The shortest distance between the lines $\frac{x-2}{3} = \frac{y+1}{2} = \frac{z-6}{2}$ and $\frac{x-6}{3} = \frac{1-y}{2} = \frac{z+8}{0}$ is equal to _______.
Step-by-Step Solution
Key Concept: Use the shortest distance formula for skew lines: $SD = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$.
$(\vec{a_2}-\vec{a_1}) = (4,2,-14)$. $\vec{b_1}\times\vec{b_2} = (-4\hat{i}+6\hat{j}-12\hat{k})$, $|\vec{b_1}\times\vec{b_2}|=14$. $SD = \frac{|(-16+12+168)|}{14} = \frac{196}{14} = 14$. Answer: 14
Correct Answer: 14