Answer to Question #93617 in Analytic Geometry for Kanha Sharma

Question

P is a point on the plane lx + my + nz = p and a point Q is taken on the line such that OP. OQ= p^2 ; show that the locus of point Q is p(lx + my + nz)= x^2 + y^2 + z^2.

Answer 1

Let P point be $(\alpha,\beta,\gamma)$ and Q point be$(x_1,y_1,z_1)$

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

Since O,Q,P are collinear, we have

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

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

So,

$l\alpha+m\beta+n\gamma=p$

or $k(x_1+my_1+nz_1)=p$ $........(2)$

Given, OP$\times$ OQ $=p^2$

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

or, $\sqrt{k^2(x_1^2+y_1^2+z_1^2)}\sqrt{(x_1)^2+(y_1)^2+(z_1^2)}=p^2$

or, $k(x_1^2+y_1^2+z_1^2)=p^2$ $......(3)$

On dividing $(2)$ by$(3)$ ,we get

$\frac{lx_1+my_1+nz_1}{x_1^2+y_1^2+z_1^2}=\frac{1}{p}$

or, $p(lx_1+my_1+nz_1)=x_1^2+y_1^2+z_1^2$

Hence the locus of point Q is $p(lx+my+nz)=x^2+y^2+z^2.$