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

જો \(x\, = a\), \(y\, = b\), \(z\, = c\) એ રેખિય સમીકરણો \(x+8y+ 7z\,= 0\) ; \(9x+ 2y+ 3z\, = 0\) ; \(x+y+z\, = 0\) નો ઉકેલ છે કે જેથી બિંદુ \((a, b, c)\) એ સમતલ \(x + 2y + z\, = 6\) પર આવલે છે તો \(2a + b + c\) મેળવો.

  1. A \(-1\)
  2. B \(0\)
  3. C \(1\)
  4. D \(2\)
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Answer & Solution

Correct Answer

(C) \(1\)

Step-by-step Solution

Detailed explanation

\(x+8 y+7 z =0 \) \(9 x+2 y+3 z =0 \) \(x+y+z =0\) \(x=\lambda \quad|y=6 \lambda| \quad z=-7 \lambda\) \(\begin{array}{*{20}{c}} {\boxed{x = \lambda }}&{\boxed{y = 6\lambda }}&{\boxed{z = - 7\lambda }} \end{array}\)…
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