Question

Let $P$ 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. $lx+my+nz-p=x^2+y^2+z^2$
• B. $lx+my+nz=p(x^2+y^2+z^2)$
• C. $p(lx+my+nz)=x^2+y^2+z^2$
• D. $x^2+y^2+z^2=p^2$

Answer 1

Answer: (C)

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

The given equation of the plane is $lx+my+nz=p$
Let $P(\alpha ,\beta ,\gamma )$ be a point on the plane and $Q(x,y,z)$ be a point on $OP$ such that $OP\cdot OQ=p^2$
$\Rightarrow l\alpha +m\beta +n\gamma =p$
The direction ratios of $OP$ are $\alpha ,\beta ,\gamma$ and $OQ$ are $x,y,z$.
Since $O,P,Q$ are collinear $\frac{\alpha }{x}=\frac{\beta }{y}=\frac{\gamma }{z}=k$ .....$\left(1\right)$
$\Rightarrow OP\cdot OQ=\sqrt{{\alpha }^2+{\beta }^2+{\gamma }^2}\cdot \sqrt{x^2+y^2+z^2}=p^2$
$\Rightarrow k(x^2+y^2+z^2)=p^2$
$\therefore l\alpha +m\beta +n\gamma =p\Rightarrow k(lx+my+nz)=p$
Hence, the locus of $Q$ is $p(lx+my+nz)=x^2+y^2+z^2$.