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

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Answer 1 · 2018-03-29T03:52:44

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°#