Question

$ABCD$ is a regular tetrahedron. $M(0,0,0)$ is the midpoint of the altitude $DN$ of the tetrahedron. Coordinates of the points $A$ and $B$ are $(0,0,-2)$ and $(b,b,0)$ respectively (with $b>0$). Also $x$ - coordinate of the point $C$ is given to be positive. Then which of the following is/are correct?

• A. Image of point $D$ in the plane generated by $ABC$ is $(2\sqrt{2},0,-2)$
• B. Image of point $D$ in the plane generated by $ABC$ is $(2,0,-2\sqrt{2})$
• C. Volume of tetrahedron $ACBM$ is $\frac{4\sqrt{2}}{3}$
• D. Volume of tetrahedron $ACBM$ is $\frac{4}{3}$

Answer 1

Answer: (A), (D)

The correct options are
A Image of point $D$ in the plane generated by $ABC$ is $(2\sqrt{2},0,-2)$
D Volume of tetrahedron $ACBM$ is $\frac{4}{3}$

$AB=BC=CD=AD=a$

$a=2\sqrt{2}$
$MA=MB=MC=2$ (Using symmetry)
$MB=2\Rightarrow 2b^2=4\Rightarrow b=\sqrt{2}$
$B(\sqrt{2},\sqrt{2},0)$
$M{A}^2+M{B}^2=A{B}^2\Rightarrow \overrightarrow{MA}\perp \overrightarrow{MB}$
Similarly $\overrightarrow{MA}\perp \overrightarrow{MC},\overrightarrow{MB}\perp \overrightarrow{MC}$
$\Rightarrow \overrightarrow{MA}.\overrightarrow{MC}\Rightarrow -2z=0\overrightarrow{MB}.\overrightarrow{MC}\Rightarrow \sqrt{2}x+\sqrt{2}y=0\Rightarrow y=-x\&z=0\therefore C(x,-x,0)$
Also $M{C}^2=2x^2=M{A}^2=4\Rightarrow x=\sqrt{2}(x>0)$
So $C(\sqrt{2},-\sqrt{2},0)$
Coordinates of $N$ are $(\frac{2\sqrt{2}}{3},0,\frac{-2}{3})$ and hence $D$ is $(\frac{-2\sqrt{2}}{3},0,\frac{2}{3}).DN$ is normal to the plane $ABC.N$ is mid point of $D{D}^{\prime }$ so $D^{\prime }=(2\sqrt{2},0,-2)$
Since $MA,MB,MC$ are orthogonal triads
Volume of tetrahedron is= $\frac{1}{6}|MA|.|MB|.|MC|=\frac{4}{3}.$