JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(S\) be the set of all values of \(\lambda\), for which the shortest distance between the lines \(\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}\) and \(\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}\) is \(13\) Then \(8\left|\sum_{\lambda \in S} \lambda\right|\) is equal to
- A \(304\)
- B \(308\)
- C \(306\)
- D \(302\)
Answer & Solution
Correct Answer
(C) \(306\)
Step-by-step Solution
Detailed explanation
Shor test distance \(=\frac{\left|\begin{array}{ccc}0 & 4 & 1 \\ 3 & -4 & 0 \\ 2 \lambda & 3 & -12\end{array}\right|}{\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 0 & 4 & 1 \\ 3 & -4 & 0\end{array}\right|}\)…
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