JEE Main 2022 (Online) 29th July Morning ShiftQuestion 1
Let a line with direction ratios $a,-4a,-7$ be perpendicular to the lines with direction ratios $3,-1,2b$ and $b,a,-2$. If the point of intersection of the line $\frac{x+1}{a^2+b^2}=\frac{y-2}{a^2-b^2}=\frac z1$ and the plane $x-y+z=0$ is $(\alpha,\beta,\gamma)$, then $\alpha+\beta+\gamma$ is equal to ________.
3D Geometry
Answer$10$
Solution
Perpendicularity: $3a+4a-14b=0\Rightarrow a=2b$; and $ab-4a^2+14=0\Rightarrow 2b^2-16b^2+14=0\Rightarrow b^2=1,\ a^2=4$. Line: $\frac{x+1}{5}=\frac{y-2}{3}=\frac z1$, point $(5s-1,\,3s+2,\,s)$. Plane $x-y+z=0$: $5s-1-3s-2+s=0\Rightarrow s=1$, point $(4,5,1)$. $\alpha+\beta+\gamma=10$.