If #A = <4 ,1 ,8 >#, #B = <6 ,7 ,3 ># and #C=A-B#, what is the angle between A and C?

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Answer 1 · 2016-02-01T11:05:02

#cos^{-1}(frac{2sqrt65}{45})#

The first step is to find #C#.

#C = <4,1,8> - <6,7,3>#

#= <-2,-6,5>#

To find the acute angle between #A# and #C#, you could use either the dot product or the cross product. I prefer to use the dot product.

#A*C = |A| |C| cos theta#

#A*C = (4)(-2) + (1)(-6) + (8)(5) = 26#

#|A| = sqrt( 4^2 + 1^2 + 8^2 ) = 9#

#|C| = sqrt( (-2)^2 + (-6)^2 + 5^2 ) = sqrt65#

Therefore,

#cos theta = frac{26}{9sqrt65} = frac{2sqrt65}{45}#

#theta = cos^{-1}(frac{2sqrt65}{45})#