Question

The position vectors of the vertices $A,B$ and $C$ of a tetrahedron $ABCD$ are $\hat{i}+\hat{j}+\hat{k},\hat{i}$ and $3\hat{i}$, respectively. The altitude from vertex $D$ to the opposite face $ABC$ meets the median line through $A$ of the $△ABC$ at a point $E$. If the length of the side $AD$ is $4$ and the volume of the tetrahedron is $\frac{2\sqrt{2}}{3}$, then find the position vector of the point $E$ for all its possible positions.

Answer 1

$F$ is mid-point of $BC$ i.e. $F=\frac{\hat{i}+3\hat{i}}{2}=2\hat{i}$ and $AE\perp DE$ [given]

Let $E$ divides $AF$ in $\lambda :1$. The position vector of $E$ is given by
$\frac{2\lambda \hat{i}+1(\hat{i}+\hat{j}+\hat{k})}{\lambda +1}=\left(\frac{2\lambda +1}{\lambda +1}\right)\hat{i}+\frac{1}{\lambda +1}\hat{j}+\frac{1}{\lambda +1}\hat{k}$
Now, volume of the tetrahedron
$=\frac{1}{3}$ (area of the base) (height)
$\Rightarrow \frac{2\sqrt{2}}{3}=\frac{1}{3}($ area of the $△ABC)(DE)$
But area of the $△ABC=\frac{1}{2}|\overrightarrow{BC}\times \overrightarrow{BA}|$
$=\frac{1}{2}|2\hat{i}\times (\hat{j}+\hat{k})|=|\hat{i}\times \hat{j}+\hat{i}\times \hat{k}|=|\hat{k}-\hat{j}|=\sqrt{2}$
$\frac{2\sqrt{2}}{3}=\frac{1}{3}(\sqrt{2})(DE)\Rightarrow DE=2$
Since, $△ADE$ is a right angle triangle, then
$=\frac{\lambda }{\lambda +1}\hat{i}-\frac{\lambda }{\lambda +1}\hat{j}-\frac{\lambda }{\lambda +1}\hat{k}$
$\Rightarrow |\overrightarrow{AE}{|}^2=\frac{1}{(\lambda +1{)}^2}\left[{\lambda }^2+{\lambda }^2+{\lambda }^2\right]=\frac{3{\lambda }^2}{(\lambda +1{)}^2}$
Therefore, $12=\frac{3{\lambda }^2}{(\lambda +1{)}^2}$
$\Rightarrow 4(\lambda +1{)}^2={\lambda }^2\Rightarrow 4{\lambda }^2+4+8\lambda ={\lambda }^2$
$\Rightarrow 3{\lambda }^2+8\lambda +4=0\Rightarrow 3{\lambda }^2+6\lambda +2\lambda +4=0$
$\Rightarrow 3\lambda (\lambda +2)+2(\lambda +2)=0$
$\Rightarrow (3\lambda +2)(\lambda +2)=0\Rightarrow \lambda =-2/3,\lambda =-2$
When $\lambda =-2/3$, position vector of $E$ is given by
$\left(\frac{2\lambda +1}{\lambda +1}\right)\hat{i}+\frac{1}{\lambda +1}\hat{j}+\frac{1}{\lambda +1}\hat{k}$
$=\frac{2+(-2/3)+1}{-2/3+1}\hat{i}+\frac{1}{-2/3+1}\hat{j}+\frac{1}{-2/3+1}\hat{k}$
$=\frac{-4/3+1}{\frac{-2+3}{3}}\hat{i}+\frac{1}{\frac{-2+3}{3}}\hat{j}+\frac{1}{\frac{-2+3}{3}}\hat{k}$
$=\frac{-4+3}{1/3}\hat{i}+\frac{1}{1/3}\hat{j}+\frac{1}{1/3}\hat{k}=-\hat{i}+3\hat{j}+3\hat{k}$
and when $\lambda =-2$, position vector of $E$ is given by,
$\frac{2\times (-2)+1}{-2+1}\hat{i}+\frac{1}{-2+1}\hat{j}+\frac{1}{-2+1}\hat{k}=\frac{-4+1}{-1}\hat{i}-\hat{j}-\hat{k}$
$=3\hat{i}-\hat{j}-\hat{k}$
Therefore, $-\hat{i}+3\hat{j}+3\hat{k}$ and $+3\hat{i}-\hat{j}-\hat{k}$ are the answer.