Quick questions on Motion as a vector function: position, velocity and acceleration vectors, speed, and dot-product analysis of the motion

The path is the curve in the plane that the particle travels along, with time stripped away. To find its Cartesian equation, eliminate $t$ between $x(t)$ and $y(t)$, exactly as for any parametric curve. For instance, if $x = 2t$ and $y = t^2$, then $t = \tfrac{x}{2}$ and $y = \tfrac{x^2}{4}$, a parabola. The path tells you the shape of the journey; the velocity and acceleration tell you the timing and dynamics along it.

The dot product turns geometric questions about these vectors into arithmetic. Recall the two formulas for vectors $\mathbf{p} = (p_1, p_2)$ and $\mathbf{q} = (q_1, q_2)$:

It is worth being deliberate about how this differs from the parametric-line work. If $\mathbf{r}(t) = \mathbf{a} + t\mathbf{d}$, then $\mathbf{v}(t) = \mathbf{d}$ is constant and $\mathbf{a}(t) = \mathbf{0}$: the particle moves in a straight line at constant speed $|\mathbf{d}|$, and "speeding up or slowing down" never arises. The moment a component is quadratic or higher (or trigonometric, or exponential) in $t$, the velocity changes, the acceleration is non-zero, the path curves, and the dot-product analysis above comes into play. A quick diagnostic: differentiate; if $\mathbf{a} = \mathbf{0}$ you are back in the constant-velocity world, and if not you are in genuine vector-calculus motion.

Rearranging the geometric formula,

Two non-zero vectors are perpendicular exactly when $\mathbf{p} \cdot \mathbf{q} = 0$ (because $\cos 90^\circ = 0$). This single fact answers "when are the velocities perpendicular?" and "when is $\angle AOB$ a right angle?", and it underlies the closest-approach result below.