JEE Main 2023 (Online) 8th April Morning ShiftQuestion 1
Let $\lambda_1,\lambda_2$ be the values of $\lambda$ for which the points $\left(\frac{5}{2},1,\lambda\right)$ and $(-2,0,1)$ are at equal distance from the plane $2x+3y-6z+7=0$. If $\lambda_1>\lambda_2$, then the distance of the point $(\lambda_1-\lambda_2,\lambda_2,\lambda_1)$ from the line $\frac{x-5}{1}=\frac{y-1}{2}=\frac{z+7}{2}$ is $\_\_\_\_\_\_$ .
3D Geometry
Answer$9$
Solution
Distances: $\frac{|5+3-6\lambda+7|}{7}=\frac{|-4-6+7|}{7}$, so $|15-6\lambda|=3\Rightarrow\lambda=2,3$. Thus $\lambda_1=3$, $\lambda_2=2$. Point $P=(1,2,3)$. Line through $A(5,1,-7)$ with direction $(1,2,2)$, $|d|=3$. $\vec{AP}=(-4,1,10)$, $\vec{AP}\times(1,2,2)=(2-20,\,10+8,\,-8-1)=(-18,18,-9)$, norm $27$. Distance $=\frac{27}{3}=9$.