3D Geometry
Point on Line at Distance — Shortest Distance to Another Line
nta_pyq_2026_jan
Grade None
Question:
Let $P(\alpha,\beta,\gamma)$ be the point on the line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=z$ at a distance $4\sqrt{14}$ from the point $(1,-1,0)$ and nearer to the origin. Then the shortest distance between the lines $\dfrac{x-\alpha}{1}=\dfrac{y-\beta}{2}=\dfrac{z-\gamma}{3}$ and $\dfrac{x+5}{2}=\dfrac{y-10}{1}=\dfrac{z-3}{1}$, is equal to
$4\sqrt{\dfrac{7}{5}}$
$2\sqrt{\dfrac{7}{4}}$
$7\sqrt{\dfrac{5}{4}}$
$4\sqrt{\dfrac{5}{7}}$
Step-by-Step Solution
Key Concept: Point on line: $(2\lambda+1,3\lambda-1,\lambda)$. Distance from $(1,-1,0)$: $\sqrt{4\lambda^2+9\lambda^2+\lambda^2}\cdot|\lambda|=4\sqrt{14}\Rightarrow\lambda=\pm4\Rightarrow\lambda=-4$ (nearer origin). $P=(-7,11,-4)$.
SD $=4\sqrt{\dfrac{7}{5}}$.
Correct Answer: 1