Question

Reduce the equation \[\vec{r} \cdot \left( \hat{i}  - 2 \hat{j}  + 2 \hat{k}  \right) + 6 = 0\] to normal form and, hence, find the length of the perpendicular from the origin to the plane.

 

Answer 1

\[ \text{ The given equation of the plane is } \]
\[ \vec{r} . \left( \hat{i}  -\text{  2 }\hat{j}  +\text{  2 }  \hat{k}  \right) + 6 = 0\]
\[ \Rightarrow \vec{r} . \left( \hat{i} -\text{  2 } \hat{j}  + \text{  2 } \hat{k}  \right) = - 6 \text{ or }  \vec{r} . \vec{n} = - 6, \text{ where }  \vec{n} = \hat{i} -\text{  2 }  \hat{j} + \text{  2 } \hat{k} \]
\[\left| \vec{n} \right| = \sqrt{1 + 4 + 4} = 3\]
\[ \text{ For reducing the given equation to normal form, we need to divide it by } \left| \vec{n} \right|. \text{ Then, we get } \]
\[ \vec{r} . \frac{\vec{n}}{\left| \vec{n} \right|} = \frac{- 6}{\left| \vec{n} \right|}\]
\[ \Rightarrow \vec{r} . \left( \frac{\hat{i} - \text{  2 } \hat{j} + \text{  2 } \hat{k}}{3} \right) = \frac{- 6}{3}\]
\[ \Rightarrow \vec{r} . \left( \frac{1}{3} \hat{i} - \frac{2}{3} \hat{j}  + \frac{2}{3} \hat{k}  \right) = - 2\]
\[ \text{ Dividing both sides by -1, we get } \]
\[ \vec{r} . \left( - \frac{1}{3} \hat{i}  + \frac{2}{3} \hat{j}  - \frac{2}{3} \hat{k}  \right) = 2 . . . \left( 1 \right)\]
\[ \text{ The equation of the plane in normal form is } \]
\[ \vec{r} . \hat{n}  = d . . . \left( 2 \right)\]
`( \text{ where d   is the distance of the plane from the origin } )`
\[\text{ Comparing (1) and (2) } ,\]
\[ \text{ length of the perpendicular from the origin to the plane = d = 2  units }\]