JEE Mains · Maths · STD 12 - 11. three dimension geometry
Line \(L_1\) passes through the point \((1,2,3)\) and is parallel to Z -axis. Line \(\mathrm{L}_2\) passes through the point \((\lambda, 5,6)\) and is parallel to \(y\)-axis. Let for \(\lambda=\lambda_1, \lambda_2, \lambda_2 \lt \lambda_1\), the shortest distance between the two lines be 3 . Then the square of the distance of the point \(\left(\lambda_1, \lambda_2, 7\right)\) from the line \(\mathrm{L}_1\) is
- A \(40\)
- B \(32\)
- C \(25\)
- D \(37\)
Answer & Solution
Correct Answer
(C) \(25\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & L_1 \equiv \frac{x-1}{0}=\frac{y-2}{0}=\frac{z-3}{1} \\ & L_2 \equiv \frac{x-\lambda}{0}=\frac{y-5}{1}=\frac{z-6}{0}\end{aligned}\)…
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