I am reading some material on the Householder reflection and it describes the projection of a vector $\vec{x}$, onto $\vec{v}$ in vector notation to be:
$\frac{\vec{x}\cdot \vec{v}}{||\vec{v}||^2}\vec{v}$
Where $a\cdot b$ is the dot product of vectors a and b.
So how is that?
I know if θ is the angle between $\vec{x}$ and $\vec{v}$, then in trigonometric terms, the projection of $\vec{x}$ on $\vec{v}$ would be $x\cos(θ)$. How are these two equivalent? Thanks.
Firstly your equation is slightly off. I'll derive the correct equation directly to show you why.
Recall that $x\cdot v = |x||v|\cos(\theta)$
$$\frac{x\cdot v}{v\cdot v}v=\frac{|x||v|\cos(\theta)}{|v|^2}v=|x|\cos(\theta)\frac{v}{|v|}$$
So this vector has length $|x|\cos(\theta)$ and it points in the direction of $v$, which is precisely what the trig should tell you.
First we should understand where the following projection formula comes from:
$$proj_{\vec v}(\vec x)=\left( \frac{\vec{x} \cdot \vec{v}}{||\vec{v}||^2} \right) \vec{v}=\left( \frac{\vec{x} \cdot \vec{v}}{\vec v \cdot \vec v} \right)\vec{v}$$
$proj_{\vec v}(\vec x)$ will be in the same direction with $\vec v$. Therefore, we can write $proj_{\vec v}(\vec x)=k \vec v$, for some scalar $k$. Additionally, if you draw these vectors, you see that the vector $(\vec x - proj_{\vec v}(\vec x))$ is orthogonal to $\vec v$. Combining these results, we find that
$$(\vec x-k\vec v) \cdot \vec v=\vec x \cdot \vec v-k(\vec v \cdot \vec v)=0$$
If we take $k$ out of this equation, we get
$$k=\frac{\vec{x} \cdot \vec{v}}{\vec v \cdot \vec v}$$
Replacing $k$ in the $proj_{\vec v}(\vec x)=k \vec v$ gives us the above formula.
The relationship between $\cos(\theta)$ and the projection is given here.
$(x.v/v.v)v$ is the projection vector of $x$ onto $v$, while, $|x|\cos\theta=|x|(x.v/|v||x|)=(x.v/|v|)$ is just the length(magnitude) of the projection vector.
projection vector($x$ on $v$) $=(x.v/v)*$ unitvector in direction of $v$ = $(x.v/|v|)(v/|v|)=(x.v/|v|^2)v=(x.v/v.v)v$
