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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $\vec{r}$.

Similar Questions

Find the projection of the vector $\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.

If $\vec{a}$ and $\vec{b}$ are unit vectors such that the vector $\vec{a} + 3\vec{b}$ is perpendicular to $7\vec{a} - 5\vec{b}$,then find the angle between $\vec{a}$ and $\vec{b}$.

If the vectors $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$,$\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}$,and $\vec{c} = \alpha\hat{i} + \hat{j} + \beta\hat{k}$ are mutually orthogonal,then $(\alpha, \beta) = $

If $\bar{a}$ and $\bar{b}$ are unit vectors and $\theta$ is the angle between them,then $\bar{a}+\bar{b}$ is a unit vector when $\theta$ is

The scalars $l$ and $m$ such that $la + mb = c,$ where $a, b$ and $c$ are given vectors,are equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module