Vector Dot Product & Geometric Projections

$$\mathbf{u}\cdot \mathbf{v}=u_x{v}_x+u_y{v}_y$$
The sum of the products of corresponding coordinate components. The dot product yields a scalar, not a vector.

$$\mathbf{u}\cdot \mathbf{v}=\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \cos \theta$$
Where $\theta$ is the angle between the two vectors ($0\le \theta \le \pi$).

The vectors are orthogonal (perpendicular). Since $\cos (90^{\circ })=0$, $\mathbf{u}\cdot \mathbf{v}=0⟺\theta =90^{\circ }$.

$$\mathbf{v}\cdot \mathbf{v}=\Vert \mathbf{v}{\Vert }^2$$
The dot product of a vector with itself equals the square of its Euclidean length (magnitude).

$${\text{scal}}_{\mathbf{u}}(\mathbf{v})=\frac{\mathbf{u}\cdot \mathbf{v}}{\Vert \mathbf{u}\Vert }=\Vert \mathbf{v}\Vert \cos \theta$$
The signed length of the orthogonal shadow cast by $\mathbf{v}$ along the line of $\mathbf{u}$.

$${\text{proj}}_{\mathbf{u}}(\mathbf{v})=\left(\frac{\mathbf{u}\cdot \mathbf{v}}{\Vert \mathbf{u}{\Vert }^2}\right)\mathbf{u}$$
A vector oriented along $\mathbf{u}$ with length equal to the scalar projection.

$$|\mathbf{u}\cdot \mathbf{v}|\le \Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert$$
Equality holds if and only if $\mathbf{u}$ and $\mathbf{v}$ are linearly dependent (collinear).

Mechanical Work ($W$):
$$W=\mathbf{F}\cdot \mathbf{d}=\Vert \mathbf{F}\Vert \Vert \mathbf{d}\Vert \cos \theta$$
Only the component of force parallel to the displacement does work.