Answer to Question #2724 in Vector Calculus for saman khan

Question

Find the angles which the vector 3i-6j+3k makes with the coordinate axes.

Answer 1

The module of the vector is √(3² + 6² + 3²) = √54
As a b = |a||b| cos α,
we can obtain
the angle with i - axis (1,0,0):
cos α1 = (3*1 -6*0 + 3*0)/√54*1 = 0.41, α1 = 66 degrees

the angle with j - axis (0,1,0):
cos α2 = (3*0 -6*1 + 3*0)/√54*1 = - 0.82 α2 = 145 degrees

the angle with k- axis (0,0,1):
cos α3 = (3*0 -6*0 + 3*1)/√54*1 = 0.41 α3 = 66 degrees