Find the foot of the perpendicular drawn from the point P ( 1 , 1 , 3 ) to the line x - 4 2 = y 1 = z - 2 - 1 .Also find the perpendicular distance of the line from the given point.

x - 4 2 = y 1 = z - 2 - 1 = λ x = 2 λ + 4 , y = λ , z = - λ + 2 The line P Q has a ratio of ( 2 λ + 3 , λ - 1 , - λ - 1 ) A B line of ratio is 2 , 1 , - 1 If two lines are perpendicular then a 1 a 2 + b 1 b 2 + c 1 c 2 = 0 4 λ + 6 + λ - 1 + λ + 1 = 0 6 λ = - 6 λ = - 1 Distance, PQ = ( 2 - 1 ) 2 + ( - 1 - 1 ) 2 + ( 3 - 3 ) 2 = 1 + 4 = 5 square units.

Find the distance of the point ( 1 , - 1 , - 10 ) from the line x - 4 1 = y + 3 - 4 = z + 1 7 measured parallel to the line x + 2 2 = y - 3 - 3 = z - 4 8 .

Given that, the point is ( 1 , − 1 , - 10 ) And the equation of line is x - 4 1 = y + 3 4 = z + 1 7 = k say Thus, any general point on the line is x = k + 4 , y = - 4 k - 3 , z = 7 k - 1 → 1 And also, the equation of parallel line is x + 2 2 = y - 3 - 3 = z - 4 8 = λ say Thus, any general point on the line is x = 2 λ - 2 , y = - 3 λ + 3 , z = 8 λ + 4 → 2 From equation 1 & 2 , we have: k + 4 = 2 λ - 2 → 3 - 4 k - 3 = - 3 λ + 3 → 4 7 k - 1 = 8 λ + 4 → 5 On solving equations 3 , 4 and 5 , we get: k = 6 5 & λ = 18 5 Thus, x = 26 5 , y = - 39 5 , z = 37 5 Therefore, the required distance = 26 5 - 1 2 + - 39 5 + 1 2 + 37 5 + 10 2 = 21 5 2 + - 34 5 2 + 87 5 2 = 9166 5 units Hence, the required distance = 9166 5 units.

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The equation of the perpendicular from the point (α, β, γ) to the plane $a x + b y + c z + d = 0$ is

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Important Questions on Three Dimensional Geometry

HARD

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

The perpendicular distance of the point

(2, 4, - 1)

from the line

\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9}

MEDIUM

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

What will be the distance of point

(2, 4, - 1)

from the straight line given by

(x + 5) = (y + 3) / 4 = (6 - z) / 9 ?

HARD

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

P (1, 2, 3)

is the given point and

P N

is the length of perpendicular from

P

on the given line

\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}

, then

P N =

HARD

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

The image of the point

(1, 6, 3)

in the line

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

HARD

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

Consider the line

L

given by the equation

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

Let

Q

be the mirror image of the point

(2, 3, - 1)

with respect to

L .

Let a plane

P

be such that it passes through

Q,

and the line

L

is perpendicular to

P .

Then which of the following points is on the plane

P ?

EASY

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

Find the foot of the perpendicular drawn from the point $P (1, 1, 3)$ to the line $\frac{x - 4}{2} = \frac{y}{1} = \frac{z - 2}{- 1}$ .Also find the perpendicular distance of the line from the given point.

MEDIUM

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

The distance of line

3 y - 2 z - 1 = 0 = 3 x - z + 4

from the point

(2, - 1, 6)

is :

MEDIUM

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

If the foot of the perpendicular drawn from the point

(1, 0, 3)

on a line passing through

(α, 7,1)

(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})

, then

α

is equal to

HARD

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

(a, b, c)

is the image of the point

(1, 2, - 3)

in the line,

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

then

a + b + c

is equal to:

MEDIUM

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

Let the foot of perpendicular from a point

P (1, 2, - 1)

to the straight line

L : \frac{x}{1} = \frac{y}{0} = \frac{z}{- 1}

N

. Let a line be drawn from

P

parallel to the plane

x + y + 2 z = 0

which meets

L

at point

Q

. If

α

is the acute angle between the lines

P N

and

P Q

, then

\cos α

is equal to .

HARD

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

Find the distance of the point $(1, - 1, - 10)$ from the line $\frac{x - 4}{1} = \frac{y + 3}{- 4} = \frac{z + 1}{7}$ measured parallel to the line $\frac{x + 2}{2} = \frac{y - 3}{- 3} = \frac{z - 4}{8}$ .

EASY

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

Let

a, b \in R .

If the mirror image of the point

P (a, 6, 9)

with respect to the line

\frac{x - 3}{7} = \frac{y - 2}{5} = \frac{z - 1}{- 9}

(20, b, - a - 9),

then

| a + b |

is equal to:

HARD

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

The vertices

B

and

C

of a

Δ A B C

lie on the line,

\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}

such that

B C = 5

units. Then the area (in sq. units) of this triangle, given the point

A (1, - 1, 2),

MEDIUM

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

Find the image of the point

(3, - 2, 1)

in the plane

3 x - y + 4 z = 2

MEDIUM

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

The length of the perpendicular from the point

(2, - 1, 4)

on the straight line

\frac{x + 3}{10} = \frac{y - 2}{- 7} = \frac{z}{1}

MEDIUM

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

If the length of the perpendicular from the point

(β, 0, β), (β \neq 0)

to the line,

\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{- 1}

\sqrt{\frac{3}{2}}

, then

β

is equal to

HARD

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

The distance of the point having position vector

- \hat{i} + 2 \hat{j} + 6 \hat{k}

from the straight line passing through the point

(2, 3, - 4)

and parallel to the vector,

6 \hat{i} + 3 \hat{j} - 4 \hat{k}

MEDIUM

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

The length of the perpendicular from the point

(1, 2, 3)

to the line

\frac{- x + 6}{- 3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}

is _________.

HARD

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

The shortest distance between

A (1, 0, 2)

and the line

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

is given by line joining

A

and

B

, then

B

in the line is

MEDIUM

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

The image of the point

(1, 6, 3)

in the line

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

The equation of the perpendicular from the point (α, β, γ) to the plane ax+by+cz+d=0 is

Important Questions on Three Dimensional Geometry

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The equation of the perpendicular from the point (α, β, γ) to the plane $a x + b y + c z + d = 0$ is