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  4. The centre of sphere inscribed in a tetrahedron bounded by the planes vecr⋅ hati=0= vecr⋅ hatj= vecr⋅ hatk and vecr⋅( hati+ hatj+ hatk)=a is

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Q. 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

Three Dimensional Geometry

Solution:

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 $\bot$ 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
$\left(\alpha, \alpha, \alpha\right)=\left(\frac{a}{3-\sqrt{3}}, \frac{a}{3-\sqrt{3}}, \frac{a}{3-\sqrt{3}}\right)$