JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the shortest distance between the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}\) is \(\frac{5}{\sqrt{6}}\), then the sum of all possible values of \(\alpha\) is
- A \(\frac{3}{2}\)
- B \(-\frac{3}{2}\)
- C 3
- D -3
Answer & Solution
Correct Answer
(D) -3
Step-by-step Solution
Detailed explanation
\begin{aligned} & L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \\ & L_1: \frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1} \\ & \vec{x}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & \alpha & 1\end{array}\right|=\hat{\mathrm{i}}(3-4…
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