SATHEE: 3D Geometry 3 Question 40

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)

Show Answer

Answer:

Correct Answer: 40. (c)

Solution:

Given planes are

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

$∴$ Distance between two planes

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

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