If #A = <2 ,4 ,-1 >#, #B = <3 ,8 ,2 ># and #C=A-B#, what is the angle between A and C?

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Answer 1 · 2016-09-30T19:24:17

#theta ~~ 2.437# radians

#C = <2-3, 4-8, -1-1>#

#C = <-1, -4, -2>#

The rectangular definition of the dot-product is:

#A•C = (A_x)(C_x) + (A_y)(C_y) + (A_z)(C_z)#

#A•C = (2)(-1) + (4)(-4) + (-1)(-2)#

#A•C = -2 + -16 + 2#

#A•C = -16#

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

#|A| = sqrt(21)#

#|C| = sqrt((-1)^2 + (-4)^2 + (-2)^2)#

#|C| = sqrt(1 + 16 + 4)#

#|C| = sqrt(21)#

The polar definition of the dot-product is:

#A•C = |A||C|cos(theta)#

where #theta# is the angle between the vectors.

Substituting what we have into the above:

#-16 = sqrt21sqrt21cos(theta)#

Solve for #theta#:

#theta = cos^-1(-16/21)#

#theta ~~ 2.437# radians