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

### Question 1

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

#### SOLUTION

Solution :C

The perpendicular distance of the origin (0, 0, 0) from the plane x + y + z = p is ∣∣∣−p√1+1+1∣∣∣=|p|√3

If the coordinates of P are (x, y, z), then we must have

∣∣∣x+y+z−p√3∣∣∣=|p|√3⇒|x+y+z−p|=|p|

Which is satisfied by (c)

### Question 2

The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x2+y2+z2+4x−2y−6z=155 is

#### 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+9−327√144+16+9∣∣∣=33813=26

So the plane is outside the sphere and the required distance is equal to 26 - 13 = 13.

### Question 3

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

#### SOLUTION

Solution :A

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

### Question 4

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

#### 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)=12√2+2cosθ[Q cos θ=l1l2+m1m2+n1n2]=12√2(1+cosθ)=cos(θ2)

∴ d.c's of OC are

l1+l22(OC),m1+m22(OC),n1+n22(OC)

### Question 5

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

#### SOLUTION

Solution :C

Let the line segment beAB, then as given

ABcosα=3,ABcosβ=4,ABcosγ=5⇒AB2(cos2α+cos2β+cos2γ)=32+42+52AB=√9+16+25=5√2

where α,β and γ are the angles made by the line with the axes.

### Question 6

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

#### 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=p4n∵l2+m2+n2=1⇒p216x21+p216y21+p216z21=1

or x21y21+y21z21+z21x21=16/p2x21y21z21

∴ Locus is x2y2+y2z2+z2x2=16p2x2y2z2

∴k=16p2

### Question 7

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

#### SOLUTION

Solution :B

The straight line joining the points (1, 1, 2) and (3, -2, 1) is x−12=y−1−3=z−2−1=r (say)

∴ Point is (2r + 1, 1 -3r, 2 –r )

Which lies on 3x + 2y + z = 6

∴3(2r+1)+2(1−3r)+2−r=6

∴r=1

Required point is (3, -2, 1).

### Question 8

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

#### SOLUTION

Solution :A

Equation of planes passing through intersection of the planes 3x-y-4z=0 and x+3y+6=0 is, (3x−y−4z)+λ(x+3y+6)=0.....(i)

Given, distance of plane (i) from origin is 1.

∴6λ√(3+λ)2+(3λ−1)2+42=1

or 36λ2=10λ2+26 or λ=±1

Put the value of λ in (i),

∴(3x−y−4z)±(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.

### Question 9

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

#### 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.

### Question 10

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

#### SOLUTION

Solution :D

Here, d1=d cos(90∘−α)

d2=d cos(90∘−β)

and d3=d cos(90∘−γ)

∴d1=dsinα

d2=dsinβ

and d3=dsinγ

∴d21+d22+d23=kd2

⇒d2(sin2α+sin2β+sin2γ)=k(d)2

∴k=2

### Question 11

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

#### SOLUTION

Solution :B

cos2α+cos2β+cos2γ=2cos2α−1+2cos2β−1+2cos2γ−1=2(cos2α+cos2β+cos2γ)−3=2−3=−1

### Question 12

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

#### SOLUTION

Solution :A

Perpendicular distance of (x,y,z) from y axis is given by =√x2+z2.

=> Required distance =√32+52=√34.

### Question 13

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

#### 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=6√2√13

### Question 14

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

#### 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

a−29=b27=c22

∴ Equation of the required plane is

– 29(x – 2) + 27(y + 1) + 22(z) = 0

⇒−29x+27y+22z+85=0

⇒29x−27y−22z=85

### Question 15

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 :

#### SOLUTION

Solution :A

Line parallel to the line x2=y2=z−6 and passing through (1,-2, 3) is

x−12=y+23=z−3−6=r

A = (2r + 1, 3r – 2, – 6r + 3) is a point on this line.

A lies in the plane x – y + z = 5

∴2r+1−3r+2−6r+3=5⇒r=17∴A=(97,−117,157)

Distance of A from (1, –2, 3) is

√(97−1)2+(−117+2)2+(157−3)2=17√4+9+36=77=1