Angle between two planes

Question

I have two triangles in 3d space which share 1 edge and I would like to determine if the angle between their normal vectors is "uphill" or "downhill". In other words, if you set one flat, would the angle between them be acute (uphill) or obtuse (downhill). Hopefully that makes sense.

So far I have the normal vectors of each triangle and have calculated the angle between the two triangles, however this angle is always >= 90 degrees.

Vector3 v1 = triangle1.Normal;
Vector3 v2 = triangle2.Normal;
float cosAngle = Vector3.Dot(v1, v2) / (v1.magnitude * v2.magnitude);
float degAngle = Mathf.Acos(cosAngle) * Mathf.Rad2Deg;

Is there a trick I am missing that lets me determine which direction this angle is in?

Answer 1

Here is my suggestion. You just need two vectors.

• in the first triangle, choose the normal vector, n1
• in the second triangle, choose a vector e2 from a point on the shared edge to the point not on the shared edge.

Then compute their dot product: n1 . e2. If it’s positive, the angle is acute. If it’s negative, the angle is obtuse.

Answer 2

Consider the plane parallel to the normal of the first triangle and containing the shared edge. You can find a vector normal to this plane by crossing the shared edge with the normal of the first triangle. Then take the dot product of this vector with the second triangle normal - the sign of the dot product will tell you whether the shared edge forms a hill or a valley (with zero indicating the triangles are co-planar to within the available precision).

A picture should help explain why this is so. I'm visualizing it with quads because it's easier to show perspective, but it applies equally to triangles. Note that your signs may be reversed based on various choices of coordinate system/order of terms.