The equation of the plane that contains the point (1,-1,2) and is perpendicular to each of the planes 2 x+3 y-2 z=5 and x+2 y-3 z=8, is

Question

The equation of the plane that contains the point (1,-1,2) and is perpendicular to each of the planes 2 x+3 y-2 z=5 and x+2 y-3 z=8, is (A) 5 x-4 y=7 (B) 5 x-4 y-z=0 (C) 5 x-4 y-z=7 (D) 5 x+4 y+z=7

Tardigrade Admin · Accepted Answer

The equation of the plane containing the given point is

A (x - 1) + B (y + 1) + C (z - 2) = 0

...(i)
Applying the condition of perpendicularly to the plane given in Eq. (i) with the planes

2 x + 3 y - 2 z = 5 and x + 2 y - 3 z = 8, we have

2 A + 3 B - 2 C = 0 and A + 2 B - 3 C = 0

Solving these equations, we find

A = - 5 C

and

B = 4 C

.
Hence, the required equation is

- 5 C (x - 1) + 4 C (y + 1) + C (z - 2) = 0

i.e.,

5 x - 4 y - z = 7

Q. The equation of the plane that contains the point $(1, - 1, 2)$ and is perpendicular to each of the planes $2 x + 3 y - 2 z = 5$ and $x + 2 y - 3 z = 8$ , is

$5 x - 4 y = 7$

$5 x - 4 y - z = 0$

$5 x - 4 y - z = 7$

$5 x + 4 y + z = 7$

Q. The equation of the plane that contains the point (1,−1,2) and is perpendicular to each of the planes 2x+3y−2z=5 and x+2y−3z=8, is

5x−4y=7

5x−4y−z=0

5x−4y−z=7

5x+4y+z=7

Q. The equation of the plane that contains the point $(1, - 1, 2)$ and is perpendicular to each of the planes $2 x + 3 y - 2 z = 5$ and $x + 2 y - 3 z = 8$ , is

$5 x - 4 y = 7$

$5 x - 4 y - z = 0$

$5 x - 4 y - z = 7$

$5 x + 4 y + z = 7$