The shortest distance between lines
\( L_{1}:\ \displaystyle\frac{x-1}{2}=\frac{2-y}{-3}=\frac{z-3}{4} \) and \( L_{2}:\ \displaystyle\frac{x-2}{3}=\frac{y-4}{4}=\frac{5-z}{-5} \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{1}{\sqrt{6}} \)
In standard form, the numerators must have +y and +z:
\( \displaystyle \frac{2-y}{-3}=\frac{y-2}{3} \)
\( \displaystyle \frac{5-z}{-5}=\frac{z-5}{5} \)
So
\( \displaystyle \vec{a}_{1}=(1,2,3),\ \vec{b}_{1}=(2,3,4) \)
\( \displaystyle \vec{a}_{2}=(2,4,5),\ \vec{b}_{2}=(3,4,5) \)
For the shortest distance between skew lines:
\( \displaystyle \vec{b}_{1}\times\vec{b}_{2}=(-1,2,-1) \)
\( \displaystyle |\vec{b}_{1}\times\vec{b}_{2}|=\sqrt{6} \)
\( \displaystyle \vec{a}_{2}-\vec{a}_{1}=(1,2,2) \)
\( \displaystyle d=\frac{|(1)(-1)+(2)(2)+(2)(-1)|}{\sqrt{6}} \)
\( \displaystyle =\frac{1}{\sqrt{6}} \)