JEE Main 2020 (Online) 3rd September Evening ShiftQuestion 1
Let a plane $P$ contain two lines $\vec r=\hat i+\lambda(\hat i+\hat j)$, $\lambda\in\mathbf{R}$ and $\vec r=-\hat j+\mu(\hat j-\hat k)$, $\mu\in\mathbf{R}$. If $Q(\alpha,\beta,\gamma)$ is the foot of the perpendicular drawn from the point $M(1,0,1)$ to $P$, then $3(\alpha+\beta+\gamma)$ equals ________.
3D Geometry
Answer$5$
Solution
Direction vectors $(1,1,0)$ and $(0,1,-1)$; normal $=(1,1,0)\times(0,1,-1)=(-1,1,1)$. Point $(1,0,0)$ lies on the plane: $-x+y+z=-1$. (The second line's point $(0,-1,0)$ gives $-1$ also, so the lines are coplanar.) At $M(1,0,1)$: $-x+y+z+1=1$. Foot $Q=M-\dfrac{1}{3}(-1,1,1)=(\tfrac43,-\tfrac13,\tfrac23)$. $3(\alpha+\beta+\gamma)=3\cdot\tfrac53=5$.