Question

Find the distance between the planes \[\vec{r} \cdot \left( \hat{i}  + 2 \hat{j}  + 3 \hat{k}  \right) + 7 = 0 \text{ and } \vec{r} \cdot \left( 2 \hat{i}  + 4 \hat{j}  + 6 \hat{k}  \right) + 7 = 0 .\]

 

Answer 1

\[ \text{ The given planes are }\]
\[ \vec{r} . \left( \hat{i} + 2 \hat{j} + 3 \hat{k}  \right) = - 7\]
\[ \Rightarrow x + 2y + 3z = - 7 \]
\[ \text{ Multiplying this equation of the plane by 2, we get } \]
\[2x + 4y + 6z = - 14...................... (1)\]
\[ \text{ and } \]
\[ \vec{r} . \left( 2 \hat{i} + 4 \hat{j}  + 6 \hat{k}  \right) = - 7\]
\[ \Rightarrow 2x + 4y + 6z = - 7.......................\left( 2 \right)\]
\[ \text{ We know that the distance between two planes ax + by }+ cz = d_{ 1} \text{ and } ax + by + cz = d_{2 } \text{ is } \frac{\left| d_{2 }- d_1 \right|}{\sqrt{a^2 + b^2 + c^2}}\]
\[\text{ So, the required distance } \]

\[=\frac{\left| - 7 - \left( - 14 \right) \right|}{\sqrt{2^2 + 4^2 + 6^2}}\]
\[=\frac{\left| 7 \right|}{\sqrt{4 + 16 + 36}}\]
\[ = \frac{7}{\sqrt{56}} \text{ units } \]