Question:
The shortest distance between the lines \(\frac{x + 3}{1} = \frac{y-2}{2} = \frac{z+4}{3} \space and \space \frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}\) is: =
The shortest distance between the lines \(\frac{x + 3}{1} = \frac{y-2}{2} = \frac{z+4}{3} \space and \space \frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}\) is: =
Updated On: Oct 10, 2024
- \(\frac{197}{\sqrt{179}}\)
- \(\frac{197}{\sqrt{191}}\)
- \(\frac{197}{\sqrt{189}}\)
- \(\frac{197}{\sqrt{237}}\)
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The Correct Option is D
Solution and Explanation
The correct option is (D):
\(\frac{197}{\sqrt{237}}\)
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