JEE Mains · Maths · STD 12 - 11. three dimension geometry
For real numbers \(\alpha\) and \(\beta \neq 0\), if the point of intersection of the straight lines \(\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3}\) and \(\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3}\) lies on the plane \(x+2 y-z=8\), then \(\alpha-\beta\) is equal to:
- A \(5\)
- B \(3\)
- C \(7\)
- D \(9\)
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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