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

रेखा \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\) तथा समतल \(x-y+z=16\) के प्रतिच्छेद बिंदु की, बिंदु \((1,0,2)\) से दूरी है

  1. A \(13\)
  2. B \(2\sqrt {14} \)
  3. C \(8\)
  4. D \(3\sqrt {21} \)
Verified Solution

Answer & Solution

Correct Answer

(A) \(13\)

Step-by-step Solution

Detailed explanation

Let \(\frac{{x - 2}}{3} = \frac{{y + 1}}{4} - \frac{{z - 2}}{{12}} = \lambda \) \( \Rightarrow x = 3\lambda + 2,y = 4\lambda - 1,z = 12\lambda + 2\) The coorodinates of any point on ve lies are given by \(\left( {3\lambda + 2,4\lambda - 1,12\lambda + 2} \right).\) The ponit of…
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