If #A = <2 ,3 ,-4 >#, #B = <6 ,1 ,2 ># and #C=A-B#, what is the angle between A and C?

1 Answer
Jan 3, 2016

#56.912^@#

Explanation:

#C=A-B#
#=(2,3,-4)-(6,1,2)#
#=(-4,2,-6)#.

Now to find the angle #theta# between any 2 vectors A and C, we may use either of the following 2 theorems involving algebraic products :

  1. #AxxC=||A|| * ||C|| * sin theta#
  2. #A*B=||A||*||B||*cos theta#

Since it is quicker to evaluate a Euclidean inner product as opposed to a vector cross product, I shall solve this problem using the latter theorem.

#therefore cos theta = (A*C)/(|A||*||C||#

#therefore theta = cos^(-1)((-8+6+24)/(sqrt(2^2+3^2+4^2)*sqrt(4^2+2^2+6^2)))#

#=cos^(-1)(22/(sqrt29*sqrt56))#

#=56.912^@#.