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

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(B) Let $E$ be the foot of the perpendicular from vertex $B$ to the side $AD$. The vector $\vec{AE}$ is the projection of $\vec{AB}$ onto $\vec{AD}$.T

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $E$ be the foot of the perpendicular from vertex $B$ to the side $AD$. The vector $\vec{AE}$ is the projection of $\vec{AB}$ onto $\vec{AD}$.Thus,$\vec{AE} = 	ext{proj}_{\vec{p}} \vec{q} = \left( \frac{\vec{q} \cdot \vec{p}}{\vec{p} \cdot \vec{p}} ight) \vec{p}$.In $	riangle ABE$,by the triangle law of vector addition,we have $\vec{AB} + \vec{BE} = \vec{AE}$.Given $\vec{AB} = \vec{q}$ and $\vec{BE} = \vec{r}$,we have $\vec{q} + \vec{r} = \vec{AE}$.Therefore,$\vec{r} = \vec{AE} - \vec{q}$.Substituting the expression for $\vec{AE}$,we get $\vec{r} = \left( \frac{\vec{q} \cdot \vec{p}}{\vec{p} \cdot \vec{p}} ight) \vec{p} - \vec{q}$.

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 the vector that coincides with the altitude directed from the vertex $B$ to the side $AD$,then $\vec{r}$ is given by:

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