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If $(α, β, γ)$ be the foot of the perpendicular from the point $(1, 2, 3)$ on the line $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ then $19 (α + β + γ)$ will be

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Important Questions on Three Dimensional Geometry

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

If vector equation of the line

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

, is

\vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + λ (2 \hat{i} - \frac{3}{2} \hat{j} + p \hat{k})

, then

p

is equal to

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

A plane

P

meets the coordinate axes at

A, B

and

C

respectively. The centroid of

∆ ABC

is given to be

(1, 1, 2)

. Then the equation of the line through this centroid and perpendicular to the plane

P

is :

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

The foot of the perpendicular drawn from the point

(4, 2, 3)

to the line joining the points

(1, - 2, 3)

and

(1, 1, 0)

lies on the plane

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

Let

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

be the position vector of a point

B

. Let

A

be a point on the line which is passing through

B

and parallel to the vector

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

. If

|\vec{B A}| = 18

, then the position vector of

A

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

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}

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

The length of perpendicular drawn from the point

(2, 3, 4)

to line

\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3}

, is

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

(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:

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

If the position vectors of three points

A, B, C

respectively are

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

and

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

, then the perpendicular distance of the point

C

from the line

A B

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

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>Lines in 3D

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

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

The distance of the point

(1, 2, - 4)

from the line

\frac{x - 3}{2} = \frac{y - 3}{3} = \frac{z + 5}{6}

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

If the points

(- 1, 3, 2), (- 4, 3, - 2)

and

(5, l, m)

lie on a straight line then

l

and

m

are

EASY

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

The parametric equations of the line passing through

A (3, 4, - 7), B (1, - 1, 6)

are

(λ \in R)

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

The image of the point

(1, 6, 3)

in the line

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

EASY

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

The equation of the line passing through the point

(2, 3, - 4)

and perpendicular to the

X O Z

plane is

(λ \in R)

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

The length of perpendicular from the point

\hat{i} + 2 \hat{j} + 3 \hat{k}

to the line

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

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

A point

P

lies on a line through

Q (1, - 2, 3)

and is parallel to the line

\frac{x}{1} = \frac{y}{4} = \frac{z}{5} .

P

lies on the plane

2 x + 3 y - 4 z + 22 = 0,

then segment

P Q

equals

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

The distance of the point

(1, 2, 1)

from the line

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

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

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),

EASY

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Lines in 3D

The equation of the line passing through the point

(5, 3, 2)

and perpendicular to the lines

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

and

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

If α,β,γ be the foot of the perpendicular from the point 1,2,3 on the line x+35=y-12=z+43 then 19α+β+γ will be

Important Questions on Three Dimensional Geometry

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If $(α, β, γ)$ be the foot of the perpendicular from the point $(1, 2, 3)$ on the line $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ then $19 (α + β + γ)$ will be