Let ABCD be a parallelogram such that vec{AB} | vedclass

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(C) Let the origin be at $A$. Then $\vec{A} = \vec{0}$,$\vec{B} = \vec{q}$,and $\vec{D} = \vec{p}$.Let $M$ be the foot of the altitude from $B$ to the

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$\vec{r} = -\vec{q} + \frac{(\vec{p} \cdot \vec{q})}{\vec{p} \cdot \vec{p}} \vec{p}$

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(C) Let the origin be at $A$. Then $\vec{A} = \vec{0}$,$\vec{B} = \vec{q}$,and $\vec{D} = \vec{p}$.Let $M$ be the foot of the altitude from $B$ to the line $AD$. Since $M$ lies on the line $AD$,$\vec{AM} = k\vec{p}$ for some scalar $k$.The vector $\vec{BM}$ is perpendicular to $\vec{AD}$,so $\vec{BM} \cdot \vec{AD} = 0$.Since $\vec{BM} = \vec{AM} - \vec{AB} = k\vec{p} - \vec{q}$,we have $(k\vec{p} - \vec{q}) \cdot \vec{p} = 0$.$k(\vec{p} \cdot \vec{p}) - \vec{q} \cdot \vec{p} = 0$,which gives $k = \frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}}$.The vector $\vec{r}$ is the altitude vector $\vec{BM} = k\vec{p} - \vec{q} = \frac{(\vec{p} \cdot \vec{q})}{\vec{p} \cdot \vec{p}} \vec{p} - \vec{q}$.Thus,$\vec{r} = -\vec{q} + \frac{(\vec{p} \cdot \vec{q})}{\vec{p} \cdot \vec{p}} \vec{p}$.

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

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