JEE Mains · Maths · STD 12 - 11. three dimension geometry
The shortest distance between lines \(\mathrm{L}_1\) and \(\mathrm{L}_2\), where \(L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}\) and \(L_2\) is the line passing through the points \(\mathrm{A}(-4,4,3) \cdot \mathrm{B}(-1,6,3)\) and perpendicular to the line \(\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}\), is
- A \(\frac{121}{\sqrt{221}}\)
- B \(\frac{24}{\sqrt{117}}\)
- C \(\frac{141}{\sqrt{221}}\)
- D \(\frac{42}{\sqrt{117}}\)
Answer & Solution
Correct Answer
(C) \(\frac{141}{\sqrt{221}}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{L}_2=\frac{\mathrm{x}+4}{3}=\frac{\mathrm{y}-4}{2}=\frac{\mathrm{z}-3}{0} \\ & \therefore \mathrm{S} . \mathrm{D}=\frac{\left|\begin{array}{ccc}\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \mid \\ 2 & -3 & 2 \\ 3 & 2…
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