Projection is finding the length of a vectors "shadow" across another. The [[Orthogonal]] projection of $\vec u$ onto $\vec v$. $\huge\proj{\vec{v}}\pa{\vec u} = \vec v \pa{\frac{\vec u \cdot \vec v }{\norms{v}^{2}}}$ Note this works with abstract [[Vector|Vectors]] who live in an [[Inner Product Space]], $\huge \proj{\vec v}{\pa{\vec u}} = \vec v \frac{\braket{\vec v, \vec u}}{\braket{\vec v, \vec 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]] $\huge\begin{align*} \vec v \parallel \vec u \\ \op{proj}_{\vec v}\paren{\vec u} &= \vec u \\ \op{proj}_{\vec v\perp}\paren{\vec u} &= \vec 0 \end{align*}$ $\huge\begin{align*} ||\op{proj}_{\vec v}\paren{\vec u} || &= \frac{\vec u \cdot \vec v}{\parallel \vec v \parallel} \\ \op{proj}_{\vec v}\paren{\vec u} &= \frac{\vec u \cdot \vec v}{\parallel \vec v \parallel ^ {2}}\vec v \end{align*}$