JEE Main 2022 (Online) 27th July Evening ShiftQuestion 1
If the length of the perpendicular drawn from the point $P(a,4,2)$, $a>0$ on the line $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$ is $2\sqrt6$ units and $Q(\alpha_1,\alpha_2,\alpha_3)$ is the image of the point $P$ in this line, then $a+\sum_{i=1}^{3}\alpha_i$ is equal to :
3D Geometry
Solution
Line: point $B(-1,3,1)$, direction $d=(2,3,-1)$, $|d|^2=14$. $\vec{BP}=(a+1,1,1)$. $\vec{BP}\times d=(-1-3,\,2+a+1,\,3a+3-2)=(-4,a+3,3a+1)$. Perpendicular$^2=\frac{16+(a+3)^2+(3a+1)^2}{14}=24\Rightarrow10a^2+12a+26=336\Rightarrow5a^2+6a-155=0\Rightarrow a=5$ ($a>0$). $P=(5,4,2)$: $\vec{BP}\cdot d=12+3-1=14$, foot $=B+d=(1,6,0)$. Image $Q=2F-P=(-3,8,-2)$, sum $=3$. $a+\sum\alpha_i=5+3=8$. Option (B).