# What is the projection of << 3, -6, 2 >> onto << 3, 2, 1 >>?

May 31, 2016

It is <-3/14, -1/7, -1/14>.

#### Explanation:

The length of the projection of a vector ${v}_{1}$ on the vector ${v}_{2}$ is given by the dot product of ${v}_{1} \cdot {v}_{2}$.
The dot product is the sum of the products of the components:

${v}_{1} = < 3 , - 6 , 2 >$
${v}_{2} = < 3 , 2 , 1 >$

${v}_{1} \cdot {v}_{2} = 3 \cdot 3 + \left(- 6\right) \cdot 2 + 2 \cdot 1 = 9 - 12 + 2 = - 1$.

The direction of the vector is the one of ${v}_{2}$ but we need to take the unitary vector in that direction that is the vector ${v}_{2}$ divided by its length.
The length is calculated again with the scalar product, this time of ${v}_{2}$ by itself (it is like to project the vector on itself)

$| | {v}_{2} | {|}^{2} = {v}_{2} \cdot {v}_{2} = 3 \cdot 3 + 2 \cdot 2 + 1 \cdot 1 = 14$

So the unitary vector in the direction of ${v}_{2}$ is

$\setminus {\hat{v}}_{2} = {v}_{2} / 14 = < \frac{3}{14} , \frac{2}{14} , \frac{1}{14} \ge < \frac{3}{14} , \frac{1}{7} , \frac{1}{14} >$

This unitary vector gives the direction, while the length and direction was given by the initial dot product that was $- 1$.
So the final projection is

$- 1 \cdot \setminus {\hat{v}}_{2} = < - \frac{3}{14} , - \frac{1}{7} , - \frac{1}{14} >$.

May 31, 2016

I got $\left\langle- \frac{3}{14} , - \frac{1}{7} , - \frac{1}{14}\right\rangle$, and Wolfram Alpha confirms it.

PROJECTION OF $\setminus m a t h b f \left(\vec{a}\right)$ ONTO $\setminus m a t h b f \left(\vec{b}\right)$

Let $\vec{a} = \left\langle3 , - 6 , 2\right\rangle$ and $\vec{b} = \left\langle3 , 2 , 1\right\rangle$.

The projection of $\vec{a}$ onto $\vec{b}$, which yields a vector, is:

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

This is drawn as:

For this, we should define a few more things.

• The dot product of two $n$-long vectors $\vec{u}$ and $\vec{v}$:

$\setminus m a t h b f \left(\vec{u} \cdot \vec{v}\right)$

$= \left\langle{u}_{1} , {u}_{2} , . . . , {u}_{N}\right\rangle \cdot \left\langle{v}_{1} , {v}_{2} , . . . , {v}_{N}\right\rangle$

$= \setminus m a t h b f \left({u}_{1} {v}_{1} + {u}_{2} {v}_{2} + . . . + {u}_{N} {v}_{n}\right)$

• The norm of $\vec{v}$:

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

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

$= \sqrt{{v}_{1} {v}_{1} + {v}_{2} {v}_{2} + . . . + {v}_{N} {v}_{N}}$

$= \setminus m a t h b f \left(\sqrt{{v}_{1}^{2} + {v}_{2}^{2} + . . . + {v}_{N}^{2}}\right)$

COMPONENTS OF THE PROJECTION

So, what we now have to do is:

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

$= \left\langle3 , - 6 , 2\right\rangle \cdot \left\langle3 , 2 , 1\right\rangle$

$= 3 \cdot 3 + \left(- 6 \cdot 2\right) + 2 \cdot 1$

$= 9 - 12 + 2$

$= \textcolor{g r e e n}{- 1}$

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

$= \sqrt{\left\langle3 , 2 , 1\right\rangle \cdot \left\langle3 , 2 , 1\right\rangle}$

$= \sqrt{3 \cdot 3 + 2 \cdot 2 + 1 \cdot 1}$

$= \sqrt{9 + 4 + 1}$

$= \textcolor{g r e e n}{\sqrt{14}}$

FINAL CALCULATION

Finally, put it all back in the first equation.

$\textcolor{b l u e}{{\text{proj}}_{\vec{b}} \vec{a}} = \frac{\vec{a} \cdot \vec{b}}{| | \vec{b} | | \cdot | | \vec{b} | |} \vec{b}$

$= \frac{- 1}{\sqrt{14} \cdot \sqrt{14}} \left\langle3 , 2 , 1\right\rangle$

$= \textcolor{b l u e}{\left\langle- \frac{3}{14} , - \frac{1}{7} , - \frac{1}{14}\right\rangle}$