Question

A plane $\pi$ passes through the point $(1,1,1)$ and is parallel to the vectors $b=(1,0,-1)$ and $a=(-1,1,0)$, $\pi$ meets the axes in $A,B$ and $C$ forming the tetrahedron $OABC$.

Answer 1

(A) equation of the plane $\pi$ is $A(x-1)+B(y-1)+C(z-1)=0$ wherer $A+0.B-1.C=0$ and $-A+B+0.C=0$.

$\Rightarrow A=B=C$ and the equation of $\pi$ is $x+y=z=3$.

Which meets the coordinates axes at $A(3,0,0),B(0,3,0)$ and $C(0,0,3)$.

So volume of the tetrahedron $OABC=\frac{1}{6}\left|\begin{array}{cccc}0& 0& 0& 1\\ 3& 0& 0& 1\\ 0& 3& 0& 1\\ 0& 0& 3& 1\end{array}\right|=\frac{1}{6}\times 27=\frac{9}{2}$

(B) $AB=BC=CA=3\sqrt{2},$ so the area of the face $ABC$ which is an equilateral triangle $=\frac{\sqrt{3}}{4}\times {\left(3\sqrt{2}\right)}^2=\frac{9\sqrt{3}}{2}$.

Each of the face $OAB,OBC,OCA$ is an isosceles traingle with sides $3,3\sqrt{2}$

So area of each of these faces is $\frac{1}{2}\times 3\sqrt{2}\times \sqrt{9-\frac{9}{2}}=\frac{9}{2}$

(C) required distance is $\left|\frac{\frac{3\sqrt{3}}{2}+3\sqrt{3}+3-3}{\sqrt{1+1+1}}\right|=\frac{9}{2}$

(D) Distance of $O$ from the face $ABC=\left|\frac{0+0+0-3}{\sqrt{1+1+1}}\right|=\sqrt{3}$

Distance of $A,B,C$ from opposite faces $OBC,OCA,OAB$ are equal and is $3$.