The position vectors of vertices A,B,C of a tetrahedron ABCD are i+2 j+ k, i+3 j-2 k, 2 i+…

Question

The position vectors of vertices $A,B,C$ of a tetrahedron $ABCD$ are $\hat i+2\hat j+\hat k$, $\hat i+3\hat j-2\hat k$, $2\hat i+\hat j-\hat k$ respectively. The altitude from vertex $D$ to face $ABC$ meets the median from $A$ of triangle $ABC$ at point $E$. If $|AD| = \dfrac{\sqrt{110}}3$ and the volume of the tetrahedron is $\dfrac{\sqrt{805}}{6\sqrt2}$, the position vector of $E$ is:

Accepted Answer

*(This question's raw solution text was significantly scrambled by OCR; the source's key intermediate results are reported here.)* Using $\vec{AB},\vec{AC}$ derived from the vertices, the area of $\triangle ABC$ works out to $\dfrac{\sqrt{35}}2$. From Volume $=\dfrac13\cdot\text{Area}\cdot h$: $h=\sqrt{\dfrac{23}2}$ (the perpendicular distance $DE$).

With $|AD|^2=\dfrac{110}9$, $AE^2 = AD^2-DE^2 = \dfrac{13}{18}$, so $AE=\sqrt{13/18}$. Along the median direction $\dfrac{\hat i-5\hat k}{\sqrt{26}}$ from $A$:
$$\vec{AE} = \sqrt{\frac{13}{18}}\cdot\frac{\hat i-5\hat k}{\sqrt{26}} = \frac{\hat i-5\hat k}6$$
$$\text{P.V. of } E = \vec{AE}+\vec A = \frac{\hat i-5\hat k}6+(\hat i+2\hat j+\hat k) = \frac16(7\hat i+12\hat j+\hat k)$$

Solutions · 1