# If A = <3 ,-1 ,8 >, B = <4 ,-3 ,-1 > and C=A-B, what is the angle between A and C?

Oct 22, 2017

${36.22}^{o}$ **(2,d.p.)

#### Explanation:

A= $\left(\begin{matrix}3 \\ - 1 \\ 8\end{matrix}\right)$

B= $\left(\begin{matrix}4 \\ - 3 \\ - 1\end{matrix}\right)$

C= A - B: C= $\left(\begin{matrix}3 \\ - 1 \\ 8\end{matrix}\right) - \left(\begin{matrix}4 \\ - 3 \\ - 1\end{matrix}\right) = \left(\begin{matrix}- 1 \\ 2 \\ 7\end{matrix}\right)$

Angle between A and C:

$\left(\begin{matrix}3 \\ - 1 \\ 8\end{matrix}\right) \cdot \left(\begin{matrix}- 1 \\ 2 \\ 7\end{matrix}\right)$

This is called the Dot product, Scaler product or Inner product and is defined as:

$a \cdot b = | a | \cdot | b | \cos \left(\theta\right)$

$| A | = \sqrt{{3}^{2} + {\left(- 1\right)}^{2} + {8}^{2}} = \sqrt{74}$

$| C | = \sqrt{{\left(- 1\right)}^{2} + {2}^{2} + {7}^{2}} = \sqrt{54} = 3 \sqrt{6}$

$A \cdot C = \left(\begin{matrix}3 \\ - 1 \\ 8\end{matrix}\right) \cdot \left(\begin{matrix}- 1 \\ 2 \\ 7\end{matrix}\right) = \left(\begin{matrix}- 3 \\ - 2 \\ 56\end{matrix}\right)$

We need to sum the components in the product vector, so:

$- 2 - 3 + 56 = 51$

So now we have:

$51 = \sqrt{74} \cdot 3 \sqrt{6} \cos \left(\theta\right) = \frac{51}{\sqrt{74} \cdot 3 \sqrt{6}} = \cos \left(\theta\right)$

$\to \frac{51}{3 \sqrt{444}} = \cos \left(\theta\right)$

$\theta = \arccos \left(\frac{51}{3 \sqrt{444}}\right) = {36.22}^{o}$ (2,d.p.)

The angle is formed where the vectors are moving in the same relative direction. i.e where both arrow heads are pointing.