MHT CET · Maths · Vector Algebra
Let p and q be the position vectors of P and Q respectively, with respect to O and \(|\overline{\mathrm{p}}|=\mathrm{p},|\overline{\mathrm{q}}|=\mathrm{q}\). The points R and S divide PQ internally and externally in the ratio \(2: 3\) respectively. If \(O R\) and OS are perpendiculars, then
- A \(9 \mathrm{p}^2=4 \mathrm{q}^2\)
- B \(4 \mathrm{p}^2=9 \mathrm{q}^2\)
- C \(9 \mathrm{p}=4 \mathrm{q}\)
- D \(4 \mathrm{p}=9 \mathrm{q}\)
Answer & Solution
Correct Answer
(A) \(9 \mathrm{p}^2=4 \mathrm{q}^2\)
Step-by-step Solution
Detailed explanation
Let \(\overline{\mathrm{r}}\) and \(\overline{\mathrm{s}}\) be the position vectors of points R and \(S\) respectively.
\(\therefore \quad \overline{\mathrm{r}}=\frac{3 \mathrm{p}+2 \mathrm{q}}{3+2}\) and \(\overline{\mathrm{s}}=\frac{3 \mathrm{p}-2 \mathrm{q}}{3-2}\)
As OS and OR are perpendicular, we get \(\overline{\mathrm{r}} \cdot \overline{\mathrm{s}}=0\)
\(\begin{aligned}
& \therefore \quad\left(\frac{3 p+2 q}{5}\right)\left(\frac{3 p-2 q}{1}\right)=0 \\
& \therefore \quad 9 p^2-6 p q+6 p q-4 q^2=0 \\
& \therefore \quad 9 p^2=4 q^2
\end{aligned}\)
\(\therefore \quad \overline{\mathrm{r}}=\frac{3 \mathrm{p}+2 \mathrm{q}}{3+2}\) and \(\overline{\mathrm{s}}=\frac{3 \mathrm{p}-2 \mathrm{q}}{3-2}\)
As OS and OR are perpendicular, we get \(\overline{\mathrm{r}} \cdot \overline{\mathrm{s}}=0\)
\(\begin{aligned}
& \therefore \quad\left(\frac{3 p+2 q}{5}\right)\left(\frac{3 p-2 q}{1}\right)=0 \\
& \therefore \quad 9 p^2-6 p q+6 p q-4 q^2=0 \\
& \therefore \quad 9 p^2=4 q^2
\end{aligned}\)
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