Projection is finding the length of a vectors "shadow" across another.
The [[Orthogonal]] projection of $\vec u$ onto $\vec v$.
$\huge\proj{\vec{v}}\pa{\vec u} = \vec v \pa{\frac{\vec u \cdot v }{\norms{v}}}$
### Orthogonal Complement of the Projection
The to get the vector of $\vec u$ projecting down to $\vec v$.
$\huge \op{proj}_{\vec v\perp}\paren{\vec u} $
$ \huge \projperp{\vec{v}}\pa{\vec u} = \vec u - \proj{\vec v}\pa{\vec u}$
### Illustration
![[../../00 Excalidraw/Vector Projection .excalidraw.dark.svg]]
## Properties
$\huge{\op{proj}_{\vec v}\paren{\vec u}} \parallel \vec v $
*The projection is parallel to the vector being projected onto*
$\huge{\op{proj}_{\vec v \perp}\paren{\vec u }} \perp \vec v $
*The vector from the projection to vector being projected is perpendicular to the vector being projected onto*
$\huge \op{proj}_{\vec v}\paren{\vec u} +
\op{proj}_{\vec v\perp}\paren{\vec u} = \vec u \\
$
*The sum of the shadow and direction of u to the shadow is u*
### Relations to the Zero Vector
If the two vectors are [[Orthogonal]], then the projection will equal the [[Zero Vector]].
$\huge\begin{align*}
\vec v \perp \vec u \\
\op{proj}_{\vec v}\paren{\vec u} &= \vec 0 \\
\op{proj}_{\vec v\perp}\paren{\vec u} &= -\vec u
\end{align*}$
If the two vectors are parallel, then the projection is the vector, and the complement is the [[Zero Vector]]
$\huge\begin{align*}
\vec v \parallel \vec u \\
\op{proj}_{\vec v}\paren{\vec u} &= \vec u \\
\op{proj}_{\vec v\perp}\paren{\vec u} &= \vec 0
\end{align*}$
$\huge\begin{align*}
\parallel\op{proj}_{\vec v}\paren{\vec u} \parallel&= \frac{\vec u \cdot \vec v}{\parallel \vec v \parallel} \\
\op{proj}_{\vec v}\paren{\vec u} &= \frac{\vec u \cdot \vec v}{\parallel \vec v \parallel ^ {2}}\vec v
\end{align*}$
```functionplot
---
title:
xLabel:
yLabel:
bounds: [-10,10,-10,10]
disableZoom: false
grid: true
---
f(x)=acos(x)
```