JEE Mains · Maths · STD 12 - 11. three dimension geometry
The shortest distance between the lines \(\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}\) and \(\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}\) is
- A \(7 \sqrt{3}\)
- B \(5 \sqrt{3}\)
- C \(6 \sqrt{3}\)
- D \(4 \sqrt{3}\)
Answer & Solution
Correct Answer
(C) \(6 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
Shortest distance between two lines \(\frac{x-x_1}{a_1}=\frac{y-y_1}{a_2}=\frac{z-z_1}{a_3}\) \(\frac{x-x_2}{b_1}=\frac{y-y_2}{b_2}=\frac{z-z_2}{b_3}\) is given as…
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