Question

The position vectors of the vertices $A,B,C$ 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 triangle $ABC$ at a point $E$. If the length of side $AD$ is $4$ and volume of the tetrahedron is $2\sqrt{2}/3$ then find the all possible vectors of the point $E$.

Answer 1

Given

$A(1,1,1),B(1,0,0),C(3,0,0),D(x,y,z)$

$AD=4$

$\sqrt{(x-1{)}^2+(y-1{)}^2+(z-1{)}^2}=4$

$(x-1{)}^2+(y-1{)}^2+(z-1{)}^2=16$

Possible value of $x,y,z$ are $4,1,1$

$D(4,1,1)$

$AB=\hat{i}-\hat{i}-\hat{j}-\hat{k}$

$AB=-\hat{j}-\hat{k}$

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

$AC=2\hat{i}-\hat{j}-\hat{k}$

$AB\times AC=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& -1& -1\\ 2& -1& -1\end{array}\right|$

$AB\times AC=\hat{i}\left(0\right)-\hat{j}\left(2\right)+\hat{k}\left(2\right)$

$AB\times AC=-2\hat{j}+2\hat{k}$

$|AB\times AC|=\sqrt{(-2{)}^2+2^2}$

$|AB\times AC|=\sqrt{8}$

$|AB\times AC|=2\sqrt{2}$

Volume $=\frac{2\sqrt{2}}{3}$

$\frac{1}{3}\times \frac{|AB\times AC|}{2}height=\frac{2\sqrt{2}}{3}$

$\frac{2\sqrt{2}}{2}height=2\sqrt{2}$

$\frac{1}{2}height=1$

$height=2=DF$

$DF=2$

$\sqrt{(x-4{)}^2+(y-1{)}^2+(z-1{)}^2}=2$

$(x-4{)}^2+(y-1{)}^2+(z-1{)}^2=4$

Possible coordinates of $F(x,y,z)$ are $F(2,1,1)$

eq of altitude

$\frac{x-4}{-2}=\frac{y-1}{0}=\frac{z-1}{0}$

let H is the mid point of line BC

$H(\frac{3+1}{2},0,0)$

$H(2,0,0)$

eq of median from A

$\frac{x-1}{1}=\frac{y-1}{-1}=\frac{z-1}{-1}$

Altitude meets at E on median so

eq of altitude

$\frac{x-4}{-2}=\frac{y-1}{0}=\frac{z-1}{0}=\lambda$

$x=-2\lambda +4$

$y=1$

$z=1$

Putting $x,y,z$ in eq of median

$\frac{-2\lambda +4-1}{1}=\frac{1-1}{-1}=\frac{1-1}{-1}$

$\frac{-2\lambda +3}{1}=0=0$

$-2\lambda +3=0$

$\lambda =\frac{3}{2}$

putting $\lambda$ in x

$x=-2\frac{3}{2}+4$

$x=-3+4$

$x=1$

So, the point $E(1,1,1)$.