Find the angles which the vector 3i-6j+3k makes with the coordinate axes.
1
Expert's answer
2011-06-01T14:41:01-0400
The module of the vector is √(32 + 62 + 32) = √54 As ab = |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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