1.Let ABCD be a parallelogram such that AB→=p⃗\overrightarrow{AB} = \vec{p}AB=p, AD→=q⃗\overrightarrow{AD} = \vec{q}AD=q and ∠BAD\angle BAD∠BAD be an acute angle. If r⃗\vec{r}r is the vector that coincides with the altitude directed from the vertex B to the side AD, then r⃗\vec{r}r is given bya.r⃗=3q⃗−3(p⃗⋅q⃗)∣p⃗∣2p⃗\vec{r} = 3\vec{q} - \frac{3(\vec{p}\cdot\vec{q})}{|\vec{p}|^2}\vec{p}r=3q−∣p∣23(p⋅q)pb.r⃗=−q⃗+p⃗⋅q⃗∣p⃗∣2p⃗\vec{r} = -\vec{q} + \frac{\vec{p}\cdot\vec{q}}{|\vec{p}|^2}\vec{p}r=−q+∣p∣2p⋅qpc.r⃗=q⃗−p⃗⋅q⃗∣p⃗∣2p⃗\vec{r} = \vec{q} - \frac{\vec{p}\cdot\vec{q}}{|\vec{p}|^2}\vec{p}r=q−∣p∣2p⋅qpd.r⃗=−3q⃗+3(p⃗⋅q⃗)∣p⃗∣2p⃗\vec{r} = -3\vec{q} + \frac{3(\vec{p}\cdot\vec{q})}{|\vec{p}|^2}\vec{p}r=−3q+∣p∣23(p⋅q)pLogin to continueOnly logged in users canattempt or see the solution.