Find the angle between each of these pairs of vectors:

Solution 47E a) Two vectors are given as, A = 2 i + 6 j B = 2 i 3 j We know the dot product of two vectors can be written as, A. B = AxB x A y -y------------------(1) Where A xB axd A ,y aye the components of the two vectors. And it can also be shown as, A. B = | || |s ---------------(2) Where is the angle between two vectors. The magnitude of a vector we can get by the following rule, | |= A + A 2 = ( 2) + 6 = 4 + 36 = 40 = 6.32 . x y | |= Bx2+ B y2= (2) + 3 = 4 + 9 = 13 = 3.60 From equation (1) and (2), A. B = A B cos = A B + A B | || | x x y y 6.32 × 3.60 × cos = ( 2) × 2 + 6 × ( 3) 22.78 × cos = 4 18 = 22 cos = 22 / 22.78 = 0.965 1 0 = cos ( 0.965) = 165 So the actual angle between these two vectors is