JEE MATHEMATICS MCQ: 3D PLANES

1 Easy

1 of 1

Application +4 | -1

The distance of the point

(1,-2,4)

from the plane passing through the point

(1,2,2)

and perpendicular to the planes

x-y+2 z=3

and

2 x-2 y+z+12=0

, is :

View answer and explanation

Correct option: C

2 \sqrt{2}

Option C (

2 \sqrt{2}

) is correct because it is the distance calculated from the point to the plane, where the plane is perpendicular to the given planes.

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