Question

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

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

Answer 1

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

Let $P$ be the point $(x_1,y_1,z_1)$ on the given plane, then

$l{x}_1+m{y}_1+n{z}_1=p$ ....(1)

Let $Q$ be $(\alpha ,\beta ,\gamma )$. Since $O,P,Q$ are collinear, therefore

$\frac{x_1}{\alpha }=\frac{y_1}{\beta }=\frac{z_1}{\gamma }=k$(say) ....(2)

Now, $OP.OQ=p^2\Rightarrow \sqrt{{x_1}^2+{y_1}^2+{z_1}^2}\sqrt{{\alpha }^2+{\beta }^2+{\gamma }^2}=p^2$

$\Rightarrow k({\alpha }^2+{\beta }^2+{\gamma }^2)=p^2$ ...(3)

Also from (1) and (2), we get

$k(l\alpha +m\beta +n\gamma )=p$

From (3) and (4), we have

$p(l\alpha +m\beta +n\gamma )=({\alpha }^2+{\beta }^2+{\gamma }^2)$

$\therefore$ locus of $Q\equiv (\alpha ,\beta ,\gamma )$ is

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