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

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Answer 1 · 2016-03-11T18:31:14

#20.438^@#

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

#=(10,12,17)#.

Hence the angle #theta# between #A and C# may be given in terms of the Euclidean inner product as follows

#theta=cos^(-1)((A*C)/(||A||||C||))#

#=cos^(-1)((1xx10+6xx12+9xx17)/(sqrt(1^2+6^2+9^2)(sqrt(10^2+12^2+17^2))))#

#=cos^(-1)((235)/(sqrt118sqrt533))#

#=20.438^@#