Question

The direction ratios of the perpendicular from the origin to a plane are 12, −3, 4 and the length of the perpendicular is 5. Find the equation of the plane. 

Answer 1

\[ \text{It is given that the direction ratios of the normal vector }  \vec{n}  \text{ are 12, -3, 4. }\]
\[So, \vec{n} = \text{ 12 } \hat{i}  - \text{ 3 }\hat{j}  + \text{ 4 }\hat{k} \]
\[\left| \vec{n} \right| = \sqrt{{12}^2 + \left( - 3 \right)^2 + 4^2} = \sqrt{144 + 9 + 16} = \sqrt{169} = 13\]
\[ \text{ Now } , \hat{n}  = \frac{\vec{n}}{\left| \vec{n} \right|} = \frac{12 \hat{i}  - 3 \hat{j}  + 4 \hat{k} }{13} = \frac{12}{13} \hat{i}  - \frac{3}{13} \hat{j}  + \frac{4}{13} \hat{k}  \]
\[ \text{ Length of the perpendicular from the origin to the plane,d= 5 } \]
\[ \text{ Equation of the plane in normal form is } \]
\[ \vec{r} . \hat{n}  = d\]
\[ \Rightarrow \vec{r} . \left( \frac{12}{13} \hat{i }  - \frac{3}{13} \hat{j }  + \frac{4}{13} \hat{k }  \right) = 5\]