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Question
The distance between the planes 2x + 2y − z + 2 = 0 and 4x + 4y − 2z + 5 = 0 is
Options
\[\frac{1}{2}\]
\[\frac{1}{4}\]
\[\frac{1}{6}\]
None of these
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Solution
\[\frac{1}{6}\]
\[\text{ Multiplying the first equation of the plane by } \]
\[4x + 4y - 2z + 4 = 0 \]
\[4x + 4y - 2z = - 4 . . . \left( 1 \right)\]
\[\text{ The second equation of the plane is } \]
\[4x + 4y - 2z + 5 = 0\]
\[4x + 4y - 2z = - 5 . . . \left( 2 \right)\]
\[\text{ We know that the distance between two planes } ax + by + cz = d_1 \text{ and }ax + by + cz = d_2 \text{ is }\frac{\left| d_2 - d_1 \right|}{\sqrt{a^2 + b^2 + c^2}}\]
\[\text{ So, the required distance } \]
\[=\frac{\left| - 5 + 4 \right|}{\sqrt{4^2 + 4^2 + \left( - 2 \right)^2}}\]
\[=\frac{\left| - 1 \right|}{\sqrt{16 + 16 + 4}}\]
\[ = \frac{1}{\sqrt{36}}\]
\[ = \frac{1}{6} \text{ units } \]
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