# If A = <2 ,2 ,5 >, B = <5 ,-7 ,-1 > and C=A-B, what is the angle between A and C?

Dec 17, 2017

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#### Explanation:

$A = \left(\begin{matrix}2 \\ 2 \\ 5\end{matrix}\right)$

$B = \left(\begin{matrix}5 \\ - 7 \\ - 1\end{matrix}\right)$

$C = \left(\begin{matrix}2 \\ 2 \\ 5\end{matrix}\right) - \left(\begin{matrix}5 \\ - 7 \\ - 1\end{matrix}\right) = \left(\begin{matrix}- 3 \\ 9 \\ 6\end{matrix}\right)$

Angle between vectors can be found by using the dot product:

$A \cdot C = | | A | | \cdot | | C | | \cos \left(\theta\right)$

$A \cdot C = - 6 + 18 + 30 = 42$

$| | A | | = \sqrt{{\left(2\right)}^{2} + {\left(2\right)}^{2} + {\left(5\right)}^{2}} = \sqrt{33}$

$| | C | | = \sqrt{{\left(- 3\right)}^{2} + {\left(9\right)}^{2} + {\left(6\right)}^{2}} = \sqrt{126} = 3 \sqrt{14}$

$42 = \sqrt{33} \cdot 3 \sqrt{14} \cos \theta$

$\cos \theta = \frac{42}{\sqrt{33} \cdot 3 \sqrt{14}} \implies \theta = \arccos \left(\frac{42}{\sqrt{33} \cdot 3 \sqrt{14}}\right) = {49.36}^{\circ}$

2 .d.p.