Q. 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 $\frac{\sqrt{110}}3$ and the volume of the tetrahedron is $\frac{\sqrt{805}}{6\sqrt2}$, then the position vector of $E$ is
A) $\frac1{12}\left(7\hat i+4\hat j+3\hat k\right)$
B) $\frac12\left(\hat i+4\hat j+7\hat k\right)$
C) $\frac16\left(12\hat i+12\hat j+\hat k\right)$
D) $\frac16\left(7\hat i+12\hat j+\hat k\right)$
Explanation
[Solution by Dr. Abubacker] $\vec{AB}\times\vec{AC}=(-5,-3,-1)$, so area$(ABC)=\frac{\sqrt{35}}2$. The altitude is $h=\frac{3V}{\text{area}}$, $h^2=\frac{23}2$. In the right triangle $ADE$: $AE^2=AD^2-h^2=\frac{110}9-\frac{23}2=\frac{13}{18}$. The median from $A$ goes to $M\left(\frac32,2,-\frac32\right)$ with $\vec{AM}=\left(\frac12,0,-\frac52\right)$, $AM^2=\frac{13}2$, so $AE=\frac13AM$. Hence $E=A+\frac13\vec{AM}=\left(\frac76,2,\frac16\right)=\frac16(7\hat i+12\hat j+\hat k)$. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.