The shortest distance between the lines
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$
$\frac{x-3}{4}=\frac{y-3}{6}=\frac{z+5}{12}$ is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{\sqrt{293}}{7}$
$\vec{a_1}=\hat i+2\hat j-4\hat k$
$\vec{a_2}=3\hat i+3\hat j-5\hat k$
($L_2$ is parallel to as $4\hat i+6\hat j+12\hat k$ ⇒ $l_2$ parallel to $2\hat i+3\hat j+6\hat k$)
so $\vec b=2\hat i+3\hat j+6\hat k$
so distance = $\frac{\vec b×(\vec{a_2}-\vec{a_1})}{|\vec b|}$
$=\frac{|(2\hat i+3\hat j+6\hat k)×(2\hat i+\hat j-\hat k)|}{7}$
$=\frac{|-9\hat i+14\hat j-4\hat k|}{7}$
$=\frac{\sqrt{293}}{7}$ units