# What is the projection of < 3 , -7, 0> onto < -1, -4 , 6 >?

Mar 5, 2018

The vector projection is $= < - \frac{25}{53} , - \frac{100}{53} , \frac{150}{53} >$

#### Explanation:

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

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

$\vec{a} = < - 1 , - 4 , 6 >$

$\vec{b} = < 3 , - 7 , 0 >$

The dot product is

$\vec{a} . \vec{b} = < - 1 , - 4 , 6 > . < 3 , - 7 , 0 > = \left(- 1\right) \cdot \left(3\right) + \left(- 4\right) \cdot \left(- 7\right) + \left(6\right) \cdot \left(0\right)$

$= - 3 + 28 + 0 = 25$

The modulus of $\vec{a}$ is

$| | \vec{a} | | = | | < - 1 , - 4 , 6 > | | = \sqrt{{\left(- 1\right)}^{2} + {\left(- 4\right)}^{2} + {\left(6\right)}^{2}}$

$= \sqrt{1 + 16 + 36} = \sqrt{53}$

Therefore,

$p r o {j}_{\vec{a}} \vec{b} = \frac{25}{\sqrt{53}} ^ 2 \cdot < - 1 , - 4 , 6 >$

$= \frac{25}{53} < - 1 , - 4 , 6 >$

$= < - \frac{25}{53} , - \frac{100}{53} , \frac{150}{53} >$