A plane passes through (1,-2,1) and is perpendicular to two planes 2 x-2 y+z=0 and x-y+2 z=4, then the distance of the plane from the point (1,2,2) is.

Question

A plane passes through (1,-2,1) and is perpendicular to two planes 2 x-2 y+z=0 and x-y+2 z=4, then the distance of the plane from the point (1,2,2) is. (A) 0 (B) 1 (C) √2 (D) 2 √2

Tardigrade Admin · Accepted Answer

D.R. of the normal to the plane that that is perpendicular to the plane

2 x - 2 y + z = 0

and

x - y +

2 z = 4

is given by

∣ ∣ \hat{i} 21 \hat{j} - 2 - 1 \hat{k} 12 ∣ ∣ = - 3 \hat{i} - 3 \hat{j}

or

\hat{i} + \hat{j}

So the required plane passing through

(1, - 2, 1)

is

(x - 1) + (y + 2) = 0

or

x + y + 1 = 0

Hence the distance of

(1, 2, 2)

form this plane is

d = \frac{4}{2} = 22

units

Q. A plane passes through $(1, - 2, 1)$ and is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , then the distance of the plane from the point $(1, 2, 2)$ is.

$0$

$1$

$2$

$22$

Q. A plane passes through (1,−2,1) and is perpendicular to two planes 2x−2y+z=0 and x−y+2z=4, then the distance of the plane from the point (1,2,2) is.

0

1

2​

22​

Q. A plane passes through $(1, - 2, 1)$ and is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , then the distance of the plane from the point $(1, 2, 2)$ is.

$0$

$1$

$2$

$22$