Three Dimensional Geometry(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

THREE DIMENSIONAL GEOMETRY MCQs

Total Questions : 30 | Page 1 of 3 pages

The plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ 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

$(a, \frac{b}{4}, \frac{c}{4})$
$(\frac{a}{4}, b, \frac{c}{4})$
$(\frac{a}{4}, \frac{b}{4}, c)$
$(\frac{a}{2}, \frac{b}{4}, \frac{c}{4})$ .

Discuss Question

Answer: Option D. ->

(\frac{a}{2}, \frac{b}{4}, \frac{c}{4})

.
:
D

A (a, 0, 0), B (0, b, 0), C (0, 0, c), D (\frac{a}{2}, \frac{b}{2}, 0), E (\frac{a}{2}, 0, \frac{c}{2})

So midpoint of DE is

(\frac{a}{2}, \frac{b}{4}, \frac{c}{4})

Question 2.

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

$\frac{3}{4}$
$\frac{4}{3}$
$\frac{7}{5}$
$1$

Discuss Question

Answer: Option C. ->

\frac{7}{5}

:
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 3.

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 4.

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then k =

$\frac{2}{9}$
$\frac{9}{2}$
0
3

Discuss Question

Answer: Option B. ->

\frac{9}{2}

:
B
Any point on

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

is,

(2 λ + 1, 3 λ - 1, 4 λ + 1); λ \in R

Any point on

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

is,

(μ + 3, 2 μ + k, μ); μ \in R

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

λ

and

μ)

2 λ + 1 = μ + 3 . . . . (i)

3 λ - 1 = 2 μ + k . . . . . (i i)

4 λ + 1 = μ . . . . (i i i)

has a unique solution.
Solving (i) and (iii), we get

λ = \frac{- 3}{2}, μ = - 5

From (ii), we get

\frac{- 9}{2} - 1 = - 10 + k \Rightarrow k = \frac{9}{2}

Question 5.

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

$\sqrt{107}$
$\sqrt{14}$
$\frac{\sqrt{107}}{\sqrt{14}}$
$10$

Discuss Question

Answer: Option A. ->

\sqrt{107}

:
A
Centroid

\equiv (\frac{\sum x}{4}, \frac{\sum y}{4}, \frac{\sum z}{4}) = (1, 2, - 1)

\Rightarrow a = 1, b = 5, c = - 9; ∴ \sqrt{a^{2} + b^{2} + c^{2}} = \sqrt{107}

Question 6.

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

Rectangle
Square
Parallelogram
Trapezium

Discuss Question

Answer: Option B. -> Square
:
B
Let A=(1,1,1);B=(-2,4,1);C=(-1,5,5) & D=(2,2,5)

A B = \sqrt{9 + 9 + 0} = 3 \sqrt{2}, B C = \sqrt{1 + 1 + 16} = 3 \sqrt{2}

and

C D = 3 \sqrt{2}

and

A D = 3 \sqrt{2}

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

Question 7.

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 8.

A straight line, makes an angle of $60^{\circ}$ with each of y and z-axis,then the inclination of the line with x-axis is

$60^{\circ}$
$30^{\circ}$
$45^{\circ}$
$75^{\circ}$

Discuss Question

Answer: Option C. ->

45^{\circ}

:
C
If

α

is a single at which the straight line is inclined to x-axis, then

c o s^{2} α + c o s^{2} 60^{\circ} + c o s^{2} 60^{\circ} = 1

\Rightarrow c o s^{2} α = \frac{1}{2}

\Rightarrow α = 45^{\circ} o r 135^{\circ}

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 :