Question

Let P be any point on the plane lx + my + nz = p and Q be a point on the line OP such that $OP\cdot OQ=p^2$. 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

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 line
$l{x}_1+m{y}_1+n{z}_1=p.....\left(i\right)$
Let Q be $(\alpha ,\beta ,\gamma )$ and O, P, Q are collinear, so
$\frac{x_1}{\alpha }=\frac{y_1}{\beta }=\frac{z_1}{\gamma }=k\left(say\right)$ ...(ii)
Now, $OP\cdot OQ=p^2$
$\Rightarrow \sqrt{(x_1^2+y_1^2+z_1^2)}\sqrt{{\alpha }^2+{\beta }^2+{\gamma }^2}=p^2$
$\Rightarrow k({\alpha }^2+{\beta }^2+{\gamma }^2)=p^2$ ...(iv)
From Eqs. (iii) and (iv).
$p(l\alpha +m\beta +n\gamma )={\alpha }^2+{\beta }^2+{\gamma }^2$
$\therefore$ Locus of Q $=(\alpha ,\beta ,\gamma )$ is $p(lx+my+nz)=x^2+y^2+z^2$