I am currently studying this answer on physics.stackexchange to a question I asked on the law of refraction in vector form. The answer (by the user JEB) proceeds as follows:
Given two vectors $\vec a$ and $\vec b$, the vector projection of $\vec a$ onto $\vec b$ is:
$$\vec a_{\parallel \vec b}= \frac{\vec a \cdot \vec b}{\vec b \cdot \vec b} \vec b$$
The component of $\vec a$ that is perpendicular to this is called the vector rejection on $\hat b$:
$$ \vec a_{\perp \vec b}=\vec a - \vec a_{\parallel \vec b}=\vec a - \frac{\vec a \cdot \vec b}{\vec b \cdot \vec b} $$
If $\vec b=\hat b$ is a unit vector:
$$ \vec a_{\perp \vec b}=\vec a -(\vec a\cdot \hat b)\hat b$$
How is it that $\vec a -(\vec a\cdot \hat b)\hat b = \vec a - \frac{\vec a \cdot \vec b}{\vec b \cdot \vec b}$? If we say that $\vec b = \hat b$, then we have $\vec a - \frac{\vec a \cdot \hat b}{\hat b \cdot \hat b}$, but I don't see how that is $\vec a -(\vec a\cdot \hat b)\hat b$.
There is an obvious typographical error. Note that $\frac{\vec a\cdot \vec b}{\vec b\cdot \vec b}$ is a scalar, not a vector, while $(\vec a\cdot \hat b)\hat b$ is a vector that is parallel to $\vec b$.
Instead, using $\hat b=\frac{\vec b}{|\vec b|}$, we have
$$\begin{align} (\vec a\cdot \hat b)\hat b&=\left(\vec a\cdot \frac{\vec b}{|\vec b|}\right)\frac{\vec b}{|\vec b|}\\\\ &=\left(\frac{\vec a\cdot \vec b}{|\vec b|^2}\right) \vec b\\\\ &=\left(\frac{\vec a\cdot \vec b}{\vec b\cdot \vec b}\right) \vec b\\\\ \end{align}$$