JEE Main 2025 (Online) 23rd January Morning ShiftQuestion 1
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
3D Geometry
Solution
Area of $ABC=\frac12|(0,1,-3)\times(1,-1,-2)|=\frac{\sqrt{35}}2$, so $DE=\frac{3V}{\text{area}}=\sqrt{\frac{23}{2}}$ and $AE^2=\frac{110}9-\frac{23}2=\frac{13}{18}$. With $M=\left(\frac32,2,-\frac32\right)$, $AM^2=\frac{13}2$, so $E=A+\frac13(M-A)=\frac16(7\hat i+12\hat j+\hat k)$.