In a quadrilateral PQRS, A divides SR in the | vedclass

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(D) Let the position vectors of $P, Q, R, S, A, B$ be $\vec{p}, \vec{q}, \vec{r}, \vec{s}, \vec{a}, \vec{b}$ respectively. Given that $A$ divides $SR$

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let the position vectors of $P, Q, R, S, A, B$ be $\vec{p}, \vec{q}, \vec{r}, \vec{s}, \vec{a}, \vec{b}$ respectively. Given that $A$ divides $SR$ in the ratio $1:3$, we have $\vec{a} = \frac{3\vec{s} + 1\vec{r}}{1+3} = \frac{3\vec{s} + \vec{r}}{4}$. Since $B$ is the mid-point of $PR$, we have $\vec{b} = \frac{\vec{p} + \vec{r}}{2}$. The given expression is $3\vec{SR} - \vec{QR} - 3\vec{PS} - \vec{PQ} = k\vec{AB}$. Substituting the vectors: $3(\vec{r} - \vec{s}) - (\vec{r} - \vec{q}) - 3(\vec{s} - \vec{p}) - (\vec{q} - \vec{p}) = k(\vec{b} - \vec{a})$ $3\vec{r} - 3\vec{s} - \vec{r} + \vec{q} - 3\vec{s} + 3\vec{p} - \vec{q} + \vec{p} = k\left(\frac{\vec{p} + \vec{r}}{2} - \frac{3\vec{s} + \vec{r}}{4}\right)$ Combining like terms on the left side: $(3\vec{r} - \vec{r}) + (-3\vec{s} - 3\vec{s}) + (3\vec{p} + \vec{p}) + (\vec{q} - \vec{q}) = k\left(\frac{2\vec{p} + 2\vec{r} - 3\vec{s} - \vec{r}}{4}\right)$ $2\vec{r} - 6\vec{s} + 4\vec{p} = k\left(\frac{2\vec{p} + \vec{r} - 3\vec{s}}{4}\right)$ Multiply both sides by $4$: $8\vec{r} - 24\vec{s} + 16\vec{p} = k(2\vec{p} + \vec{r} - 3\vec{s})$ $8\vec{r} - 24\vec{s} + 16\vec{p} = k\vec{r} - 3k\vec{s} + 2k\vec{p}$ Comparing the coefficients of $\vec{p}, \vec{q}, \vec{r}, \vec{s}$ on both sides, we get $k = 8$.

In a quadrilateral $PQRS$, $A$ divides $SR$ in the ratio $1:3$ and $B$ is the mid-point of $PR$. If $3SR - QR - 3PS - PQ = kAB$, then $k=$

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