# What is the projection of < 5 ,- 3, 2 > onto < -1 , -2 , -4 >?

##### 1 Answer
Mar 1, 2018

The vector projection is $= < \frac{1}{3} , \frac{2}{3} , \frac{4}{3} >$

#### Explanation:

The vector projection of $\vec{b}$ onto $\vec{a}$ is

$p r o {j}_{\vec{a}} \vec{b} = \frac{\vec{a} . \vec{b}}{| | \vec{a} | |} ^ 2 \cdot \vec{a}$

$\vec{a} = < - 1 , - 2 , - 4 >$

$\vec{b} = < 5 , - 3 , 2 >$

The dot product is

$\vec{a} . \vec{b} = < - 1 , - 2 , - 4 > . < 5 , - 3 , 2 > = \left(- 1\right) \cdot \left(5\right) + \left(- 2\right) \cdot \left(- 3\right) + \left(- 4\right) \cdot \left(2\right)$

$= - 5 + 6 - 8 = - 7$

The modulus of $\vec{a}$ is

$| | \vec{a} | | = | | < - 1 , - 2 , - 4 > | | = \sqrt{{\left(- 1\right)}^{2} + {\left(- 2\right)}^{2} + {\left(- 4\right)}^{2}}$

$= \sqrt{1 + 4 + 16} = \sqrt{21}$

Therefore,

$p r o {j}_{\vec{a}} \vec{b} = \frac{- 7}{\sqrt{21}} ^ 2 \cdot < - 1 , - 2 , - 4 >$

$= - \frac{1}{3} < - 1 , - 2 , - 4 >$

$= < \frac{1}{3} , \frac{2}{3} , \frac{4}{3} >$