Reflection of the line x - 1 - 1 = y - 2 3 = z - 4 1 in the plane x + y + z = 7 is

x - 1 - 3 = y - 2 1 = z - 4 - 1

x - 1 - 3 = y - 2 - 1 = z - 4 1

The image of the line x - 1 3 = y - 3 1 = z - 4 - 5 in the plane 2 x - y + z +3=0 is the line

x - 3 - 3 = y + 5 - 1 = z - 2 5

x + 3 - 3 = y - 5 - 1 = z + 2 5

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A plane passing through a point $(1, 2, - 4)$ having normal perpendicular to two below lines

$\frac{x - 8}{3} = \frac{y + 19}{- 16} = \frac{z - 10}{7}$
and
$\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}$

The perpendicular distance of the point $(0, 0, 0)$ from plane is

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Important Questions on Three Dimensional Geometry

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

A plane passing through the point

(3, 1, 1)

contains two lines whose direction ratios are

1, - 2, 2

and

2, 3, - 1

respectively. If this plane also passes through the point

(α, - 3, 5)

, then

α

is equal to

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

The perpendicular distance from the origin to the plane containing the two lines,

\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}

and

\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}

, is

EASY

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

The distance of the point

(4, 2, 3)

from the plane

\vec{r} \cdot (6 \hat{i} + 2 \hat{j} - 9 \hat{k}) = 46

EASY

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

If the sum of squares of distances of a point from the planes

x + y + z = 0, x - z = 0

and

x - 2 y + z = 0

p^{2}

then locus of the point is

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

A perpendicular is drawn from a point on the line

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

to the plane

x + y + z = 3

such that the foot of the perpendicular

Q

also lies on the plane

x - y + z = 3

. Then the coordinates of

Q

are

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

The distances of the point

(1, - 5, 9)

from the plane

x - y + z = 5

measured along a straight line

x = y = z

2 \sqrt{3} k

, then the value of

k

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

Foot of the perpendicular drawn from the point

(1, 3, 4)

to the plane

2 x - y + z + 3 = 0

EASY

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

A B

and

C D

are

2

line segments, where

A (2, 3, 0), B (6, 9, 0), C (- 6, - 9, 0)

and

D

is the image of point

A

about origin.

P

and

Q

are midpoint of

A B

and

C D

, respectively and

L

is the midpoint of

P Q

. Find the distance of

L

from the plane

3 x + 4 z + 25 = 0

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

Equations of planes parallel to the plane

x - 2 y + 2 z + 4 = 0

which are at a distance of one unit from the point

(1,2, 3)

are …..

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

Let

P

be a plane passing through the points

(2, 1, 0), (4, 1, 1)

and

(5, 0, 1)

and

R

be any point

(2, 1, 6)

.Then the image of

R

in the plane

P

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

Reflection of the line

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

in the plane

x + y + z = 7

EASY

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

If the distance of points

2 \hat{i} + 3 \hat{j} + λ \hat{k}

from the plane

\vec{r} \cdot (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) = 13

5

units then

λ =

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

The image of the point

(1, - 1, 1)

in the plane

x - 2 y + 3 z + 1 = 0

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

The coordinates of the foot of the perpendicular drawn from the origin to plane

2 x - 3 y + 4 z - 6 = 0

, are

EASY

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

Find the distance (in units) of the point

(1, 2, 3)

from the plane

3 y + 4 z + 4 = 0

EASY

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

If a plane has

X

-intercept

l, Y

-intercept

m

and

Z -

intercept

n,

and perpendicular distance of plane from origin is

k,

then

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

If the mirror image of the point

(1, 3, 5)

with respect to the plane

4 x - 5 y + 2 z = 8

(α, β, γ)

, then

5 (α + β + γ)

equals :

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

The mirror image of the point

(1, 2, 3)

in a plane is

(- \frac{7}{3}, - \frac{4}{3}, - \frac{1}{3}) .

Which of the following points lies on this plane?

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

Perpendiculars are drawn from points on the line

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

to the plane

x + y + z = 3

. The feet of perpendiculars lie on the line

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Plane and a Point

The image of the line

\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{- 5}

in the plane

2 x - y + z +3=0

is the line

A plane passing through a point (1,2,-4) having normal perpendicular to two below lines x-83=y+19-16=z-107 and x-153=y-298=z-5-5 The perpendicular distance of the point (0,0,0) from plane is

Important Questions on Three Dimensional Geometry

Learn and Prepare for any exam you want !

A plane passing through a point $(1, 2, - 4)$ having normal perpendicular to two below lines

$\frac{x - 8}{3} = \frac{y + 19}{- 16} = \frac{z - 10}{7}$
and
$\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}$

The perpendicular distance of the point $(0, 0, 0)$ from plane is