JEE Main 2025 (Online) 4th April Morning ShiftQuestion 1
Let $A$ and $B$ be two distinct points on the line $L:\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2\sqrt{17}$ from the foot of perpendicular drawn from the point $(1,2,3)$ on the line $L$. If $O$ is the origin, then $\overrightarrow{OA}\cdot\overrightarrow{OB}$ is equal to
3D Geometry
Solution
Direction $(3,2,-2)$, $|d|^2=17$. Foot of perpendicular from $(1,2,3)$ on $L$: $(6+3t,7+2t,7-2t)$, need $(5+3t,5+2t,4-2t)\cdot(3,2,-2)=0$: $15+9t+10+4t-8+4t=0\Rightarrow t=-1$, foot $F=(3,5,9)$. $A,B=F\pm\frac{2\sqrt{17}}{\sqrt{17}}(3,2,-2)=F\pm2(3,2,-2)$, i.e. $(9,9,5)$ and $(-3,1,13)$. $\vec{OA}\cdot\vec{OB}=-27+9+65=47$. Option (B).