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NDA 2010
Exam
NDA
Year
2010
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2010 · Mathematics · Three Dimensional Geometry

The planes p x+2 y+2 z-3=0 and 2 x-y+z+2=0 intersect at an angle π...

The planes px+2y+2z−3=0p x+2 y+2 z-3=0 and 2x−y+z+2=02 x-y+z+2=0 intersect at an angle π/4\pi / 4. What is the value of p2p^2 ?

Options

  1. A24
  2. B12
  3. C6
  4. D3

Correct answer

A. 24

Step-by-step solution

cos⁡π/4=2p−2+2p2+4+4×4+1+1⇒12=2pp2+8×6or 12=4p26p2+48or 6p2+48=8p2or p2=24\begin{aligned} & \cos \pi / 4=\frac{2 p-2+2}{\sqrt{p^2+4+4} \times \sqrt{4+1+1}} \\ & \Rightarrow \frac{1}{\sqrt{2}}=\frac{2 p}{\sqrt{p^2+8} \times \sqrt{6}} \\ & \text{or } \frac{1}{2}=\frac{4 p^2}{6 p^2+48} \\ & \text{or } 6 p^2+48=8 p^2 \\ & \text{or } p^2=24\end{aligned}
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