3D Geometry
Ratio of Shortest Distances
nta_pyq_2024_jan
Grade 12
Question:
If $d_1$ is the shortest distance between the lines $x+1=2y=-12z$, $x=y+2=6z-6$ and $d_2$ is the shortest distance between the lines $\dfrac{x-1}{2}=\dfrac{y+8}{-7}=\dfrac{z-4}{5}$, $\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-6}{-3}$, then the value of $\dfrac{32\sqrt{3}\,d_1}{d_2}$ is:
Step-by-Step Solution
Key Concept: Convert lines to standard form and apply SD formula for each pair. The second pair of lines are parallel (same direction $(2,1,-3)$... check: $(2,-7,5)$ vs $(2,1,-3)$ — not parallel). Compute $d_1$ and $d_2$ separately.
$32\sqrt3\cdot d_1/d_2=16$.
Correct Answer: 16