I am just having a difficult time processing this. Suppose I have a line which is orthogonal to a plane meaning that it is parallel to the normal vector of the plane. I just don't understand why a line cannot be orthogonal to two non-parallel planes while a 3rd plane can be orthogonal. Can the line not simply point the same direction as to the normal vector of the 3rd plane which is orthogonal to the 2 non-parallel planes. Like for example, one of the questions I got was: Find $P_3$ if it is possible between 2 planes with equations:

$P_1 : −2x + y − 4z = 2$,

$P_2 : x + 2y = 7$.

I simply found the normal vectors of $(-2,1,-4)$ and $(1,2,0)$ and made normal vector of $P_3$ be $(x,y,z)$ with the dot product $n_3 . n_1 = 0$ and $n_2 . n_1 = 0$ Used matrices, got 1 free variable and got $x = -2t$, $y = t$ , and $z = 5t/4$ where t can be any real number and found a possible general equation equating $t = 4$ to get $P_3 = -8x + 4y + 5z$ from normal vector of $P_3 (-8, 4, 5)$

Find a line which is orthogonal to both $P_1$ and $P_2$. Give an equation if it is possible and explain why it is or is not possible: I simply don't know how to explain why a line is or is not possible though or visualize this. So I need help understanding how this works.

Any help is appreciated. Thanks.

  • 2
    $\begingroup$ What definitions are you using? A line is orthogonal to a plane if it is parallel to the normal vector, i.e., if it is orthogonal to any vector in the plane. $\endgroup$
    – pancini
    Sep 21, 2019 at 4:50
  • $\begingroup$ sorry thats what I meant, I wrote it wrong. $\endgroup$ Sep 21, 2019 at 5:40

1 Answer 1


A line, if orthogonal to two planes, must be parallel to both normal vectors of the two planes. When you set $n_1 \cdot n_3 = 0, n_2 \cdot n_3 = 0$ this means that the line spanned by $n_3$ is perpendicular to both $n_1$ and $n_2.$ But you want parallel. If you think about it, a line cannot be parallel to simultaneously to two intersecting vectors. That is why you need a plane - specifically a plane spanned by the normals to the two intersecting planes.

  • $\begingroup$ Oh Thank you, I think I get it now. I was just confusing myself. Thanks for the answer. $\endgroup$ Sep 21, 2019 at 5:43

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