Question

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

• A. $\overrightarrow{r}=3\sqrt{q}-\frac{3(\overrightarrow{p}\cdot \overrightarrow{q})}{(\overrightarrow{p}\cdot \overrightarrow{p})}\overrightarrow{p}$
• B. $\overrightarrow{r}=-\overrightarrow{q}+\frac{(\overrightarrow{p}\cdot \overrightarrow{q})}{(\overrightarrow{p}\cdot \overrightarrow{p})}\overrightarrow{p}$
• C. $\overrightarrow{r}=\overrightarrow{q}-\frac{(\overrightarrow{p}\cdot \overrightarrow{q})}{(\overrightarrow{p}\cdot \overrightarrow{p})}\overrightarrow{p}$
• D. $\overrightarrow{r}=-3\overrightarrow{q}-\frac{3(\overrightarrow{p}\cdot \overrightarrow{q})}{(\overrightarrow{p}\cdot \overrightarrow{p})}\overrightarrow{p}$

Answer 1

The correct option is B $\overrightarrow{r}=-\overrightarrow{q}+\frac{(\overrightarrow{p}\cdot \overrightarrow{q})}{(\overrightarrow{p}\cdot \overrightarrow{p})}\overrightarrow{p}$

ABCD be parallelogram

A the one gram

$$\begin{array}{c}\overrightarrow{AB}=\overrightarrow{q},\\ \overrightarrow{AD}=\overrightarrow{P},\\ \overrightarrow{BE}=\overline{r}\\ \overrightarrow{AE}=orthogonalprojectionof\overrightarrow{AB}=\overrightarrow{q},on\overrightarrow{AD}=\overrightarrow{P},\\ \frac{(\overline{q}.\overline{p})\overline{p}}{{\overline{p}}^2}\\ \overrightarrow{AB}+\overrightarrow{BE}=\overrightarrow{AC}\\ \overrightarrow{BE}=-\overrightarrow{AB}+\overrightarrow{AE}=-\overline{q}+\frac{(\overline{q}.\overline{p})\overline{p}}{\left|{\overline{p}}^2\right|}\end{array}$$

so, the correct option is B.