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Find an equation of a plane that passes through $p(1,5,1)$ and is perpendicular to planes $2x+y-2z = 2$ and $x+3z=4$.

I basically need the 2 other points to make the vector and perform the cross product.

Since $ax+by+cz+d=0$ is the form of a plane. Can I obtain the points as $(a,b,c)$ of the 2 planes given? Also, how can I set them up to obtain the points?

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you have two vectors(normals of the given planes) $$u=(2,1,-2), v=(1,0,3)$$ then $n=u×v$
The plane equation is then $$[(x,y,z)-P].n=0$$

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  • $\begingroup$ I got $n = 5i-8j-k$, but what is $(x,y,z)-P$ suppose to be? and what is the period on $n$? $\endgroup$ – John Feb 4 '14 at 18:48
  • $\begingroup$ The equation of the plane is $[r-r_0].n=0$ where r is a general point on the plane and $r_0$ is a fixed point on the plane and so $r-r_0$ would be perpendicular to $n$ and their scalar product is zero $\endgroup$ – Semsem Feb 4 '14 at 18:51
  • $\begingroup$ oh i see. from the $n$ i obtained above. I get: $5(x-1)-8(y-5)-(z-1)=0$ $\endgroup$ – John Feb 4 '14 at 18:53
  • $\begingroup$ @John I think $n$ should be $3i-8j-k$ $\endgroup$ – Pratyush Nov 28 '15 at 6:30

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