Question

Find the distance of the point  \[2 \hat{i} - \hat{j} - 4 \hat{k}\]  from the plane  \[\vec{r} \cdot \left( 3 \hat{i}  - 4 \hat{j}  + 12 \hat{k}  \right) - 9 = 0 .\]

Answer 1

\[ \text{ We know that the perpendicular distance of a pointPof position vector } \vec{a} \text{ from the plane } \vec{r} . \vec{n} = \text{ d is given by } \]
\[p = \frac{\left| \vec{a} . \vec{n} - d \right|}{\left| \vec{n} \right|}\]
\[ \text{ Here }, \vec{a} = 2 \hat{i}  - \hat{j}  - 4 \hat{k}  ; \vec{n} = 3 \hat{i}  - 4 \hat{j}  + 12  \hat{k}  ; d = 9\]
\[ \text{ So, the required distance,p } \]
\[ = \frac{\left| \left( 2 \hat{i} - \hat{j}  - 4 \hat{k}  \right) . \left( 3 \hat{i}  - 4 \hat{j}  + 12 \hat{k}  \right) - 9 \right|}{\left| 3 \hat{i} - 4 \hat{j} + 12 \hat{k}  \right|}\]
\[ = \frac{\left| 6 + 4 - 48 - 9 \right|}{\sqrt{9 + 16 + 144}}\]
\[ = \frac{\left| -47 \right|}{13}\]
\[ = \frac{47}{13}\text{  units }\]