Q. Shortest distance between the lines $\frac{x - 1}{1} = \frac{y - 1}{1} = \frac{\mathcal{z} - 1}{1}$ and $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{\mathcal{z} - 4}{1}$ is equal to
A) $\sqrt{14}$
B) $\sqrt{7}$
C) $\sqrt{2}$
D) None of these
Explanation
(c)
Since the given lines are parallel
[image:12_image6.emf]
From the figure, we get
$BC = \frac{(2 - 1)1}{\sqrt{3}} + \frac{(3 - 1)1}{\sqrt{3}} + \frac{(4 - 1)1}{\sqrt{3}} = \frac{1 + 2 + 3}{\sqrt{3}} = 2\sqrt{3}$
$AB = \sqrt{1 + 4 + 9} = \sqrt{14}$
Shortest distance $= AC = \sqrt{14 - 12} = \sqrt{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.