JEE Mains · Maths · STD 12 - 11. three dimension geometry
The sum of all values of \( \alpha \), for which the shortest distance between the lines \( \frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha} \) and \( \frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2\alpha} \) is \( \sqrt{2} \), is
- A 8
- B -6
- C 6
- D -8
Answer & Solution
Correct Answer
(B) -6
Step-by-step Solution
Detailed explanation
\(\sqrt{2}=\frac{\left|\begin{array}{ccc}-1 & 1 & 3 \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2 \alpha\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & j & k \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2 \alpha\end{array}\right|}\)…
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