Find the distance between the parallel planes 2x − y + 3z − 4 = 0 and 6x − 3y + 9z + 13 = 0.

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Solution

$Multiplying the first equation of the plane by 3, we get 6 x - 3 y + 9 z - 12 = 0 6 x - 3 y + 9 z = 12 . . . (1) The second equation of the plane is 6 x - 3 y + 9 z = - 13 . . . (2) We know that the distance between two planes a x + b y + c z = d_{1} and a x + b y + c z = d_{2} is \frac{|d_{2} - d_{1}|}{\sqrt{a^{2} + b^{2} + c^{2}}} So, the required distance = \frac{|- 13 - 12|}{\sqrt{6^{2} + {(- 3)}^{2} + 9^{2}}} = \frac{|- 25|}{\sqrt{36 + 9 + 81}} = \frac{25}{\sqrt{126}} = \frac{5}{3 \sqrt{14}} units$

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