What is the angle between the vectors A = 2.00i + 6.00j, and B=2.00i - 3.00j?

1 Answer
Trevor Ryan.
Jul 30, 2016

#127.875^@#.

Explanation:

We may use the Euclidean inner product in #RR^2# in this case to find the angle #theta# between the vectors #A and B# as follows :

#costheta=(A*B)/(||A|| ||B||)#

#therefore theta = cos^(-1) (((2xx2)+(6xx-3))/(sqrt(2^2+6^2)xxsqrt(2^2+(-3)^2)))#

#=cos^(-1) ((-14)/(sqrt40sqrt13))#

#=127.875^@#.

Alternatively we could also have used the vector cross product to obtain the same answer, where
#sintheta=(A xx B)/(||A||||B||)#

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