ExamBro
ExamBro
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

  1. A 8
  2. B -6
  3. C 6
  4. D -8
Verified Solution

Answer & Solution

Correct Answer

(B) -6

Step-by-step Solution

Detailed explanation

\(\sqrt{2}=\frac{\left|\begin{array}{ccc}-1 & 1 & 3 \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2 \alpha\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & j & k \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2 \alpha\end{array}\right|}\)…