# What is the projection of  <-3,0,1> onto <1,2,-4>?

Oct 28, 2017

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

#### Explanation:

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

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

Here,

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

$\vec{b} = < - 3 , 0 , 1 >$

The dot product is

$\vec{a} . \vec{b} = < 1 , 2 , - 4 > . < - 3 , 0 , 1 \ge \left(1\right) \times \left(- 3\right) + \left(2\right) \times \left(0\right) + \left(- 4\right) \times \left(1\right) = - 3 - 4 = - 7$

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,

The projection is

$p r o {j}_{\vec{a}} \vec{b} = \frac{- 7}{21} < 1 , 2 , - 4 > = - \frac{1}{3} < 1 , 2 , - 4 >$

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