Question

The position vectors of the vertices $A,B$ and $C$ of a tetrahedron are $(1,1,1),$ $(1,0,0)$ and $(3,0,0)$ respectively. The altitude from the vertex $D$ to the opposite face $ABC$ meets the median line through $A$ of the $\Delta ABC$ at a point $E.$ If the length of side $AD$ is $4$ units and volume of the tetrahedron is $\frac{2\sqrt{2}}{3}$ cubic units, then the CORRECT statement(s) is (are)

• A. The length of altitude from the vertex $D$ is $2$.
• B. There is exactly one position for the point $E$.
• C. There can be two positions for the point $E$.
• D. Vector $\hat{j}-\hat{k}$ is normal to the plane $ABC$.

Answer 1

Answer: (A), (C), (D)

Vector $\hat{j}-\hat{k}$ is normal to the plane $ABC$.

Given, $V=\frac{2\sqrt{2}}{3}$
Let $h$ be the length of altitude drawn from $D$ to the face $ABC.$ Then
$\frac{1}{3}\cdot \frac{1}{2}\left|\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 1& 1\\ 2& -1& -1\end{array}\right|\right|h=\frac{2\sqrt{2}}{3}$
$\Rightarrow h\left|\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 1& 1\\ 2& -1& -1\end{array}\right|\right|=4\sqrt{2}$

[Note, $ABC$ is a right angled triangle with area $=\frac{1}{2}\left(2\right)\left(\sqrt{2}\right)=\sqrt{2}]$
$h\left|\hat{i}\right(-1+1)+2(\hat{j}-\hat{k}\left)\right|=4\sqrt{2}$
$\Rightarrow h|\hat{j}-\hat{k}|=2\sqrt{2}\Rightarrow h=2$

Let $E$ divides $AM$ in the ratio $\lambda :1$
Then coordinates of $E$ are $(\frac{2\lambda +1}{\lambda +1},\frac{1}{\lambda +1},\frac{1}{\lambda +1}).$

Now, $(AE{)}^2+(DE{)}^2=(AD{)}^2$
$\Rightarrow {(\frac{2\lambda +1}{\lambda +1}-1)}^2+{(1-\frac{1}{\lambda +1})}^2+{(1-\frac{1}{\lambda +1})}^2+4=16$
$\Rightarrow {\left(\frac{\lambda }{\lambda +1}\right)}^2+2{\left(\frac{\lambda }{\lambda +1}\right)}^2=12$
$\Rightarrow {\left(\frac{\lambda }{\lambda +1}\right)}^2=4$
$\Rightarrow \frac{\lambda }{\lambda +1}=2$ or $\frac{\lambda }{\lambda +1}=-2$
$\Rightarrow \lambda =-2$ or $\lambda =-\frac{2}{3}$

$\therefore$ There are two positions for $E$ which are $(-1,3,3)$ and $(3,-1,-1)$