# What is the projection of < 2 , -6, 0> onto < 5, -3, 8>?

May 5, 2016

Let $\vec{a} = \left\langle2 , - 6 , 0\right\rangle$ and $\vec{b} = \left\langle5 , - 3 , 8\right\rangle$.

The projection of $\vec{a}$ onto $\vec{b}$ is written as:

$\setminus m a t h b f \left({\text{proj}}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{| | \vec{b} | | \cdot | | \vec{b} | |} \vec{b}\right)$

So first, we have to find the dot product between $\vec{a}$ and $\vec{b}$.

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

$= \left\langle2 , - 6 , 0\right\rangle \cdot \left\langle5 , - 3 , 8\right\rangle$

$= 2 \cdot 5 + - 6 \cdot - 3 + 0 \cdot 8$

$= 10 + 18 = \textcolor{g r e e n}{28}$

And the norm of $\vec{b}$ is:

$\textcolor{g r e e n}{| | \vec{b} | |}$

$= \sqrt{\vec{b} \cdot \vec{b}}$

$= \sqrt{\left\langle5 , - 3 , 8\right\rangle \cdot \left\langle5 , - 3 , 8\right\rangle}$

$= \sqrt{5 \cdot 5 + - 3 \cdot - 3 + 8 \cdot 8}$

$= \sqrt{25 + 9 + 64}$

$= \sqrt{98} = \textcolor{g r e e n}{7 \sqrt{2}}$

Thus, the projection of $\vec{a}$ onto $\vec{b}$ is:

$\textcolor{b l u e}{{\text{proj}}_{\vec{b}} \vec{a}}$

$= \frac{28}{7 \sqrt{2} \cdot 7 \sqrt{2}} \left\langle5 , - 3 , 8\right\rangle$

$= \frac{28}{98} \left\langle5 , - 3 , 8\right\rangle$

$= \frac{2}{7} \left\langle5 , - 3 , 8\right\rangle$

$= \textcolor{b l u e}{\left\langle\frac{10}{7} , - \frac{6}{7} , \frac{16}{7}\right\rangle}$ <-- let this be $\vec{c}$.

At this point, to check that this worked, you should perform the following to determine whether the angle between the vectors is indeed ${0}^{\circ}$:

$\vec{b} \cdot \vec{c} = | | \vec{b} | | \cdot | | \vec{c} | | \cos \theta$

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

$= \arccos \left(\frac{5 \cdot \frac{10}{7} + - 3 \cdot - \frac{6}{7} + 8 \cdot \frac{16}{7}}{\sqrt{{5}^{2} + {\left(- 3\right)}^{2} + {8}^{2}} \cdot \sqrt{{\left(\frac{10}{7}\right)}^{2} + {\left(- \frac{6}{7}\right)}^{2} + {\left(\frac{16}{7}\right)}^{2}}}\right)$

$= \arccos \left(\frac{\frac{50 + 18 + 128}{7}}{7 \sqrt{2} \cdot \sqrt{\frac{100 + 36 + 256}{49}}}\right)$

$= \arccos \left(\frac{\frac{196}{7}}{7 \sqrt{2} \cdot \sqrt{8}}\right)$

$= \arccos \left(\frac{28}{28}\right)$

$= \textcolor{b l u e}{{0}^{\circ}}$