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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2019 · Mathematics · Three Dimensional Geometry

The distance between the parallel planes 4 x-2 y+4 z+9=0 and 8 x-4...

The distance between the parallel planes 4x−2y+4z+9=04 x-2 y+4 z+9=0 and 8x−4y+8z+21=08 x-4 y+8 z+21=0 is

Options

  1. A14\frac{1}{4}
  2. B12\frac{1}{2}
  3. C32\frac{3}{2}
  4. D74\frac{7}{4}

Correct answer

A. 14\frac{1}{4}

Step-by-step solution

Given planes : 4x−2y+4z+9=0⇒8x−4y+8z+18=04 x-2 y+4 z+9=0\Rightarrow 8 x-4 y+8 z+18=0...(1) 8x−4y+8z+21=08 x-4 y+8 z+21=0...(2) Distance =∣d1−d2∣a2+b2+c2=∣18−21∣64+16+64=3144=312=14=\frac{\left|d_1-d_2\right|}{\sqrt{a^2+b^2+c^2}}=\frac{|18-21|}{\sqrt{64+16+64}}=\frac{3}{\sqrt{144}}=\frac{3}{12}=\frac{1}{4}
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