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

• A. $PR+PS=2$
• B. $PR\times PS=\frac{2}{3}$
• C. $PR=PS$
• D. $PR+PS=RS$

Answer 1

Answer: (A), (B), (D)

$PR\times PS=\frac{2}{3}$
$PR+PS=RS$
$PR+PS=2$

Clearly the given line $x=y=z$ meets the plane where
$3x=1$ or $x=\frac{1}{3}=y=z,$ i.e. $P$ is $(\frac{1}{3},\frac{1}{3},\frac{1}{3}).$ It meets the
sphere $x^2+y^2+z^2=1$ where $3x^2=1=3y^2=3z^2$
$\therefore x=y=z=\frac{1}{\sqrt{3}}$ or $-\frac{1}{\sqrt{3}}$
$\therefore$ R is $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ and $S$ is $(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}).$
$\therefore (PR{)}^2={(\frac{1}{\sqrt{3}}-\frac{1}{3})}^2(1+1+1)$
$\therefore PR=\left(\frac{1}{\sqrt{3}-\frac{1}{3}}\right)\sqrt{3}=(1-\frac{1}{\sqrt{3}})$
Similarly, $PS=1+\frac{1}{\sqrt{3}}$ and $RS=\frac{2}{\sqrt{3}}.\sqrt{3}=2$
$\therefore PR+PS=2\Rightarrow \left(a\right)$
$\therefore PR.PS=1-\frac{1}{3}=\frac{2}{3}\Rightarrow \left(b\right)$
$\therefore PR+PS=RS\Rightarrow \left(d\right)$