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^@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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