Question

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

Answer 1

\[ \text{ Multiplying the first equation of the plane by 3, we get } \]
\[6x - 3y + 9z - 12 = 0 \]
\[6x - 3y + 9z = 12 . . . \left( 1 \right)\]
\[ \text{ The second equation of the plane is } \]
\[6x - 3y + 9z = - 13 . . . \left( 2 \right)\]
\[ \text{ We know that the distance between two planes ax + by }+ cz = d_{ 1} \text{ and } ax + by + cz = d_{2 } \text{ is } \frac{\left| d_{2 }- d_1 \right|}{\sqrt{a^2 + b^2 + c^2}}\]
\[\text{ So, the required distance } \]
\[=\frac{\left| - 13 - 12 \right|}{\sqrt{6^2 + \left( - 3 \right)^2 + 9^2}}\]
\[=\frac{\left| - 25 \right|}{\sqrt{36 + 9 + 81}}\]
\[ = \frac{25}{\sqrt{126}}\]
\[ = \frac{5}{3 \sqrt{14}} \text{ units } \]