JEE Mains · Maths · STD 12 - 11. three dimension geometry
The shortest distance between the lines \(\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}\) and \(\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}\) is
- A \(3 \sqrt{6}\)
- B \(6 \sqrt{3}\)
- C \(6 \sqrt{2}\)
- D \(2 \sqrt{6}\)
Answer & Solution
Correct Answer
(A) \(3 \sqrt{6}\)
Step-by-step Solution
Detailed explanation
\(S _{ d }=\left|\frac{(\overrightarrow{ a }-\overrightarrow{ b }) \times\left(\overrightarrow{ n }_1 \times \overrightarrow{ n }_2\right)}{\left|\overrightarrow{ n }_1 \times \overrightarrow{ n }_2\right|}\right|\) \(\overline{ a }=(4,-2,-3)\) \(\overline{ b }=(1,3,4)\)…
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