If #A = <4 ,9 ,-7 >#, #B = <5 ,8 ,-3 ># and #C=A-B#, what is the angle between A and C?

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Answer 1 · 2017-11-22T11:12:13

The angle between #vecA and vecC# is #49.93^0#

#vec C=vecA- vecB = (< 4,9,-7 >) - (< 5,8,-3 >)#

#=(< 4-5,9-8,-7+3 >) =< -1,1,-4 >#

#vecA =<4,9, -7> and vecC =<-1,1,-4> ; theta# be the

angle between them ; then we know

#cos theta= (vecA*vecC)/(||vecA||*||vecC||)#

#=((4.-1)+(9*1)+(-7*-4))/(sqrt(4^2+9^2+(-7)^2)* (sqrt((-1)^2+1^2+(-4)^2))#

#= 33/(sqrt146*sqrt18)=33/51.26 ~~0.6437#

#:. theta=cos^-1(0.6437)~~49.93^0#

The angle between #vecA and vecC# is #49.93^0# [Ans]