Find the angle between the diagonal of the back and left faces of a cube with one vertex on the origin A) 60 degrees B) cos^-1(1/sqrt(6)) C) cos^-1(1/3*sqrt(2)) D) 90 degrees E) 120 degrees The two diagonals eminate from the vertex at the origin. I will write to mean the "norm" of u and u*v to be u "dot" v I know we have the formula cos(theta) = u*v/( [v]). But we do not know know any points on the cube so we can't form any vectors... But since these are the diagonals of a cube they should be the same size . It which case u = v Then u*v = u^2 and = [v] which impies [v] = u^2. Then cos(theta) = 1 which implies theta = 0 which is definitely not the case... My other thought is that if these are diagonals on the face of a cube, shouldnt they bisect the 90 degree angles of the cube into 45 degree angles? Then the angle between the two vectors would be 45 + 45 = 90 degrees choice D). Is this correct?