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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.

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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:

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

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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?

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  • $\begingroup$ 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? $\endgroup$ – user98937 Jan 18 '18 at 19:03
  • $\begingroup$ @user98937 Yes, that would work. $\endgroup$ – Y. Forman Jan 18 '18 at 19:21

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