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

वास्तविक संख्याओं \(\alpha\) तथा \(\beta \neq 0\) के लिए, यदि सरल रेखाओं \(\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3}\) तथ \(\frac{ x -4}{\beta}=\frac{ y -6}{3}=\frac{ z -7}{3}\), का प्रतिच्छेदन बिंदु, समतल \(x +2 y - z =8\) पर है, तो \(\alpha-\beta\) बराबर है 

  1. A \(5\)
  2. B \(3\)
  3. C \(7\)
  4. D \(9\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(7\)

Step-by-step Solution

Detailed explanation

First line is \((\phi+\alpha, 2 \phi+1,3 \phi+1)\) and second line is \((q \beta+4,3 q+6,3 q+7)\) For intersection \(\phi+\alpha=q \beta+4 \ldots \text { (i) }\) \(2 \phi+1=3 q+6 . \text { (i) }\) \(3 \phi+1=3 q+7 \text {. (iii) }\) for \((ii)\, \& \,(iii)\)…
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