Question

The centre of sphere inscribed in a tetrahedron bounded by the planes $\vec{r}\cdot \hat{i}=0=\vec{r}\cdot \hat{j}=\vec{r}\cdot \hat{k}$ and $\vec{r}\cdot \left(\hat{i}+\hat{j}+\hat{k}\right)=a$ is

• A. $\left(\frac{a}{\sqrt{3}},\frac{a}{\sqrt{3}},\frac{a}{\sqrt{3}}\right)$
• B. $\left(\frac{a}{3\sqrt{3}},\frac{a}{3\sqrt{3}},\frac{a}{3\sqrt{3}}\right)$
• C. $\left(\frac{a}{3-\sqrt{3}},\frac{a}{3-\sqrt{3}},\frac{a}{3-\sqrt{3}}\right)$
• D. $\left(\frac{2a}{3-\sqrt{3}},\frac{2a}{3-\sqrt{3}},\frac{2a}{3-\sqrt{3}}\right)$

Answer 1

Answer: (C)

Writing the given equation of the planes in cartesian form , we get $x=0=y=z$ and $x+y+z=a$, which means sphere touches the coordinate axes. Let the centre of sphere be $(\alpha ,\alpha ,\alpha )$ and the $\perp$ distance from centre to the plane $x+y+z=a$ is $\alpha$
$\therefore \frac{\alpha +\alpha +\alpha -\alpha }{\sqrt{3}}=\alpha$
$\Rightarrow \alpha =\frac{a}{3-\sqrt{3}}$
Required centre is
$\left(\alpha ,\alpha ,\alpha \right)=\left(\frac{a}{3-\sqrt{3}},\frac{a}{3-\sqrt{3}},\frac{a}{3-\sqrt{3}}\right)$