For any two points M and N in the XY-plane, l | vedclass

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

Question

(A) Let the position vectors of points $P, Q, R, S, E,$ and $F$ be $\vec{p}, \vec{q}, \vec{r}, \vec{s}, \vec{e},$ and $\vec{f}$ respectively.Given the

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let the position vectors of points $P, Q, R, S, E,$ and $F$ be $\vec{p}, \vec{q}, \vec{r}, \vec{s}, \vec{e},$ and $\vec{f}$ respectively.Given the equation: $\overrightarrow{SP} + 5\overrightarrow{SQ} + 6\overrightarrow{SR} = \overrightarrow{0}$.Substituting the position vectors: $(\vec{p} - \vec{s}) + 5(\vec{q} - \vec{s}) + 6(\vec{r} - \vec{s}) = \overrightarrow{0}$.$\vec{p} + 5\vec{q} + 6\vec{r} - 12\vec{s} = \overrightarrow{0} \Rightarrow \vec{s} = \frac{\vec{p} + 5\vec{q} + 6\vec{r}}{12}$.Since $E$ is the mid-point of $PR$, $\vec{e} = \frac{\vec{p} + \vec{r}}{2}$.Since $F$ is the mid-point of $QR$, $\vec{f} = \frac{\vec{q} + \vec{r}}{2}$.Vector $\overrightarrow{EF} = \vec{f} - \vec{e} = \frac{\vec{q} + \vec{r}}{2} - \frac{\vec{p} + \vec{r}}{2} = \frac{\vec{q} - \vec{p}}{2}$.Vector $\overrightarrow{ES} = \vec{s} - \vec{e} = \frac{\vec{p} + 5\vec{q} + 6\vec{r}}{12} - \frac{\vec{p} + \vec{r}}{2} = \frac{\vec{p} + 5\vec{q} + 6\vec{r} - 6\vec{p} - 6\vec{r}}{12} = \frac{5\vec{q} - 5\vec{p}}{12} = \frac{5}{12}(\vec{q} - \vec{p})$.Therefore, the ratio of the lengths is $\frac{|\overrightarrow{EF}|}{|\overrightarrow{ES}|} = \frac{|\frac{1}{2}(\vec{q} - \vec{p})|}{|\frac{5}{12}(\vec{q} - \vec{p})|} = \frac{1/2}{5/12} = \frac{1}{2} 	imes \frac{12}{5} = \frac{6}{5} = 1.2$.

Similar Questions

If $|\vec{f}|=10, |\vec{g}|=14$ and $|\vec{f}-\vec{g}|=15$,then $|\vec{f}+\vec{g}|=$

If $P$ and $Q$ are two points on the curve $y = 2^{x+2}$ in the rectangular Cartesian coordinate system such that $\overline{OP} \cdot \hat{i} = -1$ and $\overline{OQ} \cdot \hat{i} = 2$,then $\overline{OQ} - 4\overline{OP} = $

The direction cosines of the vector $\vec{a} = -2 \hat{i} + \hat{j} - 5 \hat{k}$ are

In the figure,a vector $x$ satisfies the equation $x - w = v$. Then $x =$

In $\triangle OAC$,if $B$ is the mid-point of side $AC$ and $\vec{OA}=\vec{a}, \vec{OB}=\vec{b}$,then $\vec{OC}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module