How to find equation of plane given one vector and two points?

Question

Find an equation of the plane that is perpendicular to $x-2y+z=5$ and passes through $P_1(-2,4,3)$ and $P_2(3,-5,0)$?

How to solve this question? It is needed to find another vector using cross product or just use $(1,-2,1)$ and choose any point?

Answer 1

Two things first:

• If two planes are perpendicular, their normal vectors will also be perpendicular.
• If a plane contains two points, it contains the line through those points; in particular its normal vector is perpendicular to the direction vector of the line.

So the desired plane's normal vector is perpendicular to $(1,-2,1)$ and $(3,-5,0)-(-2,4,3)=(5,-9,-3)$. The cross product yields such a perpendicular vector: $(15,8,1)$. So the equation of the plane is $15x+8y+z=k$, where $k$ is found by substituting either given point in: $15\cdot(-2)+8\cdot4+1\cdot3=5$. The final answer is $15x+8y+z=5$.