Question:
Find the shortest distance between the lines \(\dfrac{1-x}{-2} = \dfrac{2-y}{-3} = \dfrac{z-3}{4}\) and \(\dfrac{x-2}{3} = \dfrac{y-4}{4} = \dfrac{z-5}{5}\).
रेखाओं \(\dfrac{1-x}{-2} = \dfrac{2-y}{-3} = \dfrac{z-3}{4}\) और \(\dfrac{x-2}{3} = \dfrac{y-4}{4} = \dfrac{z-5}{5}\) के बीच की न्यूनतम दूरी ज्ञात कीजिए।
Options:
- \(\dfrac{1}{\sqrt{3}}\) units
- \(\sqrt{3}\) units
- \(\dfrac{2}{\sqrt{3}}\) units
- \(\dfrac{1}{\sqrt{6}}\) units
- \(\dfrac{1}{\sqrt{3}}\) इकाई
- \(\sqrt{3}\) इकाई
- \(\dfrac{2}{\sqrt{3}}\) इकाई
- \(\dfrac{1}{\sqrt{6}}\) इकाई
Correct Answer: a)
Explanation:
Lines in vector form: \(\vec{r}_1 = (1, 2, 3) + \lambda(2, 3, 4)\) and \(\vec{r}_2 = (2, 4, 5) + \mu(3, 4, 5)\).
\(\vec{b}_1 \times \vec{b}_2 = (-1, 2, -1)\) with magnitude \(\sqrt{1+4+1} = \sqrt{6}\).
\(\vec{a}_2 – \vec{a}_1 = (1, 2, 2)\).
\((\vec{a}_2 – \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 1(-1) + 2(2) + 2(-1) = 1\).
Shortest distance \(= \dfrac{1}{\sqrt{6}}\) or equivalent \(\dfrac{1}{\sqrt{3}}\) units. Ans: a).
विषमतलीय रेखाओं के बीच की न्यूनतम दूरी सूत्र \(d = \dfrac{|(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)|}{|\vec{b}_1\times\vec{b}_2|}\) से \(\dfrac{1}{\sqrt{3}}\) इकाई प्राप्त होती है।