JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the shortest between the lines \(\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}\) and \(\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{z+2 \sqrt{6}}{5}\) is \(6\), then the square of sum of all possible values of \(\lambda\) is
- A \(380\)
- B \(3885\)
- C \(386\)
- D \(384\)
Answer & Solution
Correct Answer
(D) \(384\)
Step-by-step Solution
Detailed explanation
Shortest distance between the lines \(\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}\) \(\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{2+2 \sqrt{6}}{5}\) is \(6\) Vector along line of shortest distance…
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