JEE Mains · Maths · STD 12 - 11. three dimension geometry
If \(\left(2\alpha+1, \alpha^2-3\alpha, \dfrac{\alpha-1}{2}\right)\) is the image of \((\alpha, 2\alpha, 1)\) in the line \(\dfrac{x-2}{3}=\dfrac{y-1}{2}=\dfrac{z}{1}\), then the possible value(s) of \(\alpha\) is (are)
- A Only \(3\)
- B Only \(3\) and \(-1\)
- C Only \(3\), \(\dfrac{1}{4}\) and \(-1\)
- D Only \(3\) and \(\dfrac{1}{4}\)
Answer & Solution
Correct Answer
(A) Only \(3\)
Step-by-step Solution
Detailed explanation
Let the given point be \(P = (\alpha, 2\alpha, 1)\) and its image be \(P' = \left(2\alpha+1, \alpha^2-3\alpha, \dfrac{\alpha-1}{2}\right)\). For \(P'\) to be the image of \(P\) in the given line \(L: \dfrac{x-2}{3}=\dfrac{y-1}{2}=\dfrac{z}{1}\), two conditions must be satisfied:…
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