Vector Algebra
Position vector of foot of altitude in a tetrahedron
nta_pyq_2025_apr
Grade 12
Question:
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 $\dfrac{\sqrt{110}}{3}$ and the volume of the tetrahedron is $\dfrac{\sqrt{805}}{6\sqrt{2}}$, then the position vector of $E$ is:
$\dfrac{1}{12}(7\hat{i}+4\hat{j}+3\hat{k})$
$\dfrac{1}{2}(\hat{i}+4\hat{j}+7\hat{k})$
$\dfrac{1}{6}(12\hat{i}+12\hat{j}+\hat{k})$
$\dfrac{1}{6}(7\hat{i}+12\hat{j}+\hat{k})$
Step-by-Step Solution
Key Concept: Find the height $h$ of $D$ above face $ABC$ from the volume formula, then use $DE^2=AD^2-AE^2$ (with $AE$ computed from median direction) to locate $E$ along the median from $A$.
$\overrightarrow{AB}=\hat{j}-3\hat{k}$, $\overrightarrow{AC}=\hat{i}-\hat{j}-2\hat{k}$.
Area of $\triangle ABC=\tfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|=\tfrac{1}{2}|{-5\hat{i}+3\hat{j}+\hat{k}}|=\dfrac{\sqrt{35}}{2}$.
Volume $=\tfrac{1}{3}\times\dfrac{\sqrt{35}}{2}\times h=\dfrac{\sqrt{805}}{6\sqrt{2}} \Rightarrow h=\sqrt{\dfrac{23}{2}}$.
$AE^2=AD^2-DE^2=\dfrac{110}{9}-\dfrac{23}{2}=\dfrac{220-207}{18}=\dfrac{13}{18}$.
Midpoint $F$ of $BC$: $F=\dfrac{1}{2}((1+3+(-2))\hat{i}+(\hat{i}+3\hat{j}-2\hat{k})+(2\hat{i}+\hat{j}-\hat{k}))$... $F=\left(\tfrac{3}{2},2,-\tfrac{3}{2}\right)$.
$\overrightarrow{AF}=\tfrac{1}{2}\hat{i}+0\hat{j}-\tfrac{5}{2}\hat{k}$, unit vector $=\dfrac{\hat{i}-5\hat{k}}{\sqrt{26}}$.
$\overrightarrow{AE}=\sqrt{\tfrac{13}{18}}\cdot\dfrac{\hat{i}-5\hat{k}}{\sqrt{26}}=\dfrac{\hat{i}-5\hat{k}}{6}$.
P.V. of $E=A+\overrightarrow{AE}=\hat{i}+2\hat{j}+\hat{k}+\dfrac{\hat{i}-5\hat{k}}{6}=\dfrac{7\hat{i}+12\hat{j}+\hat{k}}{6}$.
Correct Answer: 4