# How do you find the projection of u onto v given u=<0, 3> and v=<2, 15>?

Feb 26, 2017

The vector projection is $= \frac{45}{\sqrt{229}} < 2 , 15 >$
The scalar projection is $= \frac{45}{\sqrt{229}}$

#### Explanation:

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

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

$\vec{u} . \vec{v} = < 0 , 3 > , < 2 , 15 \ge 0 + 45 = 45$

The modulus of $\vec{v}$ is

$= | | \vec{v} | | = | | \vec{v} | | = | | < 2 , 15 | |$

$= \sqrt{4 + 225} = \sqrt{229}$

The vector projection is

$= \frac{45}{\sqrt{229}} < 2 , 15 >$

The scalar projection is

$= \frac{\vec{u} . \vec{v}}{| | \vec{v} | |}$

$= \frac{45}{\sqrt{229}}$