JEE MATHEMATICS MCQ: 3D PLANES

1 Easy

1 of 1

Application +4 | -1

The distance of the point

(1,0,2)

from the point of intersection of the line

\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}

and the plane

x-y+z=16

, is

View answer and explanation

Correct option: D

Option D (13) is correct because it is the distance between the given point and the intersection point of the line and plane.

More Questions from 3D PLANES

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

p x-q y+

z=5

p, q \in \mathbb{R}

x-c y-b z=0, c x-y+a z=0

b x+a y-z=0

a^2+b^2+c^2+2 a b c

2 x-3 y+9 z=0

P

(0,1,-1)

4 x-2 y+7 z+9=0

P

(3,2,0)

\frac{x-4}{1}=\frac{y-7}{5}=\frac{z-4}{4}

a x+

=0

\alpha

z=0

\mathrm{ax}+\mathrm{by} \pm \mathrm{z} \sqrt{\mathrm{a}^2+\mathrm{b}^2} \tan \alpha+\mathrm{c}=0

(3,2,1)

x, y

z

A, B

C

y z

A

z x

B

x y

C