# How do you find the projection of u onto v given u=<3, 15> and <v=-1, 5>?

Feb 4, 2017

The vector projection is $= \frac{36}{13} < - 1 , 5 >$
The scalar projection is $= \frac{72}{\sqrt{26}}$

#### Explanation:

he vector projection of $\vec{u}$ onto $\vec{v}$ is

$= \frac{\vec{u} . \vec{v}}{| | \vec{v} | {|}^{2}} \cdot \vec{v}$

$\vec{u} . \vec{v} = < 3 , 15 > . < - 1 , 5 \ge - 3 + 75 = 72$

The modulus of $\vec{v}$ is

$| | \vec{v} | | = | | < - 1 , 5 > | | = \sqrt{1 + 25} = \sqrt{26}$

The vector projection is

$= \frac{72}{26} < - 1 , 5 > = \frac{36}{13} < - 1 , 5 >$

The scalar projection is

$\frac{\vec{u} . \vec{v}}{| | \vec{v} | |} = \frac{72}{\sqrt{26}}$