JEE Main 2025 Jan 23 Shift 1, Mathematics Q17: Applications of Vectors

Let $A=(1,2,1)$, $B=(1,3,-2)$, $C=(2,1,-1)$. The midpoint of BC is $F=\left(\frac{3}{2},2,-\frac{3}{2}\right)$, so the median from A goes along $\overrightarrow{AF}$.

$\overrightarrow{AB}=(0,1,-3)$, $\overrightarrow{AC}=(1,-1,-2)$. Cross product: $\overrightarrow{AB}\times \overrightarrow{AC}=(-5,-3,-1)$, magnitude $\sqrt{35}$.

Area of $△ABC=\frac{\sqrt{35}}{2}$.

Volume $=\frac{1}{3}\cdot \frac{\sqrt{35}}{2}\cdot h=\frac{\sqrt{805}}{6\sqrt{2}}$, giving $h=\sqrt{\frac{23}{2}}$. This is $DE$.

By Pythagoras in $△ADE$ (right-angled at E): $A{E}^2=A{D}^2-D{E}^2=\frac{110}{9}-\frac{23}{2}=\frac{220-207}{18}=\frac{13}{18}$.

$\overrightarrow{AF}=\left(\frac{1}{2},0,-\frac{5}{2}\right)=\frac{1}{2}(1,0,-5)$, with $|\overrightarrow{AF}|=\frac{\sqrt{26}}{2}$.

$\overrightarrow{AE}=|AE|\cdot \frac{(1,0,-5)}{\sqrt{26}}=\sqrt{\frac{13}{18}}\cdot \frac{(1,0,-5)}{\sqrt{26}}=\frac{(1,0,-5)}{6}$.

Position vector of E $=\overrightarrow{OA}+\overrightarrow{AE}=(1,2,1)+\frac{1}{6}(1,0,-5)=\frac{1}{6}(7,12,1)$.