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 $△ABC$ at a point $E$. If the length of side $AD$ is 4 and volume of the tetrahedron is $\frac{2\sqrt{2}}{3}$ then the correct statement(s) is/are

• A. The 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)

Given $V=\frac{2\sqrt{2}}{3}$


$[$ Note $ABC$ is a right triangle $\to$ Area $=\frac{1}{2}(2)(\sqrt{2})=\sqrt{2}]$
$h|\hat{i}(-1+1)+2(\hat{j}-\hat{k})|=4\sqrt{2}$
$\Rightarrow h|\hat{j}-\hat{k}|=2\sqrt{2}\Rightarrow h=2\Rightarrow \text{ A and }D$
Let $E$ divides $AM$ in the ratio $\lambda :1$
hence $E:\left(\frac{2\lambda +1}{\lambda +1},\frac{1}{\lambda +1},\frac{1}{\lambda +1}\right)$
Now, $(AE{)}^2+(DE{)}^2=(AD{)}^2$
${\left(\frac{2\lambda +1}{\lambda +1}-1\right)}^2+{\left(1-\frac{1}{\lambda +1}\right)}^2+{\left(1-\frac{1}{\lambda +1}\right)}^2+4=16$
${\left(\frac{\lambda }{\lambda +1}\right)}^2+2{\left(\frac{\lambda }{\lambda +1}\right)}^2=12\Rightarrow {\left(\frac{\lambda }{\lambda +1}\right)}^2=4\Rightarrow \frac{\lambda }{\lambda +1}=2\text{ or }-2$
$\therefore$ These are two positions for $E$ which are $(-1,3,3)$ and $(3,-1,-1)$