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JEE Main 20238 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2023 · 8 April · Evening Shift · Mathematics · Three Dimensional Geometry

Let P be the plane passing through the line (x - 1)/(1) = (y -...

Let PP be the plane passing through the line x−11=y−2−3=z+57\dfrac{x - 1}{1} = \dfrac{y - 2}{- 3} = \dfrac{z + 5}{7} and the point (2,4,−3)( 2, 4, - 3 ). If the image of the point (−1,3,4)( - 1, 3, 4 ) in the plane PP is (α,β,γ)( \alpha, \beta, \gamma ), then α+β+γ\alpha + \beta + \gamma is equal to

Options

  1. A1010
  2. B1212
  3. C99
  4. D1111

Correct answer

A. 1010

Step-by-step solution

Given,

PP be the plane passing through the line x−11=y−2−3=z+57\dfrac{x - 1}{1} = \dfrac{y - 2}{- 3} = \dfrac{z + 5}{7} and the point (2,4,−3)( 2, 4, - 3 ),

So, equation of plane is given by,

\left|{x−2y−4z+31−372−14−2−3−(−5)\begin{cases}x - 2 & y - 4 & z + 3 \\ 1 & - 3 & 7 \\ 2 - 1 & 4 - 2 & - 3 - \left(- 5\right)\end{cases}\right| = 0</p><p></p><p>⇒ \left|{x−2y−4z+31−37122\begin{cases}x - 2 & y - 4 & z + 3 \\ 1 & - 3 & 7 \\ 1 & 2 & 2\end{cases}\right| = 0</p><p></p><p>⇒ \left(x - 2\right)\left(- 20\right) - \left(y - 4\right)\left(- 5\right) + \left(z + 3\right)\left(5\right) = 0</p><p></p><p>⇒ 4x - y - z - 7 = 0</p><p>Nowfindingtheimage(α,β,γ)</p><p>Now finding the image\left(\alpha, \beta, \gamma\right) of the point (−1,3,4)\left(- 1, 3, 4\right) with respect to plane 4x−y−z−7=04x - y - z - 7 = 0 we get,

x+14=y−3−1=z−4−1=−2(−4−3−4−716+1+1)=2\dfrac{x + 1}{4} = \dfrac{y - 3}{- 1} = \dfrac{z - 4}{- 1} = - 2\left(\dfrac{- 4 - 3 - 4 - 7}{16 + 1 + 1}\right) = 2

⇒x=7=α,y=1=β,z=2=γ⇒ x = 7 = \alpha, y = 1 = \beta, z = 2 = \gamma

Hence, α+β+γ=10\alpha + \beta + \gamma = 10

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