# Finding the equation of a plane that passes that contains two points, and is perpendicular to another plane?

So I need an equation of a plane that passes through $P(0,-2,5)$ and $Q(-1,3,1)$, and is perpendicular to the plane $\pi_1: 2z=5x+4y$.

I'm not too sure how to solve it; I guess the solution would involve the vector $\vec{PQ}=<-1,5,-4>$, but the more I try to visualise it, the more confused I get.

HINT

The condition of perpendicularity with $5x+4y-2z=0$ means that the normal vector $(5,4,-2)$ is parallel to the plane.

Let indicate with $ax+by+cz=1$ the plane we are looking for, we have three condition and thus we can find $a,b,c$.

The plane equation should be:

$$-6x+22y+29z=101$$

Whenever you're looking for a plane, you should think about looking for its normal vector. That's much easier to visualize.

The normal vector you seek has to be perpendicular to the normal vector of $\pi_1$, and also perpendicular to the vector $PQ$ (because $PQ$ is in the plane). Does thinking about it this way help?

• So I guess I should be finding the cross product $\vec{PQ} \times \pi_1$, which will give me the normal vector for the plane in question? – user98937 Jan 18 '18 at 19:03
• @user98937 Yes, that would work. – Y. Forman Jan 18 '18 at 19:21