Question

The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, −4, 3). Find the equation of the plane.

 

Answer 1

\[\text{ The normal is passing through the points A(0, 0, 0) and B(12, -4, 3). So },\]
\[ \vec{n} = \vec{AB} = \vec{OB} - \vec{OA} =\left( \text{ 12 } \hat{i} -  \text{ 4 }\hat{j} + \text{ 3 } \hat{k} \right) - \left( \text{  0 }\hat{i} + \text{  0  }\hat{j} + \text{ 0 } \hat{k}  \right) = \text{  12 } \hat{i} - \text{ 4 } \hat{j} + \text{  3 } \hat{k}  \]
\[ \text{ Since the plane passes through } (12, -4, 3), \vec{a} =12 \hat{i}  - 4 \hat{j}  + 3 \hat{k}  \]
\[ \text{ We know that the vector equation of the plane passing through a point } \vec{a} \text{ and normal to  } \vec{n} \text{ is} \]
\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]
\[\text{ Substituting } \vec{a} = \hat{i}  - \hat{j}  + \hat{k}  \text{ and } \vec{n} = \text{  4 } \hat{i} + \text{ 2 } \hat{j}  - \text{  3 }\hat{k}  , \text{ we get } \]
\[ \vec{r} . \left( \text{ 12 } \hat{i} - \text{ 4 }\hat{j}  + \text{ 3 } \hat{k}  \right) = \left( \text{ 12 } \hat{i} - \text{  4 }\hat{j} + \text{ 3  }\hat{k}  \right) . \left( \text{ 12 }\hat{i} - \text{ 4 }\hat{j} + \text{ 3 }\hat{k} \right)\]
\[ \Rightarrow \vec{r} . \left( \text{ 12 } \hat{i} - \text{ 4 } \hat{j} + \text{  3 } \hat{k} \right) = 144 + 16 + 9\]
\[ \Rightarrow \vec{r} . \left( \text{ 12 } \hat{i} - \text{ 4 } \hat{j} + \text{ 3 }\hat{k} \right) = 169\]
\[ \Rightarrow \vec{r} . \left( \text{ 12 }\hat{i}  - \text{ 4 } \hat{j}  + \text{  3 } \hat{k}  \right) = 169\]
\[\text{ Substituting } \vec{r} = \text{ x  }\hat{i}  + \text{ y }\hat{j} + \text{ z }\hat{k}  \text{ in the vector equation, we get } \]
\[\left( \text{ x } \hat{i} + \text{ y } \hat{j} + \text{ z  }\hat{k} \right) . \left( \text{ 12 }\hat{i} - \text{ 4 }\hat{j}  + \text{ 3 } \hat{k}  \right) = 169\]
\[ \Rightarrow 12x - 4y + 3z = 169\]