Q.Find the angle P of the triangle whose vertices are P$(0,-1,-2)$, Q$(3,1,4)$ and R$(5,7,1)$.

Solving a.b = |a||b| cos(theta) for theta gives cos(theta) = (a.b)/(|a||b|), so the angle between any two non-zero vectors is recovered directly from their dot product and magnitudes, and the sign of a.b alone reveals whether that angle is acute, right, or obtuse. Closely related is the idea of projection: the scalar projection of b onto a is the length (a.b)/|a| (the "shadow" b casts on the line of a), and the vector projection is that scalar length carried in the direction of a, namely [(a.b)/|a|^2] a. These ideas combine to find the angle at a vertex of a triangle give …