Question

A plane passes through the point (1, −2, 5) and is perpendicular to the line joining the origin to the point

\[ \text{ 3 } \hat{i} + \hat{j} - \hat{k} .\] Find the vector and Cartesian forms of the equation of the plane.

 

Answer 1

\[\text { The normal is passing through the points } A(0, 0, 0) \text{ and } B(3, 1, -1). \text{ So } ,\]
\[ \vec{n} = \vec{OP} =\left( \text{ 3 }   \hat{i} + \hat{j}  - k \right) - \left( \text{ 0 }\hat{i}  + \text{ 0    }        \hat{j}  + \text{ 0 }\hat{k} \right) = \text{ 3 } \hat{i}  + \hat{j} - \hat{k} \]
\[\text{ Since the plane passes through }  (1, -2, 5), \vec{a} = \hat{i} - 2 \hat{j} + \text{ 5  }\hat{k}  \]
\[\text{ We know that the vector equation of the plane passing through a point } \vec{a} \text{ and normal to }\vec{n} \text{ is }\]
\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]
\[\text{ Substituting } \vec{a} = \hat{i} - \hat{j} + \hat{k} \text{ and }  \vec{n} = \text{  4 } \hat{i}  + \text{ 2 }\hat{j}  - \text{ 3 }\hat{k}  , \text{ we get } \]
\[ \vec{r} . \left( \text{ 3 }\hat{i}  + \hat{j}  - \hat{k}  \right) = \left( \hat{i} - \text{  2 }\hat{j} + \text{ 5 }\hat{k}  \right) . \left( \text{ 3 }\hat{i}  + \hat{j}  - \hat{k}  \right)\]
\[ \Rightarrow \vec{r} . \left( \text{  3 } \hat{i} + \hat{j}  - \hat{k}  \right) = 3 - 2 - 5\]
\[ \Rightarrow \vec{r} . \left( \text{ 3 } \hat{i}  + \hat{j}  - \hat{ k} \right) = - \text{ 4 }\]
\[ \Rightarrow \vec{r} . \left( \text{ 3  }\hat{i} + \hat{j}  - \hat{k} \right) = - 4\]
\[\ \text { Substituting } \vec{r} = \text{  x } \hat{i} + \text{ y }\hat{j} +  z  \hat{k}  \text { in the vector equation, we get } \]
\[\left( \text{  x } \hat{i}  + \text{ y } \hat{j}  + z \hat{k}  \right) . \left( 3 \hat{i}  + \hat{j}  - \hat{k}  \right) = - 4\]
\[ \Rightarrow 3x + y - z = - 4\]