If #A = <6 ,1 ,5 >#, #B = <4 ,6 ,-2 ># and #C=A-B#, what is the angle between A and C?

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Answer 1 · 2016-11-07T17:59:50

#52.84^o#

C=A-B=<2,-5,7>
So we have to find angle between <6,1,5> and <2,-5,7>

If that angle is #theta# then Cos #theta = (A.C)/(|A| |C|#

= #(6*2 +1*(-5) +5.7)/(sqrt(6^2+1^2+5^2) sqrt(2^2 +(-5)^2 +7^2))#

= #42/(sqrt62 sqrt78)#

=#21/(sqrt31 sqrt39#= 0.6039

#theta#= arccos 0.6039=#52.84^o#