Question

The position vectors of the four angular points of a tetrahedron $OABC$ are $(0,0,0)$,$(0,0,2),$ $(0,4,0)$ and $(6,0,0),$ respectively. A point $P$ inside the tetrahedron is at the same distance ' $r$' from the four plane faces of the tetrahedron. Find the value of $9r$.

Answer 1

The given points are $O(0,0,0),A(0,0,2),B(0,4,0)$ and $C(6,0,0)$

Here three faces of tetrahedron are $xy,yz,zx$ plane.

Since point $P$ is equidistance from $zx,xy$ and $yz$ planes, its coordinates are $P(r,r,r)$

Equation of plane $ABC$ is

$2x+3y+6z=12\left(\text{from intercept form}\right)$

$P$ is also at distance $r$ from plane $ABC$

$\Rightarrow \frac{|2r+3r+6r-12|}{\sqrt{4+9+36}}=r$

$\Rightarrow |11r-12|=7r$

$\Rightarrow -11r-12=\pm 7r$

$\Rightarrow r=\frac{12}{18},3$

$\therefore r=2/3(\mathrm{as}r<2)$

And $9r=6$