Question

If $p$ be any point on the plane $lx+my+nz=p$ and $Q$ be a point on the line $OP$ such that $OP\cdot OQ=p^2$, the locus of the point Q is

• A. $lx+my+nz=p^2$
• B. $p(lx+my+nz)=x^2+y^2+z^2$
• C. $plmn=xyz$
• D. None of these

Answer 1

The correct option is B $p(lx+my+nz)=x^2+y^2+z^2$

Let $P\equiv (\alpha ,\beta ,\gamma ),Q\equiv (x_1,y_1,z_1)$

Direction ratios of $OP$ are $\alpha ,\beta ,\gamma$ and $OQ$ are $x_1,y_1,z_1$

Since $O,Q,P$ are collinear, we have

$\frac{\alpha }{x_1}=\frac{\beta }{y_1}=\frac{\gamma }{z_1}=k$(say) ...$\left(1\right)$

As $(\alpha ,\beta ,\gamma )$ lies on the plane $lx+my+nz=p$

$\Rightarrow l\alpha +m\beta +n\gamma =p\Rightarrow k(l{x}_1+m{y}_1+n{z}_1)=p$ ...$\left(2\right)$

Given $OP.OQ=p^2$

$\therefore \sqrt{{\alpha }^2+{\beta }^2+{\gamma }^2}.\sqrt{x_1^2+y_1^2+z_1^2}=p^2\Rightarrow \sqrt{k^2(x_1^2+y_1^2+z_1^2)}.\sqrt{x_1^2+y_1^2+z_1^2}=p^2$

$\Rightarrow k(x_1^2+y_1^2+z_1^2)=p^2$ ...$\left(3\right)$

On dividing $\left(2\right)$ by $\left(3\right)$, we get

$\frac{l{x}_1+m{y}_1+n{z}_1}{x_1^2+y_1^2+z_1^2}=\frac{1}{p}\Rightarrow p(l{x}_1+m{y}_1+n{z}_1)=x_1^2+y_1^2+z_1^2$

Hence, the locus of point Q is

$p(lx+my+nz)=x^2+y^2+z^2$