The shortest distance between the lines $\dfrac{x-3}{4}=\dfrac{y+7}{-11}=\dfrac{z-1}{5}$ and $\dfrac{x-5}{3}=\dfrac{y-9}{-6}=\dfrac{z+2}{1}$ is:
$\dfrac{178}{\sqrt{563}}$
$\dfrac{187}{\sqrt{563}}$
$\dfrac{185}{\sqrt{563}}$
$\dfrac{179}{\sqrt{563}}$
Step-by-Step Solution
Key Concept: $a_1=(3,-7,1)$, $b_1=(4,-11,5)$, $a_2=(5,9,-2)$, $b_2=(3,-6,1)$. $\overrightarrow{AB}=(2,16,-3)$. $\vec{n}=b_1\times b_2=(4,-11,5)\times(3,-6,1)=(-11+30,15-4,-24+33)=(19,11,9)$.
SD $=187/\sqrt{563}$.
Correct Answer: 2