Let $$ABCD$$ be a parallelogram such that $$\vec{AB} = \vec{q}$$, $$\vec{AD} = \vec{p}$$ and $$\angle BAD$$ be an acute angle. If $$\vec{r}$$ is the vector that coincides with the altitude directed from the vertex $$B$$ to the side $$AD$$, then $$\vec{r}$$ is given by

Place the origin at vertex $$A$$. Then the position vectors of the remaining vertices of the parallelogram are

$$\vec{B}= \vec{q}, \qquad \vec{D}= \vec{p}, \qquad \vec{C}= \vec{p}+\vec{q}.$$

The side $$AD$$ lies on the line through the origin in the direction of $$\vec{p}$$. Let $$H$$ be the foot of the perpendicular drawn from $$B$$ to $$AD$$. Because $$H$$ is on $$AD$$, its position vector must be a scalar multiple of $$\vec{p}$$:

$$\vec{H}=t\,\vec{p}\qquad\text{for some real }t.$$

The altitude $$BH$$ is perpendicular to $$AD$$, so

$$\vec{BH}\cdot\vec{AD}=0 \;\Longrightarrow\; (\vec{H}-\vec{B})\cdot\vec{p}=0.$$

Substituting $$\vec{H}=t\vec{p}$$ and $$\vec{B}=\vec{q}$$ gives

$$\bigl(t\vec{p}-\vec{q}\bigr)\cdot\vec{p}=0 \;\;\Longrightarrow\;\; t\,(\vec{p}\cdot\vec{p})-(\vec{q}\cdot\vec{p})=0.$$ Hence

$$t=\frac{\vec{p}\cdot\vec{q}}{\vec{p}\cdot\vec{p}}.$$

Therefore the foot of the perpendicular is

$$\vec{H}=\frac{\vec{p}\cdot\vec{q}}{\vec{p}\cdot\vec{p}}\;\vec{p}.$$

The required vector coinciding with the altitude, directed from $$B$$ to $$AD$$, is

$$\vec{r}=\vec{BH}= \vec{H}-\vec{B} =\frac{\vec{p}\cdot\vec{q}}{\vec{p}\cdot\vec{p}}\;\vec{p}-\vec{q} = -\vec{q}+\left(\frac{\vec{p}\cdot\vec{q}}{\vec{p}\cdot\vec{p}}\right)\vec{p}.$$

Thus

Option B which is: $$\vec{r}= -\vec{q} + \left(\dfrac{\vec{p}\cdot\vec{q}}{\vec{p}\cdot\vec{p}}\right)\vec{p}.$$