# how best to draw two planes intersecting at an angle which isn't $\pi /2$?

What's the best way to draw two planes intersecting at an angle that isn't $\pi /2$?

If I make them both vertical and vary the angle between them, the diagram always looks as though our viewpoint has changed but the planes are still intersecting at $\pi /2$.

I can't quite work out how to draw one or both of them non-vertical in such a way as to make the angle between them appear to be obviously not a right angle.

Thanks for any help with this!

• One good method is to take the dot product of their unit normal vectors, and take the arc cosine of that to get the angle between the planes, as in this related question. In particular, the planes are perpendicular iff the dot product of their normal vectors is zero. Also, the plane $ax+by+cz=d$ has normal vector $(a,b,c)$. Apr 17 '12 at 9:38
• If you were to look at the intersection from the line of intersection, the planes would clearly appear to intersect at an angle other than 90 degrees(provided they don't intersect at 90 degrees).
– Ben
Apr 17 '12 at 9:42
• @bgins - apologies for causing confusion - I meant to ask about drawing them, not 'showing' non-orthogonality in the mathematical sense. I've now amended the title and question to make this clearer Apr 17 '12 at 9:55
• @BenEysenbach - unfortunately I can't do that, because I need to show two distinct points on the line of intersection Apr 17 '12 at 9:56
• One way would be to take an acute triangle and extend the larger sides into planes, sometthing like here. Another would be to draw several intersecting radial lines and extend them all to planes, perhaps using color, something like here or here. Lastly, you might try drawing a parallelopiped (like here) and refer to the planes of the faces. Apr 17 '12 at 10:07