# Answer to Question #2724 in Vector Calculus for saman khan

Question #2724

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

Expert's answer

The module of the vector is √(3

As

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

^{2}+ 6^{2}+ 3^{2}) = √54As

**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

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