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Question
Write the distance between the parallel planes 2x − y + 3z = 4 and 2x − y + 3z = 18.
Solution
\[\text{ The given equations are } \]
\[2x - y + 3z = 4 . . . \left( 1 \right)\]
\[\text{ The second equation of the plane is} \]
\[2x - y + 3z = 18 . . . \left( 2 \right)\]
\[\text{ We know that the distance between two planes } ax + by + cz = d_1 \text{ and } ax + by + cz = d_2 \text{ is} \frac{\left| d_1 - d_2 \right|}{\sqrt{a^2 + b^2 + c^2}}\]
\[\text{ So, the required distance is } \]
\[ \frac{\left| 18 - 4 \right|}{\sqrt{2^2 + \left( - 1 \right)^2 + 3^2}}\]
\[=\frac{\left| 14 \right|}{\sqrt{4 + 1 + 9}}\]
\[ = \frac{14}{\sqrt{14}}\]
\[ = \sqrt{14} \text{ units } \]
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