Geometry(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

GEOMETRY MCQs

Coordinate Geometry, Coordinate Geometry (10th Grade), Three Dimensional Geometry (10th Grade)

Total Questions : 133 | Page 7 of 14 pages

Question 61. The equation of the plane passing through the points (2, –4, 0), (3, –4, 5) and parallel to the line 2x = 3y = 4z is :

125x – 90y – 79z = 340
32x – 21y – 36z = 85
73x + 61y – 22z = 85
29x – 27y – 22z = 85

Discuss Question

Answer: Option D. -> 29x – 27y – 22z = 85
:
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.,

\frac{x}{6} = \frac{y}{4} = \frac{z}{3} . . . (3)

∴

Normal to the plane (1) is perpendicular to the line (3).

∴

6a + 4b + 3c = 0 . . .(4)
From (2) and (4), we get

\frac{a}{- 29} = \frac{b}{27} = \frac{c}{22}

∴

Equation of the required plane is
– 29(x – 2) + 27(y + 1) + 22(z) = 0

\Rightarrow - 29 x + 27 y + 22 z + 85 = 0

\Rightarrow 29 x - 27 y - 22 z = 85

Question 62. The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line

(\frac{1}{2}) x = (\frac{1}{3}) y = (\frac{- 1}{6}) z

is :

Discuss Question

Answer: Option A. -> 1
:
A
Line parallel to the line

\frac{x}{2} = \frac{y}{2} = \frac{z}{- 6}

and passing through (1,-2, 3) is

\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{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

∴ 2 r + 1 - 3 r + 2 - 6 r + 3 = 5 \Rightarrow r = \frac{1}{7} ∴ A = (\frac{9}{7}, \frac{- 11}{7}, \frac{15}{7})

Distance of A from (1, –2, 3) is

\sqrt{{(\frac{9}{7} - 1)}^{2} + {(\frac{- 11}{7} + 2)}^{2} + {(\frac{15}{7} - 3)}^{2}} = \frac{1}{7} \sqrt{4 + 9 + 36} = \frac{7}{7} = 1

Question 63. If a plane passes through the point (1,1,1) and is perpendicular to the line

\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4}

then its perpendicular distance from the origin is

Discuss Question

Answer: Option C. -> 75
:
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 plane

3 x + 4 z - 7 = 0

from (0, 0, 0) is

\frac{7}{\sqrt{3^{2} + 4^{2}}} = \frac{7}{5}

units

Question 64. The ratio in which the plane 2x - 1 = 0 divides the line joining (-2,4,7) and (3, -5, 8) is

2 : 3
4 : 5
7 : 8
1 : 1

Discuss Question

Answer: Option D. -> 1 : 1
:
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

(\frac{3 k - 2}{k + 1}, \frac{- 5 k + 4}{k + 1}, \frac{8 k + 7}{k + 1})

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

(\frac{3 k - 2}{k + 1} = \frac{1}{2} \Rightarrow k = 1)

and thus the required ratio as 1:1.

Question 65. Reflection of the line

\frac{x - 1}{- 1} = \frac{y - 2}{3} = \frac{z - 4}{1}

in the plane x +y +z =7 is :

x−13=y−21=z−41
x−1−3=y−2−1=z−41
x−1−3=y−21=z−4−1
x+13=y−21=z+41

Discuss Question

Answer: Option C. -> x−1−3=y−21=z−4−1
:
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

\frac{α}{2} = \frac{β + 5}{2} = \frac{γ + 5}{2} = 7 \Rightarrow α + β + γ = 4

Also

P Q ⊥

to the plane, i.e., parallel to the normal of the plane.

∴ \frac{α - 0}{1} = \frac{β - 5}{1} = \frac{γ - 5}{1} = λ \Rightarrow α = λ, β = λ + 5, γ = λ + 5 ∴ λ + λ + 5 λ + 5 = 4 \Rightarrow λ = - 2

∴

Q is (-2,3,3)
Now AQ will be the reflection of the given line, equation of AQ is

\frac{x - 1}{- 3} = \frac{y - 2}{1} = \frac{z - 4}{- 1}

Question 66. If the projections of a line on the axes are 9, 12 and 8. Then the length of the line is

Discuss Question

Answer: Option B. -> 17
:
B
Given

r c o s α = 9, r c o s β = 12

and

r c o s γ = 8

∴ r^{2} (c o s^{2} α + c o s^{2} β + c o s^{2} γ) = 81 + 144 + 64

r^{2} .1 = 289

r = 17

Question 67. 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 :

7x – 7y + 8z + 3 = 0
7x – 8y + 9z + 3 = 0
5x – 5y + 6z + 2 = 0
None of these

Discuss Question

Answer: Option D. -> None of these
:
D
The planes through given lines of intersection are

(x + 2 y - z - 3) + λ (3 x - y + 2 z - 1) = 0 . . . . (1)

and

(2 x - 2 y + 3 z) + μ (x - y + z + 1) = 0 . . . . (2)

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

\frac{1 + 3 λ}{2 + μ} = \frac{2 - λ}{- 2 - μ} = \frac{- 1 + 2 λ}{3 + μ} = \frac{- 3 - λ}{μ}

Solving for

λ, μ

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

Question 68. The three lines drawn from O with direction ratios, (1, –1, k), (2, –3, 0) and (1, 0, 3) are coplanar. Then k =

1
0
no such k exists
7

Discuss Question

Answer: Option A. -> 1
:
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

\Rightarrow ∣ ∣∣ ∣ \begin{matrix} 1 & - 1 & k 2 & - 3 & 0 1 & 0 & 3 \end{matrix} ∣ ∣∣ ∣ = 0 \Rightarrow k = 1

Question 69. 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

(p, 2 p, 0)
(0, 2p, -p)
(2p, p, -p)
(2p, -p, 2p)

Discuss Question

Answer: Option C. -> (2p, p, -p)
:
C
The perpendicular distance of the origin (0, 0, 0) from the plane x + y + z = p is

∣ ∣ ∣ \frac{- p}{\sqrt{1 + 1 + 1}} ∣ ∣ ∣ = \frac{| p |}{\sqrt{3}}

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

∣ ∣ ∣ \frac{x + y + z - p}{\sqrt{3}} ∣ ∣ ∣ = \frac{| p |}{\sqrt{3}} \Rightarrow | x + y + z - p | = | p |

Which is satisfied by (c)

Question 70. 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

2x -4y +3z-8 =0
2x-4y-3z+8 =0
2x+4y+3z+8 =0
x-3y+z-5=0

Discuss Question

Answer: Option A. -> 2x -4y +3z-8 =0
:
A
l+2m+2n=0, 3l+3m+2n=0,

l^{2} + m^{2} + n^{2} = 1

, we get l, m, nfrom 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.