Subject: Mathematics — Vectors
Question: The position vectors of the vertices $A,B$ and $C$ of a
tetrahedron $ABCD$ are $\widehat{i} + \widehat{j} + \widehat{k}$,
$\widehat{i}$ and $3\widehat{i}$, 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 $\frac{2\sqrt{2}}{3}$, find the
position vector of the point $E$ for all its possible positions.
Answer: $( - 1,3,3)\quad\text{or}\quad(3, - 1, - 1).$
Solution: Given that $AD = 4$.
Let $DE = h$.
$$\text{Volume of tetrahedron} = \frac{2\sqrt{2}}{3}.$$
$$\Rightarrow \frac{1}{3}Ar( \bigtriangleup ABC)h = \frac{2\sqrt{2}}{3}.$$
$$\therefore\frac{1}{2}\left| \overrightarrow{BA} \times \overrightarrow{BC} \right|h = 2\sqrt{2}.$$
$$\frac{1}{2}\left| \left( \widehat{j} + \widehat{k} \right) \times 2\widehat{i} \right|h = 2\sqrt{2}\quad\text{or}\quad\left| \widehat{j} - \widehat{k} \right|h = 2\sqrt{2}.$$
$$\text{or}\quad\sqrt{2}h = 2\sqrt{2}\quad\therefore h = 2.$$
[IMAGE] Let point $E$ divides median $AF$ in the ratio $\lambda:1$.
$$\therefore\overrightarrow{OE} = \frac{\lambda \cdot 2\widehat{i} + \left( \widehat{i} + \widehat{j} + \widehat{k} \right)}{\lambda + 1}\quad\quad\cdots(ii)$$
$$\therefore\overrightarrow{AE} = \text{P.V. of }E - \text{P.V. of }A = \frac{\lambda}{\lambda + 1}\left( \widehat{i} - \widehat{j} - \widehat{k} \right).$$
$$\therefore\left| \overrightarrow{AE} \right|^{2} = AE^{2} = \left( \frac{\lambda}{\lambda + 1} \right)^{2} \cdot 3\quad\quad\cdots(iii)$$
Now,
$$h^{2} + AE^{2} = AD^{2}.$$
$$\Rightarrow 4 + \left( \frac{\lambda}{\lambda + 1} \right)^{2} \cdot 3 = 16.$$
$$\Rightarrow 3\left( \frac{\lambda}{\lambda + 1} \right)^{2} = 12.$$
$$\Rightarrow \left( \frac{\lambda}{\lambda + 1} \right) = \pm 2 \Rightarrow \lambda = \pm (2\lambda + 2).$$
$$\therefore\lambda = - 2\quad\text{or}\quad - \frac{2}{3}.$$
Putting the value of $\lambda$ in $(ii)$, we get the possible
positions of $E$ as
$$( - 1,3,3)\quad\text{or}\quad(3, - 1, - 1).$$