What is the angle between <6 , 3 , 2 >  and  < 3 , -4 , 8 > ?

Nov 2, 2016

$\theta$= arc cos 0.3331 = 70.54 degrees

Explanation:

Let vector u be <6,3,2> and v be <3,-4,8>. If $\theta$ is the angle between these two vectors, then
cos theta= (u.v)/(||u|| ||v|| = $\frac{6 \cdot 3 - 3 \cdot 4 + 8 \cdot 2}{\sqrt{{6}^{2} + {3}^{2} + {2}^{2}} . \sqrt{{3}^{2} + {\left(- 4\right)}^{2} + {8}^{2}}}$

=$\frac{22}{7. \sqrt{89}}$ =0.3331

Hence $\theta$= arc cos 0.3331 = 70.54 degrees