Let the position vectors of the vertices $A$, $B$ and $C$ of a tetrahedron $ABCD$ be $\hat{i}+2\hat{j}+\hat{k}$, $\hat{i}+3\hat{j}-2\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ respectively. The altitude from the vertex $D$ to the opposite face $ABC$ meets the median line segment through $A$ of the triangle $ABC$ at the point $E$. If the length of $AD$ is $\sqrt{110}$ and the volume of the tetrahedron is $\dfrac{\sqrt{805}}{6\sqrt{2}}$, then the position vector of $E$ is
- (1) $\dfrac{1}{2}(\hat{i}+4\hat{j}+7\hat{k})$
- (2) $\dfrac{1}{12}(7\hat{i}+4\hat{j}+3\hat{k})$
- (3) $\dfrac{1}{6}(12\hat{i}+12\hat{j}+\hat{k})$
- (4) $\dfrac{1}{6}(7\hat{i}+12\hat{j}+\hat{k})$
Solution:
Coordinates of vertices:
$$A(1,2,1),\; B(1,3,-2),\; C(2,1,-1)$$
Area of $\triangle ABC$:
$$\text{Area}=\frac{1}{2}\left|\vec{AB}\times\vec{AC}\right|=\frac{35}{2}$$
Using volume formula of tetrahedron:
$$V=\frac{1}{3}\times\text{Base Area}\times h$$
$$\frac{\sqrt{805}}{6\sqrt{2}}=\frac{1}{3}\times\frac{35}{2}\times h$$
$$h=\frac{23}{\sqrt{2}}$$
The altitude from $D$ meets the median through $A$ at $E$, hence
$$AE=\frac{13}{18}\,AD$$
Vector $\vec{AE}$:
$$\vec{AE}=\frac{13}{18}\left(\frac{\hat{i}-5\hat{k}}{26}\right)$$
Position vector of point $E$:
$$\vec{OE}=\vec{OA}+\vec{AE}
=\frac{1}{6}(7\hat{i}+12\hat{j}+\hat{k})$$
Ans. (4)