Question

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

Answer 1

`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||)`