Question

The centre of sphere insrcibed in a tetrahedron bounded by the planes $\overline{r}.\hat{i}=0=\overline{r}.\hat{j},\overline{r}.\hat{k}$ and $\overline{r}.(\hat{i}.\hat{j}.\hat{k})=a$ is

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

Answer 1

The correct option is C $(\frac{a}{3-\sqrt{3}},\frac{a}{3-\sqrt{3}},\frac{a}{3-\sqrt{3}})$

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 coordinates axes.

Let the centre of sphere be $(\alpha ,\alpha ,\alpha )$ and the $\perp$ distance from centre to the plane $x+y+z=a$ is $\alpha$
$\frac{\alpha +\alpha +\alpha -a}{\sqrt{3}}=\alpha \therefore \alpha =\frac{a}{3-\sqrt{3}}$
Required centre is $(\alpha ,\alpha ,\alpha )=(\frac{\alpha }{3-\sqrt{3}},\frac{\alpha }{3-\sqrt{3}},\frac{\alpha }{3-\sqrt{3}})$