Question

If `A = <4 ,-9 ,5 >`, `B = <1 ,-3 ,-8 >` and `C=A-B`, what is the angle between A and C?

Answer 1

`C=<3,-6,13> -> cos theta=(<4,-9,5>*<3,-6,13>)/(sqrt(4^2+(-9)^2+5^2) sqrt(3^2+(-6)^2+13^2)`
`cos theta = (12+54+65)/((sqrt122)(sqrt(214)) ) ->theta =cos^-1 (131/sqrt26108 ) = 35.83^@`

Explanation:

First find the vector C by subtracting vector B from vector . That is
`C=<4-1,-9-(-3),5-(-8)> =<3,-6,13>` Then find the dot product between vector A and C which is `A*C=4(3)+(-9)(-6)+(5)(13)`. And divide that by the magnitude of both vector A and C and then take the cosine inverse to find the angle