JEE Main 2025 (Online) 4th April Evening ShiftQuestion 1
Let $A$ be the point of intersection of the lines $L_1:\frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}$ and $L_2:\frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$. Let $B$ and $C$ be the points on the lines $L_1$ and $L_2$ respectively such that $AB=AC=\sqrt{15}$. Then the square of the area of the triangle $ABC$ is :
3D Geometry
Solution
Intersection $A=(7,5,3)$. $\cos\theta=\frac{(1,0,-1)\cdot(3,4,5)}{\sqrt2\cdot5\sqrt2}=-\frac15$, $\sin\theta=\frac{2\sqrt6}5$. Area $=\frac12\cdot15\cdot\frac{2\sqrt6}5=3\sqrt6$, square $=54$.