JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the shortest distance between the lines \(\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}\) and \(\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}\) is \(\frac{6}{\sqrt{5}}\), then the sum of all possible values of \(\lambda\) is :
- A \(5\)
- B \(8\)
- C \(7\)
- D \(10\)
Answer & Solution
Correct Answer
(B) \(8\)
Step-by-step Solution
Detailed explanation
\( \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3} \) \( \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}\) the shortest distance between the lines…
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