Question

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

Answer 1

`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^@`.