Question

Find the distance between the parallel planes $\overline r \cdot \left( {2\overline i - 3\overline j + 6\overline k } \right) = 5$ and $\overline r \cdot \left( {6\overline i - 9\overline j + 18\overline k } \right) + 20 = 0$.

Answer 1

First we will rewrite the equations in the general forms to find the distance between two parallel planes $\overline r \cdot \left( {a\overline i + b\overline j + c\overline k } \right) + {d_1} = 0$ and $\overline r \cdot \left( {a\overline i + b\overline j + c\overline k } \right) + {d_2} = 0$, using the formula, $\dfrac{{\left| {{d_1} - {d_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$.

Complete step-by-step answer:
We are given that the planes $\overline r \cdot \left( {2\overline i - 3\overline j + 6\overline k } \right) = 5$ and $\overline r \cdot \left( {6\overline i - 9\overline j + 18\overline k } \right) + 20 = 0$ are parallel.
Rewriting the equation of the plane $\overline r \cdot \left( {6\overline i - 9\overline j + 18\overline k } \right) + 20 = 0$ by dividing both sides by 3, we get

$$\Rightarrow \dfrac{{\overline r \cdot \left( {6\overline i - 9\overline j + 18\overline k } \right)}}{3} + \dfrac{{20}}{3} = 0 \\\ \Rightarrow \overline r \cdot \left( {2\overline i - 3\overline j + 6\overline k } \right) + \dfrac{{20}}{3} = 0{\text{ .......(1)}} \\\$$

Subtracting the equation $\overline r \cdot \left( {2\overline i - 3\overline j + 6\overline k } \right) = 5$ by 5 on both sides, we get

$$\Rightarrow \overline r \cdot \left( {2\overline i - 3\overline j + 6\overline k } \right) - 5 = 5 - 5 \\\ \Rightarrow \overline r \cdot \left( {2\overline i - 3\overline j + 6\overline k } \right) - 5 = 0{\text{ ......(2)}} \\\$$

We know that the distance between two parallel planes $\overline r \cdot \left( {a\overline i + b\overline j + c\overline k } \right) + {d_1} = 0$ and $\overline r \cdot \left( {a\overline i + b\overline j + c\overline k } \right) + {d_2} = 0$ is calculated using the formula, $\dfrac{{\left| {{d_1} - {d_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$.
Finding the values of a,b ,c ,$d_1$ , and $d_2$ from comparing the equations (1) and (2) to the above equations, we get
$a = 2$
$b = - 3$
$c = 6$
${d_1} = \dfrac{{20}}{3}$
${d_2} = - 5$

Substituting these values in the above formula of distance, we get

$$\Rightarrow \dfrac{{\left| {\dfrac{{20}}{3} - \left( { - 5} \right)} \right|}}{{\sqrt {{2^2} + {{\left( { - 3} \right)}^2} + {6^2}} }} \\\ \Rightarrow \dfrac{{\left| {\dfrac{{20 + 15}}{3}} \right|}}{{\sqrt {4 + 9 + 36} }} \\\ \Rightarrow \dfrac{{35}}{{3\sqrt {99} }} \\\ \Rightarrow \dfrac{{35}}{{9\sqrt {11} }} \\\$$

Thus, the distance is $\dfrac{{35}}{{9\sqrt {11} }}$.

Note: In solving these types of questions, students need to know that the distance between two planes implies the perpendicular distance, and the perpendicular distance is also the shortest distance. If the perpendicular distance is given between two planes then the plane must be parallel to each other as this distance always measures in two planes, which are parallel.