Question

The line $x=y=z$ meets the plane $x+y+z=1$ at the point $P$ and the sphere $x^2+y^2+z^2=1$ at the points $R$ and $S$, then the value of $PS.PR$ is equal to:

• A. $\frac{1}{3}$
• B. $\frac{1}{9}$
• C. $\frac{2}{3}$
• D. $\sqrt{3}$

Answer 1

The correct option is C $\frac{2}{3}$

The coordinate of $P$ where the line $x=y=z$ meets the plane $x+y+z=1$ are $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ and the coordinates of $R$ and $S$ where the line meets the sphere $x^2+y^2+z^2=1$ are $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ and $(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})$

so that $PR=\sqrt{3}|\frac{1}{3}-\frac{1}{\sqrt{3}}|=\sqrt{3}(\frac{1}{\sqrt{3}}-\frac{1}{3})$

and $PS=\sqrt{3}(\frac{1}{3}+\frac{1}{\sqrt{3}})$

$\Rightarrow PS.PR=3(\frac{1}{3}-\frac{1}{9})=\frac{2}{3}$