JEE Mains · Maths · STD 12 - 11. three dimension geometry
The shortest distance between the lines \(x+1=2 y=-\) \(12 z\) and \(x=y+2=6 z-6\) is
- A \(2\)
- B \(3\)
- C \(\frac{5}{2}\)
- D \(\frac{3}{2}\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(\frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}}\) and \(\frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}}\) \(\Rightarrow\) Shortest distance…
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