Find the equation of the plane passing through the point $(1, - 1, 2)$ and perpendicular to the planes $2 x + 3 y - 3 z = 2$ and $5 x - 4 y + z = 6$ .

Open in App

Solution

Let the equation of the plane containing the given point $(1, - 1, 2)$ is
$\begin{matrix} A (x - x_{1}) + B (y - y_{1}) + C (z - z_{1}) = 0 (i . e) A (x - 1) + B (y + 1) + C (z - 2) = 0 \end{matrix}$

Applying the condition of perpendicular to the plane in equation (1) with the plane
$2 x + 3 y - 2 z = 5$ and $x + 2 y - 3 z = 8$

We have, $2 A + 3 B - 2 C = 0$ and $A + 2 B - 3 C = 0$

Let us solve three two equations, we get
$A = - 5 C$
$B = 4 C$

The required equation is
$\begin{matrix} - 5 C (x - 1) + 4 C (y + 1) + c (z - 2) = 0 5 x - 4 y - z - 7 = 0 5 x - 4 y - z = 7 \end{matrix}$

Hence, this is the answer.

Suggest Corrections

Similar questions

(1, 1 - 1)

x + 2 y + 3 z - 7 = 0

2 x - 3 y + 4 z = 0

(1, 1, 1)

x + 2 y + 3 z = 7

2 x - 3 y + 4 z = 0

x + 2 y + 3 z = 4

2 x + y - z = - 5,

5 x + 3 y + 6 z + 8 = 0

Join BYJU'S Learning Program