Let vectors A = (2,1,-4), B = (-3,0,1), and C = (-1,-1,2)

What is the angle θAB between vector A and B?

The answer: to find the angle between the two vectors, the following method can be used: first, computing the scalar product between the two vectors A and B, after, their length. the next is to use the main formula cos T = A*B / / / A / / / / B//, where A*B is scalar product T = teta Practice: A*B = (2, 1,-4) * (-3, 0,1) = (2x-3) + (1x0) + (-4x1) = - 6-4 = - 10 / / A / / = sqrt (2² + 1² + 4²) = sqrt (4+1+16) = sqrt (21) = 4.58 / / B / / = sqrt (10) = 3.16 so, cosT = - 10 / 3.16*4.58 = - 10/14.48 = - 0.69, cosT = - 0.69, therefore, T = arccos (-0.69) = 46,33°