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JEE Mains · Maths · STD 12 - 11. three dimension geometry

रेखा \(l_1\) बिंन्दु \((2,6,2)\) से होकर जाती है तथा समतल \(2 \mathrm{x}+\mathrm{y}-2 \mathrm{z}=10\) पर लंबवत है, तो \(\mathrm{l}_1\) तथा रेखा \(\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}\) के बीच न्यूनतम दूरी है :

  1. A \(7\)
  2. B \(\frac{19}{3}\)
  3. C \(\frac{19}{3}\)
  4. D \(9\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(9\)

Step-by-step Solution

Detailed explanation

Line \(\ell\), is given by \(L_1: \frac{x-2}{2}=\frac{y-6}{1}=\frac{z-2}{-2}\) Given, \(L _2: \frac{ x +1}{2}=\frac{ y +4}{-3}=\frac{ z }{2}\) \(\text { Shortest distance }=\left|\frac{\overline{ AB } \cdot \overline{ MN }}{ MN }\right|\)…
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