(A) The position vectors of $R$ and $S$ are $\frac{3p + 2q}{5}$ and $3p - 2q$ respectively.Therefore,$\overrightarrow{OR} = \frac{3p + 2q}{5}$ and $\o

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(A) The position vectors of $R$ and $S$ are $\frac{3p + 2q}{5}$ and $3p - 2q$ respectively.Therefore,$\overrightarrow{OR} = \frac{3p + 2q}{5}$ and $\overrightarrow{OS} = 3p - 2q.$Since $\overrightarrow{OR} \perp \overrightarrow{OS},$ their dot product is zero,so $\overrightarrow{OR} \cdot \overrightarrow{OS} = 0.$$\left( \frac{3p + 2q}{5} \right) \cdot (3p - 2q) = 0$$9|p|^2 - 6(p \cdot q) + 6(q \cdot p) - 4|q|^2 = 0$$9|p|^2 - 4|q|^2 = 0$$9|p|^2 = 4|q|^2$Given $|p| = p$ and $|q| = q,$ we get $9p^2 = 4q^2.$

Class 12 Mathematics Vector Algebra Scalar or Dot product of two vectors and its applications

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Let $p$ and $q$ be the position vectors of $P$ and $Q$ respectively with respect to $O$ and $|p| = p, |q| = q.$ The points $R$ and $S$ divide $PQ$ internally and externally in the ratio $2 : 3$ respectively. If $\overrightarrow{OR}$ and $\overrightarrow{OS}$ are perpendicular,then:

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A
$9p^2 = 4q^2$
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$4p^2 = 9q^2$
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$9p = 4q$
D
$4p = 9q$

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Let $p$ and $q$ be the position vectors of $P$ and $Q$ respectively with respect to $O$ and $|p| = p, |q| = q.$ The points $R$ and $S$ divide $PQ$ internally and externally in the ratio $2 : 3$ respectively. If $\overrightarrow{OR}$ and $\overrightarrow{OS}$ are perpendicular,then:

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