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

Jan 17, 2018

${26.54}^{\circ}$

#### Explanation:

$A = \left[\begin{matrix}3 \\ 8 \\ - 1\end{matrix}\right]$

$B = \left[\begin{matrix}4 \\ - 3 \\ - 1\end{matrix}\right]$

$C = A - B = \left[\begin{matrix}3 \\ 8 \\ - 1\end{matrix}\right] - \left[\begin{matrix}4 \\ - 3 \\ - 1\end{matrix}\right] = \left[\begin{matrix}3 - 4 \\ 8 - \left(- 3\right) \\ - 1 - \left(- 1\right)\end{matrix}\right] = \left[\begin{matrix}- 1 \\ 11 \\ 0\end{matrix}\right]$

We can find the angle between vectors using the Dot Product

The dot product states that for vectors a and b:

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

The dot product is sometimes called the inner product, because of the way the vectors a and b are multiplied and summed.

In algebra we are used to multiplying brackets in the following way.

$\left(a + b\right) \left(c + d\right) = a c + a d + b c + b d$

In the case of the dot product we multiply the vectors in the following way.

$\left(a + b + c\right) \cdot \left(d + e + f\right) = a d + b e + c f$

So we are just multiplying corresponding components and then adding them together.

Let $a = \left[\begin{matrix}x \\ y \\ z\end{matrix}\right]$

Magnitude of $a = | | a | |$

$| | a | | = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$

Now to our example:

First find the product of:

$A \cdot C$

$\left[\begin{matrix}3 \\ 8 \\ - 1\end{matrix}\right] \cdot \left[\begin{matrix}- 1 \\ 11 \\ 0\end{matrix}\right] = \left[\begin{matrix}3 \times - 1 \\ 8 \times 11 \\ - 1 \times 0\end{matrix}\right]$

$= \left[\begin{matrix}- 3 \\ 88 \\ 0\end{matrix}\right] = - 3 + 88 + 0 = 85$

Now find the magnitudes of A and C:

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

$| | C | | = \sqrt{{\left(- 1\right)}^{2} + {\left(11\right)}^{2} + {\left(0\right)}^{2}} = \sqrt{122}$

So we have for:

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

$85 = \sqrt{74} \cdot \sqrt{122} \cdot \cos \left(\theta\right)$

$\cos \left(\theta\right) = \frac{85}{\sqrt{74} \cdot \sqrt{122}}$

$\theta = \arccos \left(\cos \left(\theta\right)\right) = \arccos \left(\frac{85}{\sqrt{74} \cdot \sqrt{122}}\right) = {26.54}^{\circ}$
( 2 .d.p.)

So the angle between vectors A and C is ${26.54}^{\circ}$

Notice from the diagram that the angle found by the dot product, is the angle between the vectors where they are pointing in the same relative direction. 