – 0) = 0  . . . (1)
It passes through the point (3, –4, 5)
a – 3b + 5c = 0  . . . (2)
The plane (1) is parallel to the line
2x = 3y = 4z i.e.,x6=y4=z3...(3)
Normal to the plane (1) is perpendicular to the line (3).
6a + 4b + 3c = 0      . . .(4)
From (2) and (4), we get
a29=b27=c22
Equation of the required plane is
– 29(x – 2) + 27(y + 1) + 22(z) = 0
29x+27y+22z+85=0
29x27y22z=85

The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line (12)x=(13)y=(16)z is :

A. 1
B. 2
C. 12
D. 4

#### SOLUTION

Solution : A

Line parallel to the line x2=y2=z6 and passing through (1,-2, 3) is
x12=y+23=z36=r
A = (2r + 1, 3r – 2, – 6r + 3) is a point on this line.
A lies in the plane x – y + z = 5
2r+13r+26r+3=5r=17A=(97,117,157)
Distance of A from (1, –2, 3) is
(971)2+(117+2)2+(1573)2=174+9+36=77=1