Question

FIND EQUATION OF PLANE PASSING THROUGH POINTS (1,-1,2) AND (2,-2,2) AND WHICH IS PERPENDICULAR TO THE PLANE 6x-2y+2z=9 ?

Answer 1

Find equation of plane passing through points $(1,-1,2)$ and $(2,-2,2)$ and which is perpendicular to the plane $6x-2y+2z=9$?

Solution.

Plane which is perpendicular to the given plane should be parallel to the normal vector of this plane.

Find coordinates of normal vector from the equation $6x-2y+2z=9$: $\vec{n} = \{6, -2, 2\}$

Let's find the equation of a plane passing through points $(1,-1,2)$ and $(2,-2,2)$ and parallel to $\vec{n}$:

$$\left| \begin{array}{cccc} x - 1 & y + 1 & z - 2 \\ 2 - 1 & -2 + 1 & 2 - 2 \\ 6 & -2 & 2 \end{array} \right| = 0,$$ $$-2(x - 1) - 2(z - 2) + 0 \cdot 6(y + 1) + 6(z - 2) - 0 \cdot (-2)(x - 1) - 2(y + 1) = 0,$$ $$-2x + 2 - 2z + 4 + 6z - 12 - 2y - 2 = 0,$$ $$-2x - 2y + 4z - 8 = 0,$$ $$x + y - 2z + 4 = 0.$$

Answer.

$$x + y - 2z + 4 = 0.$$