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Question: A plane has $-4y + 6z - 4 = 0$ as its Cartesian equation. Determine the Cartesian equation of a plane that is perpendicular to and contain the point $P(-3, -10, 4)$.

I tried doing this question on my own but I messed up and I don't understand how I'm supposed to find the answer to this question. To solve the question, I tried to use the cross product but I got even more confused when doing it. Am I supposed to use the cross product? Or do I use another method? i would appreciate if anyone can help me out.

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  • $\begingroup$ If a plane is $n.(x,y,z)^T+d=0$ then $n$ is the plane normal vector, it's perpendicular to the plane. If you cross product the vector with an arbitrary vector $a$: $[n\times a]$ -- the resulting vector $n_1$ should work as the normal vector of the new plane (why?) and you can get the plane equation in the form $n_1.((x,y,z)^T-P)=0$ $\endgroup$ – Alexey Burdin May 26 at 23:47
  • $\begingroup$ Start by choosing a suitable equation for a plane, one that lends itself to the key words “perpendicular” and “contain the point”.... Anything ring a bell? Also, how can we get a normal for a plane just by reading its equation? That’s a skill you should keep in your toolbox. $\endgroup$ – gen-z ready to perish May 27 at 0:54
  • $\begingroup$ Do you recognize that there’s not a unique solution to this problem? $\endgroup$ – amd May 27 at 4:43
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Take $y=0$ or $1$ and then you can get two position vector. Then the direction of the plane can be found by using cross product of the two direction you had obtained.

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