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

रेखा, \(\frac{ x -3}{1}=\frac{ y -4}{2}=\frac{ z -5}{2}\) तथा समतल \(x + y + z =17\) के प्रतिच्छेदन बिन्दु की बिन्दु \((1,1,9)\) से दूरी है

  1. A \(2 \sqrt{19}\)
  2. B \(19 \sqrt{2}\)
  3. C \(38\)
  4. D \(\sqrt{38}\)
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Answer & Solution

Correct Answer

(D) \(\sqrt{38}\)

Step-by-step Solution

Detailed explanation

Let \(\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=t\) \(\Rightarrow x=3+t, y=2 t+4, z=2 t+5\) for point of intersection with \(x+y+z=17\) \(3+ t +2 t +4+2 t +5=17\) \(\Rightarrow \quad 5 t =5 \Rightarrow t =1\) \(\Rightarrow\) point of intersection is \((4,6,7)\) distance between…
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