If #A = <8 ,3 ,-7 >#, #B = <6 ,-4 ,5 >#, and #C=A-B#, what is the angle between A and C?

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Answer 1 · 2018-05-06T15:34:21

#theta = arccos (121/sqrt(24034) ) approx 38.69^circ #

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

# A = (8,3,-7)#

#C= A-B=(8,3,-7)-(6,-4,5) = (2,7,-12)#

# (8, 3, -7) cdot (2,7,-12) = |(8", "3", "-7)| \ |(2", "7 ", " -12)| \ cos theta#

#cos theta = {8(2) + 3(7) + -7(-12)}/{\sqrt{ (8^2+3^2+7^2)(2^2+7^2+12^2) }} = 121/sqrt(24034) #

#theta approx 38.69^circ #