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NDA 2012
Exam
NDA
Year
2012
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2012 · Mathematics · Three Dimensional Geometry

What is the equation to the plane through (1,2,3) parallel to 3 x+4...

What is the equation to the plane through (1,2,3)(1,2,3) parallel to 3x+4y−5z=03 x+4 y-5 z=0 ?

Options

  1. A3x+4y+5z+4=03 x+4 y+5 z+4=0
  2. B3x+4y−5z+14=03 x+4 y-5 z+14=0
  3. C3x+4y−5z+4=03 x+4 y-5 z+4=0
  4. D3x+4y−5z−4=03 x+4 y-5 z-4=0

Correct answer

C. 3x+4y−5z+4=03 x+4 y-5 z+4=0

Step-by-step solution

Let equation of plane parallel to 3x+4y−5z=03 x+4 y-5 z=0 is 3x+4y−5z+λ=0∵3 x+4 y-5 z+\lambda=0\because it passes through (1,2,3)∴3(1)+4(2)−5(3)+λ=0(1,2,3)\therefore 3(1)+4(2)-5(3)+\lambda=0⇒3+8−15+λ=0⇒λ=4\begin{aligned} \Rightarrow & & 3+8-15+\lambda & =0 \\ \Rightarrow & & \lambda & =4\end{aligned}∴\therefore equation of plane is 3x+4y−5z+4=03 x+4 y-5 z+4=0.
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