JEE Mains · Maths · STD 12 - 11. three dimension geometry
The shortest distance between the lines \(\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}\) and \(x+y+z+1=0\), \(2 x-y+z+3=0\) is
- A \(\frac{1}{2}\)
- B \(1\)
- C \(\frac{1}{\sqrt{2}}\)
- D \(\frac{1}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
Line of intersection of planes \(x+y+z+1=0\) \(2 x-y+z+3=0\) eliminate y \(3 x+2 z+4=0\) \(x=\frac{-2 z-4}{3}\) put in equaiton (1) \(z=-3 y+1\) from (3) and (4) \(\frac{3 x+4}{-2}=-3 y+1=z\)…
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