A vector is given as follows A 2 i 2 j 3 k The angle betw

A vector is given as follows: A = -2 i - 2 j - 3 k. The angle between vector A and the y-axis in degrees is

Solution

ande can be found using dot product:

A.y = |A||y^| cos@

A = (-2i - 2j - 3)

|A| = sqrt(2^2 + 2^2 + 3^2) = 4.123

y^ = j

|y^| = 1

((-2i - 2j - 3) . (j) = (4.123)(1)cos@

cos@ = - 2/ 4.123 = - 0.485

@ = 119 deg