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 13 of 14 pages

Question 121. Perpendicular distance of the point (3, 4, 5) from the y-axis, is

√34
√41
4
5

Discuss Question

Answer: Option A. -> √34
:
A
Perpendicular distance of (x,y,z) from y axis is given by

= \sqrt{x^{2} + z^{2}}

.
=>Required distance

= \sqrt{3^{2} + 5^{2}} = \sqrt{34}

Question 122. The projection of any line on co-ordinate axes be respectively 3, 4, 5 then its length is

12
50
5√2
25

Discuss Question

Answer: Option C. -> 5√2
:
C
Let the line segment be AB, then as given

A B c o s α = 3, A B c o s β = 4, A B c o s γ = 5 \Rightarrow A B^{2} (c o s^{2} α + c o s^{2} β + c o s^{2} γ) = 3^{2} + 4^{2} + 5^{2} A B = \sqrt{9 + 16 + 25} = 5 \sqrt{2}

where

α, β

and

γ

are the angles made by the line with the axes.

Question 123. 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 124. The distance between two points P and Q is d and the length of their projections of PQ on the coordinate planes are

d_{1}, d_{2}, d_{3}

. Then

d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = k d^{2}

, where k is

Discuss Question

Answer: Option D. -> 2
:
D
Here,

d_{1} = d c o s (90^{\circ} - α)

d_{2} = d c o s (90^{\circ} - β)

and

d_{3} = d c o s (90^{\circ} - γ)

∴ d_{1} = d s i n α

d_{2} = d s i n β

and

d_{3} = d s i n γ

∴ d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = k d^{2}

\Rightarrow d^{2} (s i n^{2} α + s i n^{2} β + s i n^{2} γ) = k (d)^{2}

∴ k = 2

Question 125. 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 126. The line joining the points (1, 1, 2) and (3, -2, 1) meets the plane 3x + 2y + z = 6 at the point

(1, 1, 2)
(3, -2, 1)
(2, -3, 1)
(3, 2, 1)

Discuss Question

Answer: Option B. -> (3, -2, 1)
:
B
The straight line joining the points (1, 1, 2) and (3, -2, 1) is

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

(say)

∴

Point is (2r + 1, 1 -3r, 2 –r )
Which lies on 3x + 2y + z = 6

∴ 3 (2 r + 1) + 2 (1 - 3 r) + 2 - r = 6

∴ r = 1

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

Question 127. If

(l_{1}, m_{1}, n_{1})

and

(l_{2}, m_{2}, n_{2},)

are d.c.'s of

¯ ¯¯¯¯¯¯¯¯ ¯ O A,

¯ ¯¯¯¯¯¯ ¯ O B

such that

∠ A O B = θ

where ‘O’ is the origin, then the d.c.’s of the internal bisector of the angle

∠ A O B

are

l1+l22sinθ2,m1+m22sinθ2,n1+n22sinθ2
l1+l22cosθ2,m1+m22cosθ2,n1+n22cosθ2
l1−l22sinθ2,m1−m22sinθ2,n1−n22sinθ2
l1−l22cosθ2,m1−m22cosθ2,n1−n22cosθ2

Discuss Question

Answer: Option B. -> l1+l22cosθ2,m1+m22cosθ2,n1+n22cosθ2
:
B
Let OA and OB be two lines with d.c’s

l_{1}, m_{1}, n_{1}

and

l_{2}, m_{2}, n_{2},

. Let OA = OB = 1. Then, the coordinates of A and B are

(l_{1}, m_{1}, n_{1})

and

(l_{2}, m_{2}, n_{2})

, respectively. Let OC be the bisector of

∠ A O B

.Then, C is the mid point of AB and so its coordinates are

(\frac{l_{1} + l_{2}}{2}, \frac{m_{1} + m_{2}}{2}, \frac{n_{1} + n_{2}}{2})

∴

d.r's of OC are

\frac{l_{1} + l_{2}}{2}, \frac{m_{1} + m_{2}}{2}, \frac{n_{1} + n_{2}}{2}

We have,

O C = \sqrt{{(\frac{l_{1} + l_{2}}{2})}^{2} + {(\frac{m_{1} + m_{2}}{2})}^{2} + {(\frac{n_{1} + n_{2}}{2})}^{2}} = \frac{1}{2} \sqrt{(l_{1}^{2} + m_{1}^{2} + n_{1}^{2}) + (l_{2}^{2} + m_{2}^{2} + n_{2}^{2}) + (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2})} = \frac{1}{2} \sqrt{2 + 2 c o s θ} [Q c o s θ = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}] = \frac{1}{2} \sqrt{2 (1 + c o s θ)} = c o s (\frac{θ}{2})

If (l1,m1,n1) And (l2,m2,n2,) Are D.c.'s Of ¯¯¯¯¯¯¯¯...

∴

d.c's of OC are

\frac{l_{1} + l_{2}}{2 (O C)}, \frac{m_{1} + m_{2}}{2 (O C)}, \frac{n_{1} + n_{2}}{2 (O C)}

Question 128. The perpendicular distance of the origin form the plane which makes intercepts 12, 3 and 4 on x, y, z axes respectively, is

13
11
17
6√2√13

Discuss Question

Answer: Option D. -> 6√2√13
:
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

\frac{x}{(\frac{p}{l})} + \frac{y}{(\frac{p}{m})} + \frac{z}{(\frac{p}{n})} = 1

According to question,

\frac{p}{l} = 12, \frac{p}{m} = 3, \frac{p}{n} = 4

\frac{p}{12} = l, \frac{p}{3} = m, \frac{p}{4} = n

\frac{p^{2}}{144} + \frac{p^{2}}{9} + \frac{p^{2}}{16} = l^{2} + m^{2} + n^{2} = 1

\Rightarrow p^{2} (\frac{1}{144} + \frac{1}{9} + \frac{1}{16}) = 1 p^{2} (\frac{1 + 16 + 9}{144}) = 1

p^{2} = \frac{144}{26} = \frac{72}{13}

∴ p = \frac{6 \sqrt{2}}{\sqrt{13}}

Question 129. 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