# If veca = <8 ,1 ,-5 >, vecb = <6 ,-2 ,-8 > and vecc=veca-vecb, what is the angle between veca and vecc?

May 4, 2016

I get about ${84.84}^{\circ}$ or $\text{1.48 rad}$.

The angle between $\vec{a}$ and $\vec{c}$ is ultimately determined from the following relationship:

$\setminus m a t h b f \left(\vec{a} \cdot \vec{c} = | | \vec{a} | | \cdot | | \vec{c} | | \cos \theta\right)$

Therefore, we will need to find $\vec{c}$ by subtracting $\vec{a}$ and $\vec{b}$.

$\textcolor{g r e e n}{\vec{a} - \vec{b}}$

$= \left\langle8 , 1 , - 5\right\rangle - \left\langle6 , - 2 , - 8\right\rangle$

$= \left\langle8 - 6 , 1 - \left(- 2\right) , - 5 - \left(- 8\right)\right\rangle$

$= \textcolor{g r e e n}{\left\langle2 , 3 , 3\right\rangle = \vec{c}}$

Next, the angle between $\vec{a}$ and $\vec{c}$ is gotten like so:

$\cos \theta = \frac{\vec{a} \cdot \vec{c}}{| | \vec{a} | | \cdot | | \vec{c} | |}$

$\textcolor{g r e e n}{\theta = \arccos \left(\frac{\vec{a} \cdot \vec{c}}{| | \vec{a} | | \cdot | | \vec{c} | |}\right)}$

Afterwards, we do not yet know $\vec{a} \cdot \vec{c}$, so we'll need to evaluate the dot product between $\vec{a}$ and $\vec{c}$, which is just a sum of product terms consisting of corresponding vector coordinates.

$\textcolor{g r e e n}{\vec{a} \cdot \vec{c}}$

$= \left\langle8 , 1 , - 5\right\rangle \cdot \left\langle2 , 3 , 3\right\rangle$

$= 8 \cdot 2 + 1 \cdot 3 + - 5 \cdot 3$

$= 16 + 3 - 15 = \textcolor{g r e e n}{4}$

And lastly, we'll need to know the product of the norms of $\vec{a}$ and $\vec{c}$. The norm of some arbitrary vector $\vec{v}$ is

$\setminus m a t h b f \left(| | \vec{v} | | = \sqrt{\vec{v} \cdot \vec{v}}\right) ,$

so we'll need to dot $\vec{a}$ and $\vec{c}$ with themselves and take the square root to get $| | \vec{a} | |$ and $| | \vec{c} | |$, respectively.

$\textcolor{g r e e n}{| | \vec{a} | |} = \sqrt{\vec{a} \cdot \vec{a}}$

$= \sqrt{\left\langle8 , 1 , - 5\right\rangle \cdot \left\langle8 , 1 , - 5\right\rangle}$

$= \sqrt{8 \cdot 8 + 1 \cdot 1 + \left(- 5\right) \cdot \left(- 5\right)}$

$= \sqrt{64 + 1 + 25}$

$= \textcolor{g r e e n}{\sqrt{90}}$

And now for $\vec{c}$:

$\textcolor{g r e e n}{| | \vec{c} | |} = \sqrt{\vec{c} \cdot \vec{c}}$

$= \sqrt{\left\langle2 , 3 , 3\right\rangle \cdot \left\langle2 , 3 , 3\right\rangle}$

$= \sqrt{2 \cdot 2 + 3 \cdot 3 + 3 \cdot 3}$

$= \sqrt{4 + 9 + 9}$

$= \textcolor{g r e e n}{\sqrt{22}}$

Finally, we can combine all this to get the angle between $\vec{a}$ and $\vec{c}$ as:

$\textcolor{b l u e}{\theta} = \arccos \left(\frac{\vec{a} \cdot \vec{c}}{| | \vec{a} | | \cdot | | \vec{c} | |}\right)$

$= \arccos \left(\frac{4}{\sqrt{90} \cdot \sqrt{22}}\right)$

$= \arccos \left(\frac{4}{3 \sqrt{10} \cdot \sqrt{22}}\right)$

$= \arccos \left(\frac{4}{3 \sqrt{220}}\right)$

$= \arccos \left({\cancel{4}}^{2} / \left(3 \cdot \cancel{2} \sqrt{55}\right)\right)$

$= \arccos \left(\frac{2}{3 \cdot \sqrt{55}}\right)$

$\approx$ $\textcolor{b l u e}{\text{1.48 rad}}$ or about $\textcolor{b l u e}{{84.84}^{\circ}}$.

Indeed, that should be the answer, as we see the result here to be:

$\arccos \left(\frac{2}{3 \sqrt{55}}\right)$.