JEE Main 2025 Apr 4 Shift 2, Mathematics Q25: Applications of Vectors

For any triangle with centroid $G$, the identity

$|\overrightarrow{AG}{|}^2+|\overrightarrow{BG}{|}^2+|\overrightarrow{CG}{|}^2=\frac{1}{3}\left(|\overrightarrow{AB}{|}^2+|\overrightarrow{BC}{|}^2+|\overrightarrow{CA}{|}^2\right)$

holds (each median-to-centroid distance is $\frac{2}{3}$ of the median, and the sum of squares of the medians equals $\frac{3}{4}$ of the sum of squares of the sides).

Identify the three side-vectors and their square-lengths:

$|2\hat{i}-\hat{j}+\hat{k}{|}^2=4+1+1=6$,

$|\hat{i}-3\hat{j}-5\hat{k}{|}^2=1+9+25=35$,

$|3\hat{i}-4\hat{j}-4\hat{k}{|}^2=9+16+16=41$.

Sum $=6+35+41=82$. Then

$|\overrightarrow{AG}{|}^2+|\overrightarrow{BG}{|}^2+|\overrightarrow{CG}{|}^2=\frac{82}{3}$.

$\therefore 6\cdot \frac{82}{3}=2\cdot 82=164$.