Question

If `A = <8 ,3 ,2 >`, `B = <-1 ,-1 ,5 >`, and `C=A-B`, what is the angle between A and C?

Answer 1

`C = <8-(-1),3-(-1),2-5> = <9,4,-3> -> cos theta = (A*C)/(|A||C|) =( <8,3,2>*<9,4,-3>)/(sqrt(8^2+3^2+2^2) sqrt(9^2+4^2+(-3)^2) )` `theta =cos^-1 ( ((8)(9)+(3)(4)+(2)(-3))/(sqrt77 sqrt106 ) )= cos^-1 (78/sqrt8162), theta = 30.3^@`

Explanation:

Find vector C then to find theta you take the cosine inverse of the dot product of vectors A and C divided by each of their magnitude.