If #A = <1 ,6 ,-8 >#, #B = <-9 ,4 ,1 ># and #C=A-B#, what is the angle between A and C?

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Answer 1 · 2018-04-28T13:37:11

#theta = arccos(94/sqrt{18685})approx 46.55^circ #

#A cdot C = |A||C| cos theta #

The dot product divided by the magnitudes gives the cosine of the angle.

#cos theta = { A cdot C}/{|A||C|} #

#C = A - B = (1,6,-8) -(-9,4,1) = (10, 2, -9)#

#cos theta = {(1,6,-8) cdot (10, 2, -9 ) } / {|((1,6,-8))||((10, 2, -9 )) |} #

#cos theta = {1(10) + 6(2) + -8(-9) }/{ sqrt{ (1^2+6^2+8^2 ) (10^2+2^2+9^2 ) } }#

#cos theta = 94/sqrt(18685)#

#theta = arccos(94/sqrt{18685})#

#theta approx 46.55^circ #