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

ધારોકે રેખા \(\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}\) તથા બિંદુ \((2,4,-3)\) માંથી પસાર થતો સમતલ \(P\) છે. જો બિંદુ \((-1,3,4)\) નું સમતલ \(P\) માં પ્રતિબિંબે \((\alpha, \beta, \gamma)\) હોય, તો \(\alpha+\beta+\gamma=.......\)

  1. A \(12\)
  2. B \(11\)
  3. C \(9\)
  4. D \(10\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(10\)

Step-by-step Solution

Detailed explanation

Equation of plane is given by \(\left|\begin{array}{ccc} x -1 & y -2 & z +5 \\ 1 & 2 & 2 \\ 1 & -3 & 7\end{array}\right|=0\) \(4 x-y-z=7\) \(\frac{\alpha+1}{4}=\frac{\beta-3}{-1}=\frac{\gamma-4}{-1}=\frac{-2(-4-3-4-7)}{16+1+1}=2\) \(\alpha=7, \beta=1, \gamma=2\)…
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