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 triangle $ABC$ at a point $E.$ If the length of the side $AD$ is $4$ and the volume of the tetrahedron is $\frac{2\sqrt{2}}{3}$ cubic units, then the position vector(s) of the point $E$ for all its positions is (are)

• A. $-\hat{i}+3\hat{j}+3\hat{k}$
• B. $3\hat{i}-\hat{j}-\hat{k}$
• C. $-\hat{i}+3\hat{j}-3\hat{k}$
• D. $3\hat{i}+3\hat{j}-\hat{k}$

Answer 1

Answer: (A), (B)

$3\hat{i}-\hat{j}-\hat{k}$

We are given $AD=4$
Let length $DE=p$
Volume of tetrahedron $=\frac{2\sqrt{2}}{3}$
$\Rightarrow \frac{1}{3}\left(Areaof△ABC\right)\times height=\frac{2\sqrt{2}}{3}\Rightarrow \frac{1}{3}\times \frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|p=\frac{2\sqrt{2}}{3}\Rightarrow |\overrightarrow{AB}\times \overrightarrow{AC}|p=4\sqrt{2}\Rightarrow |(-\hat{j}-\hat{k})\times (2\hat{i}-\hat{j}-\hat{k})|p=4\sqrt{2}\Rightarrow |\hat{j}-\hat{k}|p=2\sqrt{2}\Rightarrow \sqrt{2}p=2\sqrt{2}$
$\Rightarrow p=2$

We have to find the position vector of point $E.$ Let it divide median $AF$ in the ratio $\lambda :1$
$\overrightarrow{E}=\frac{2\lambda \hat{i}+(\hat{i}+\hat{j}+\hat{k})}{\lambda +1}\dots \left(1\right)\therefore \overrightarrow{AE}=\overrightarrow{E}-\overrightarrow{A}=\frac{\lambda (\hat{i}-\hat{j}-\hat{k})}{\lambda +1}\therefore {\left|\overrightarrow{AE}\right|}^2=3{\left(\frac{\lambda }{\lambda +1}\right)}^2$

Now in $△AED,$
$4+3{\left(\frac{\lambda }{\lambda +1}\right)}^2=16$
$\Rightarrow \frac{\lambda }{\lambda +1}=\pm 2$
$\Rightarrow \lambda =-2,-\frac{2}{3}$
Putting the value of $\lambda$ in $\left(1\right),$ we get the position vector of possible positions of $E$ as $-\hat{i}+3\hat{j}+3\hat{k}$ or $3\hat{i}-\hat{j}-\hat{k}$