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.

  • $\begingroup$ 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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.