Orthogonal Projections

# Orthogonal Projections

Consider a vector $\vec{u}$. This vector can be written as a sum of two vectors that are respectively perpendicular to one another, that is $\vec{u} = \vec{w_1} + \vec{w_2}$ where $\vec{w_1} \perp \vec{w_2}$.

First construct a vector $\vec{b}$ that has its initial point coincide with $\vec{u}$:

We will now construct a $\vec{w_1}$ that also has its initial point coinciding with $\vec{v}$ and $\vec{u}$. This vector will run along $\vec{b}$.

We will now drop a perpendicular vector $\vec{w_2}$ that has its initial point at the terminal point of $\vec{w_1}$, and whose terminal point is at the terminal point of $\vec{u}$. Thus we get that $\vec{u} = \vec{w_1} + \vec{w_2}$, and $\vec{w_1} \perp \vec{w_2}$ like we wanted.

The vector $\vec{w_1}$ has a special name, which we will formally define as follows.

 Definition: If $\vec{u} = \vec{w_1} + \vec{w_2}$, $\vec{w_1} \| \vec{b}$ and $\vec{w_1} \perp \vec{w_2}$, then $\vec{w_1}$ the Orthogonal Projection of $\vec{u}$ Along $\vec{b}$ denoted $w_1 = \mathrm{proj}_{\vec{b}} \vec{u} =\frac{(\vec{u} \cdot \vec{b})}{\| \vec{b} \|^2} \vec{b}$, and $\vec{w_2}$ is the Vector Component of $\vec{u}$ Orthogonal to $\vec{b}$ and $\vec{w_2} = \vec{u} - \mathrm{proj}_{\vec{b}} \vec{u}$.

In the definition above, we formally defined $\mathrm{proj}_{\vec{b}} \vec{u} = \frac{(\vec{u} \cdot \vec{b})}{\| \vec{b} \|^2} \vec{b}$. We will now prove this with the following theorem.

 Theorem 1: If $\vec{u} = \vec{w_1} + \vec{w_2}$, $\vec{w_1} \| \vec{b}$ and $\vec{w_1} \perp \vec{w_2}$ then $\vec{w_1} = \mathrm{proj}_{\vec{b}} \vec{u} = \frac{(\vec{u} \cdot \vec{b})}{\| \vec{b} \|^2} \vec{b}$.
• Proof: We note that $\vec{b}$ and $\vec{w_1}$ are parallel to each other and thus for some scalar $k$, $\vec{w_1} = k\vec{b}$. Furthermore, we know that $\vec{u} = \vec{w_1} + \vec{w_2}$, and therefore $\vec{u} = k\vec{b} + \vec{w_2}$. Taking the dot product of both sides and noting that $\vec{w_2} \cdot \vec{b} = 0$ since $\vec{b} \perp \vec{w_2}$ we get that:
(1)
\begin{align} \vec{u} \cdot \vec{b} = (k\vec{b} + \vec{w_2}) \cdot \vec{b} \\ \vec{u} \cdot \vec{b} = k(\vec{b} \cdot \vec{b}) + \vec{w_2} \cdot \vec{b} \\ \vec{u} \cdot \vec{b} = k \| \vec{b} \|^2 \\ k = \frac{\vec{u} \cdot \vec{b}}{\| \vec{b} \|^2} \end{align}
• Once again we know that $\vec{w_1} = k \vec{b}$, so substituting our value for $k$ get get that:
(2)
\begin{align} \vec{w_1} =\mathrm{proj}_{\vec{b}} \vec{u} = \frac{(\vec{u} \cdot \vec{b})}{\| \vec{b} \|^2} \vec{b} \\ \blacksquare \end{align}

Of course, we also need a formula to compute the norm of $\mathrm{proj}_{\vec{b}} \vec{u}$. The following theorem gives us a relatively nice formula to use.

 Theorem 2: If $\vec{u} = \vec{w_1} + \vec{w_2}$, $\vec{w_1} \| \vec{b}$ and $\vec{w_1} \perp \vec{w_2}$ then $\| \mathrm{proj}_{\vec{b}} \vec{u} \| = \| \vec{u} \| \| \vec{b} \| \cos \theta$.
• Proof: The proof of the formula given in theorem 2 is rather straightforward. Noting that $\vec{u} \cdot \vec{b} = \| \vec{u} \| \| \vec{b} \| \cos \theta$ we have that:
(3)
\begin{align} \| \mathrm{proj}_{\vec{b}} \vec{u} \| = \biggr \| \frac{(\vec{u} \cdot \vec{b})}{\| \vec{b} \|^2} \vec{b} \biggr \| \\ \| \mathrm{proj}_{\vec{b}} \vec{u} \| = \mathrm{abs}\left ( \frac{(\vec{u} \cdot \vec{b})}{\| \vec{b} \|^2} \right ) \| \vec{b} \| \\ \| \mathrm{proj}_{\vec{b}} \vec{u} \| = \frac{\mid \vec{u} \cdot \vec{b}\mid}{\| \vec{b} \|^2} \| \vec{b} \| \\ \| \mathrm{proj}_{\vec{b}} \vec{u} \| = \frac{\mid \vec{u} \cdot \vec{b}\mid}{\| \vec{b} \|} \\ \| \mathrm{proj}_{\vec{b}} \vec{u} \| = \frac{\mid \| \vec{u} \| \| \vec{b} \| \cos \theta \mid}{\| \vec{b} \|} \\ \| \mathrm{proj}_{\vec{b}} \vec{u} \| = \frac{\| \vec{u} \| \| \vec{b} \| \mid \cos \theta \mid}{\| \vec{b} \|} \\ \| \mathrm{proj}_{\vec{b}} \vec{u} \| = \mid \cos \theta \mid \| \vec{u} \| \quad \blacksquare \end{align}