3D Geometry 3 Question 40

####40. Distance between two parallel planes $2 x+y+2 z=8$ and $4 x+2 y+4 z+5=0$ is

(a) $\frac{3}{2}$

(b) $\frac{5}{2}$

(c) $\frac{7}{2}$

(d) $\frac{9}{2}$

(2013 Main)

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Answer:

Correct Answer: 40. (c)

Solution:

  1. Given planes are

$2 x+y+2 z-8=0$ and $2 x+y+2 z+\frac{5}{2}=0$

$\therefore$ Distance between two planes

$ =\frac{\left|c_{1}-c_{2}\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}=\left|\frac{-8-\frac{5}{2}}{\sqrt{2^{2}+1^{2}+2^{2}}}\right|=\frac{21 / 2}{3}=\frac{7}{2} $



NCERT Chapter Video Solution

Dual Pane