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

### Question 1

The plane xa+yb+zc=1 meets the coordinate axes at A, B, C respectively. D and E are the mid-points of AB and AC respectively. Coordinates of the mid-point of DE are

#### SOLUTION

Solution :D

A(a,0,0),B(0,b,0),C(0,0,c),D(a2,b2,0),E(a2,0,c2)

So midpoint of DE is (a2,b4,c4).

### Question 2

The ratio in which the plane 2x - 1 = 0 divides the line joining (-2,4,7) and (3, -5, 8) is

#### SOLUTION

Solution :D

Let the required ratio be k:1, then the coordinates of the point which divides the join of the points (-2, 4, 7) and (3, -5, 8) in this ratio are given by (3k−2k+1,−5k+4k+1,8k+7k+1)

As this point lies on the plane 2x - 1 = 0.

(3k−2k+1=12⇒k=1) and thus the required ratio as 1:1.

### Question 3

If a plane passes through the point (1,1,1) and is perpendicular to the line x−13=y−10=z−14 then its perpendicular distance from the origin is

#### SOLUTION

Solution :C

The d.r of the normal to the plane is(3,0,4) The equation of the plane is 3x + 0y + 4z + d = 0 since it passes through (1, 1, 1) so; d = --7

Now distance of the plane3x+4z−7=0 from (0, 0, 0) is 7√32+42=75 units

### Question 4

If centroid of the tetrahedron , whose vertices are given by (0,0,0), (*a*, 2, 3),(1, *b*, 2) and (2, 1, *c*) be (1, 2, –1), then distance of P(a,b,c) from origin is equal to

#### SOLUTION

Solution :A

Centroid ≡(∑x4,∑y4,∑z4)=(1,2,−1)

⇒a=1,b=5,c=−9;∴√a2+b2+c2=√107.

### Question 5

If the lines x−12=y+13=z−14 and x−31=y−k2=z1 intersect, then *k* =

#### SOLUTION

Solution :B

Any point on x−12=y+13=z−14=λ is,

(2λ+1,3λ−1,4λ+1);λ∈R

Any point on x−31=y−k2=z1=μ is,

(μ+3,2μ+k,μ);μ∈R

the given lines intersect if and only if the system of equations (in λ and μ)

2λ+1=μ+3....(i)

3λ−1=2μ+k.....(ii)

4λ+1=μ....(iii)

has a unique solution.

Solving (i) and (iii), we get λ=−32,μ=−5

From (ii), we get −92−1=−10+k⇒k=92.

### Question 6

Points (1, 1, 1), (–2, 4, 1), (–1, 5, 5) and (2, 2, 5) are the vertices of a

#### SOLUTION

Solution :B

Let A=(1,1,1);B=(-2,4,1);C=(-1,5,5) & D=(2,2,5)

AB=√9+9+0=3√2,BC=√1+1+16=3√2 and CD=3√2 and AD=3√2.

also AB.CD=0. Hence it is a square.

### Question 7

A straight line, makes an angle of 60∘ with each of y and z-axis,then the inclination of the line with x-axis is

#### SOLUTION

Solution :C

If α is a single at which the straight line is inclined to x-axis, then cos2α+cos260∘+cos260∘=1

⇒cos2α=12

⇒α=45∘or135∘.

### Question 8

If the projections of a line on the axes are 9, 12 and 8. Then the length of the line is

#### SOLUTION

Solution :B

Given r cosα=9,r cosβ=12 and r cosγ=8

∴r2(cos2α+cos2β+cos2γ)=81+144+64

r2.1=289

r=17

### Question 9

If the direction cosines of a variable line in two adjacent positions be l, m, n and l + a, m + b, n + c and the small angle between the two positions be θ, then :

#### SOLUTION

Solution :B

l2+m2+n2=1,(l+a)2+(m+b)2+(n+c)2=1⇒al+bm+cn=−12(a2+b2+c2)

cosθ=(a+l)l+m(b+m)+n(c+n)

= 1 + al + bm + cn

⇒a2+b2+c2=2(1−cosθ)=4sin2(θ2)

Since θ is small, sin(θ2)≈θ2

∴θ2=a2+b2+c2[∵sin2(θ2)≈θ24]

### Question 10

Direction cosines of the line which is perpendicular to the lines whose direction ratios are 1, –1, 2 and 2, 1, –1 are given by :

#### SOLUTION

Solution :A

If l, m, n are the d.c.’s of the line which is perpendicular to the given lines, then

l – m + 2n = 0 and 2l + m –n = 0

∴l1−2=m4+1=n1+2⇒l−1=m5=n3.

so d.r's =(−1,5,3)

### Question 11

The three lines drawn from O with direction ratios, (1, –1, k), (2, –3, 0) and (1, 0, 3) are coplanar. Then k =

#### SOLUTION

Solution :A

If lines are coplanar and l, m, n are d.c.’s of the normal to the plane, then

l – 1.m + k.n = 0

l – 3.m + 0.n = 0

l – 0.m + 3.n = 0

⇒∣∣ ∣∣1−1k2−30103∣∣ ∣∣=0⇒k=1

### Question 12

The equation of the plane containing the two lines of intersection of the two pairs of planes x + 2y – z – 3 = 0 and 3x – y + 2z – 1 = 0, 2x – 2y + 3z = 0 and x – y + z + 1 =0 is :

#### SOLUTION

Solution :D

The planes through given lines of intersection are

(x+2y−z−3)+λ(3x−y+2z−1)=0....(1)

and (2x−2y+3z)+μ(x−y+z+1)=0....(2)

Planes (1) and (2) are same. Comparing the coefficients,

we get, 1+3λ2+μ=2−λ−2−μ=−1+2λ3+μ=−3−λμ

Solving for λ,μ, these equations are inconsistent. Therefore, there exists no such plane i.e., the lines are neither parallel nor intersecting.

### Question 13

Equation of the plane perpendicular to the plane x – 2y + 5z + 1 = 0 which passes through the points (2, –3, 1) and (–1, 1, –7) is given by :

#### SOLUTION

Solution :B

Equation of the plane through (2, –3, 1) is

a(x – 2) + b(y + 3) + c(z – 1) = 0

It passes through (–1, 1, –7) and perpendicular to the plane x – 2y + 5z + 1 = 0

∴ –3a+ 4b – 8c = 0 and a – 2b + 5c = 0

∴a4=b7=c2

Hence, equation of required plane is

4x + 7y + 2z + 11 = 0

### Question 14

The foot of the perpendicular from (0, 2, 3) to the line x+35=y−12=z+43 is :

#### SOLUTION

Solution :D

x+35=y−12=z+43=λ

Let P(5λ−3,2λ+1,3λ−4) be the foot of perpendicular of A(0, 2, 3). Then AP and given line are perpendicular to each other.

∴(5λ−3)5+(2λ−1)2+(3λ−7)3=0

⇒38λ−38=0⇒λ=1.

Hence, P = (2, 3, –1)

### Question 15

Reflection of the line x−1−1=y−23=z−41 in the plane x +y +z =7 is :

#### SOLUTION

Solution :C

Given line passes through the point A(1, 2, 4) and this point also lies in the plane. To find the reflection of the line, we need one more point of the line. Clearly P(0, 5, 5) also lies in the line.

Let Q(α,β,γ)be the reflection of P in the plane x + y + z = 7

Then α2=β+52=γ+52=7⇒α+β+γ=4

Also PQ⊥ to the plane, i.e., parallel to the normal of the plane.

∴α−01=β−51=γ−51=λ⇒α=λ,β=λ+5,γ=λ+5∴λ+λ+5λ+5=4⇒λ=−2

∴ Q is (-2,3,3)

Now AQ will be the reflection of the given line, equation of AQ is

x−1−3=y−21=z−4−1