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 $r.$

The equation of a plane passing through three points $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ is $\begin{vmatrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\end{vmatrix}=0$

Using this,i found the equations of the planes $OAB,OBC,OAC$ are $x=0,z=0,y=0$ respectively and the equation of the plane $ABC$ is $2x+3y+6z-12=0$.Let the coordinates of $P$ be $(x_1,y_1,z_1)$.

Distance of $P$ from $OAB$ plane is $x_1$.
Distance of $P$ from $OBC$ plane is $z_1$.
Distance of $P$ from $OAC$ plane is $y_1$.
Distance of $P$ from $ABC$ plane is $\frac{2x_1+3y_1+6z_1-12}{7}$.

According to the question,
I solved the equations to find $x_1=y_1=z_1=3$.So $r=3$
But the answer given in the book is $\frac{2}{3}$.I dont know where have i gone wrong?Please help me.Thanks.

  • $\begingroup$ Obviously, P is the incenter of the tetrahedron. $\endgroup$
    – Lucian
    Nov 8, 2015 at 19:11

1 Answer 1


The distance of $P$ from $ABC$ plane is $$\frac{|2x_1+3y_1+6z_1-12|}{7}.$$

So, solving $$x_1=y_1=z_1=\frac{|2x_1+3y_1+6z_1-12|}{7}=r$$ gives $$(x_1,y_1,z_1,r)=(3,3,3,3),\left(\frac 23,\frac 23,\frac 23,\frac 23\right).$$

But the former is outside the tetrahedron.

  • $\begingroup$ How do we say that $(3,3,3)$ is outside the tetrahedron?,Sir.Is there any method for checking this. $\endgroup$ Nov 8, 2015 at 16:31
  • $\begingroup$ @VinodKumarPunia: Let $F(x,y,z)=2x+3y+6z−12$. We have $F(0,0,0)=−12\lt 0$. This means that we have to have at least $F(x,y,z)\lt 0$ in order for $(x,y,z)$ to be inside the tetrahedron. Now, $F(3,3,3)=21\gt 0$, which means that $(3,3,3)$ is outside the tetrahedron. $\endgroup$
    – mathlove
    Nov 8, 2015 at 17:03

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