Question

The position vectors of the vertices $A, B$ and $C$ of a tetrahedron $A B C D$ are $\hat{ i }+\hat{ j }+\hat{ k }, \hat{ i }$ and $3 \hat{ i }$, respectively. The altitude from vertex $D$ to the opposite face $A B C$ meets the median line through $A$ of the $\triangle A B C$ at a point $E$. If the length of the side $A D$ 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 $B C$ i.e. $F=\frac{\hat{ i }+3 \hat{ i }}{2}=2 \hat{ i }$ and $A E \perp D E$ [given]

Let $E$ divides $A F$ 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 $ \triangle A B C)(D E)$
But area of the $\triangle A B C=\frac{1}{2}|\overrightarrow{ B C } \times \overrightarrow{ B A }|$
$=\frac{1}{2}|2 \hat{ i } \times(\hat{ j }+\hat{ k })|=|\hat{ i } \times \hat{ j }+\hat{ i } \times \hat{ k }|=|\hat{ k }-\hat{ j }|=\sqrt{2}$
$\frac{2 \sqrt{2}}{3}=\frac{1}{3}(\sqrt{2})(D E) \Rightarrow D E=2$
Since, $\triangle A D E$ is a right angle triangle, then
$=\frac{\lambda}{\lambda+1} \hat{ i }-\frac{\lambda}{\lambda+1} \hat{ j }-\frac{\lambda}{\lambda+1} \hat{ k }$
$\Rightarrow |\overrightarrow{ A E }|^{2}=\frac{1}{(\lambda+1)^{2}}\left[\lambda^{2}+\lambda^{2}+\lambda^{2}\right]=\frac{3 \lambda^{2}}{(\lambda+1)^{2}}$
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.