If A = <8 ,1 ,4 >, B = <6 ,5 ,-8 > and C=A-B, what is the angle between A and C?

1 Answer
Mar 29, 2018

First figure out the C.

C = A - B = langle8,1,4rangle - langle6,5,-8rangle=langle8-6,1-5,4--8rangle=langle2,-4,12rangle

So now we need to find the angle between A and C. We can use this formula costheta=frac{u*v}{||u||\ ||v||}.

A*C = langle8,1,4rangle*langle2,-4,12rangle=8*2+1*-4+4*12=60

||A||=sqrt{8^2+1^2+4^2} = sqrt{81}=9
||C||=sqrt{2^2+(-4)^2+12^2} = sqrt164 =2sqrt41

So putting these values together would get you costheta=frac{60}{9*2sqrt41} = frac{60}{sqrt41} = \frac{10}{3\sqrt41}

Now take the inverse of cosine function on both sides to obtain theta = cos^{-1}(frac{10}{3sqrt41})\approx58.6289°