# If A = <6 ,1 ,5 >, B = <4 ,6 ,-2 > and C=A-B, what is the angle between A and C?

Nov 7, 2016

${52.84}^{o}$

#### Explanation:

C=A-B=<2,-5,7>
So we have to find angle between <6,1,5> and <2,-5,7>

If that angle is $\theta$ then Cos theta = (A.C)/(|A| |C|

= $\frac{6 \cdot 2 + 1 \cdot \left(- 5\right) + 5.7}{\sqrt{{6}^{2} + {1}^{2} + {5}^{2}} \sqrt{{2}^{2} + {\left(- 5\right)}^{2} + {7}^{2}}}$

= $\frac{42}{\sqrt{62} \sqrt{78}}$

=21/(sqrt31 sqrt39= 0.6039

$\theta$= arccos 0.6039=${52.84}^{o}$