The question is as follows:

Write an equation for a plane that is perpendicular to the plane $2x − y + 3z = 6$ and that passes through the origin.

The normal vector I know would be $[a, b, c]$ where the variables belong in the equation $a(x_2 - x_1) + b(x_2 - x_1) + c(x_2 - x_1) = d$. Therefore, the normal vector of the given plane is $[2, -1, 3]$. Since it is going through the origin then the equation of the plane should be $2x - y + 3z = 0$. However, when I graph this online I get two planes that looks like they're parallel to each other.

enter image description here

Is there a place that I am making an error and, if so, where? Any help will be greatly appreciated.

  • $\begingroup$ The plane that's perpendicular to the given plan will have normal vector perpendicular to the normal vector of the given plane. $\endgroup$ – Vasya Apr 7 '18 at 1:48

The two planes are perpendicular to each other. So, their normals are also perpendicular to each other. Your two normals are the same and that's why the planes are parallel.

There are infinitely many possible planes that are perpendicular to the given plane and pass through the origin.

For example, as $(2,-1,3)\cdot(1,2,0)=0$, $x+2y=0$ is one of the possible answers.

  • $\begingroup$ Thank you so much for pointing out the error! $\endgroup$ – geo_freak Apr 7 '18 at 2:01

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