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JEE Main 202431 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
31 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2024 · 31 January · Evening Shift · Mathematics · Three Dimensional Geometry

Let (α, β, γ) be mirror image of the point (2, 3, 5) in the line (x...

Let (α,β,γ)\left(\alpha, \beta, \gamma\right) be mirror image of the point (2,3,5)\left(2, 3, 5\right) in the line x−12=y−23=z−34\dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{4}. Then 2α+3β+4γ2\alpha + 3\beta + 4\gamma is equal toLet (α, β, γ) be mirror image of the point (2, 3, 5) in the line (x - 1)/(2) = (y - 2)/(3) = (z - 3)/(4) . Then 2α + 3β

Options

  1. A3232
  2. B3333
  3. C3131
  4. D3434

Correct answer

B. 3333

Step-by-step solution

Given: x−12=y−23=z−34\dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{4}

Let, P≡(2,3,5)P ≡ \left(2, 3, 5\right) and R≡(α,β,γ)R ≡ \left(\alpha, \beta, \gamma\right).

Let (α, β, γ) be mirror image of the point (2, 3, 5) in the line (x - 1)/(2) = (y - 2)/(3) = (z - 3)/(4) . Then 2α + 3β

Now, direction ratio of PR=α−2,β−3,γ−5PR = \alpha - 2, \beta - 3, \gamma - 5

And the line x−12=y−23=z−34\dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{4} will be perpendicular,

So, using perpendicular condition we get,

⇒(α−2)×2+(β−3)×3+(γ−5)×4=0⇒ \left(\alpha - 2\right) \times 2 + \left(\beta - 3\right) \times 3 + \left(\gamma - 5\right) \times 4 = 0

⇒2α−4+3β−9+4γ−20=0⇒ 2\alpha - 4 + 3\beta - 9 + 4\gamma - 20 = 0

⇒∣2α+3β+4γ∣=33⇒ \left|2\alpha + 3\beta + 4\gamma\right| = 33

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