The projection of \(\textbf{v}\) onto \(\textbf{u}\) represents the component of \(\textbf{v}\) in the direction of \(\textbf{u}\) (or the opposite direction of \(\textbf{u}\)):
Let the projection be \(\textbf{p} = \operatorname{proj}_{\textbf{u}}(\textbf{v})\). Let \(\theta\) be the angle between \(\textbf{v}\) and \(\textbf{u}\). The magnitude of this projection is:
$$ \Vert \operatorname{proj}_{\textbf{u}}(\textbf{v}) \Vert = \Vert \textbf{v} \Vert \ | \cos(\theta)|$$
From the definition of the dot product:
$$ \cos(\theta) = \frac{\textbf{v} \cdot \textbf{u}}{\Vert\textbf{v}\Vert \ \Vert\textbf{u}\Vert}$$
This means:
$$ \Vert \operatorname{proj}_{\textbf{u}}(\textbf{v}) \Vert = \Vert \textbf{v} \Vert \frac{| \textbf{v} \cdot \textbf{u} |}{\Vert\textbf{v}\Vert \ \Vert\textbf{u}\Vert} = \frac{| \textbf{v} \cdot \textbf{u} |}{ \Vert\textbf{u}\Vert}$$
If we want the vector \(\textbf{p}\), we can take the unit vector of \(\textbf{u}\) and scale (or shrink) it by the magnitude of \(\textbf{p}\). Whether \(\textbf{p}\) faces the same direction or the opposite direction of \(\textbf{u}\) depends on the sign of \(\cos(\theta)\):
$$ \operatorname{proj}_{\textbf{u}}(\textbf{v}) = \frac{\textbf{u}}{\Vert \textbf{u} \Vert} \Vert \operatorname{proj}_{\textbf{u}}(\textbf{v}) \Vert \ \operatorname{sign}(\cos(\theta))= \textbf{u} \ \frac{| \textbf{v} \cdot \textbf{u} |}{ \Vert\textbf{u}\Vert^2} \ \operatorname{sign}(\cos(\theta))$$
This means:
$$ \operatorname{proj}_{\textbf{u}}(\textbf{v}) = \textbf{u} \ \frac{| \textbf{v} \cdot \textbf{u} |}{ \Vert\textbf{u}\Vert^2} \frac{\textbf{v} \cdot \textbf{u}}{| \textbf{v} \cdot \textbf{u} |} = \textbf{u} \ \frac{ \textbf{v} \cdot \textbf{u}}{ \Vert\textbf{u}\Vert^2}$$