Given two vectors A and B A=2i +j _3k and B=3i +2j+k .find A×B, B, a.b , c,the angle between a.b is

1 answer

A x B = (2j - 3k)i + (-3i + 2k)j + (2 - 2j)k

a.b = 2(3) + 1(2) + (-3)(1) = 7

c = |A| |B| = sqrt(13) sqrt(6) = sqrt(78)

The angle between A and B = cos-1(7/sqrt(78)) = 57.3°