Question

Find the vector and Cartesian equations of a plane passing through the point (1, −1, 1) and normal to the line joining the points (1, 2, 5) and (−1, 3, 1).

 

Answer 1

\[ \text{ Since the given plane passes through the point (1, -1, 1) and is normal to the line joiningA(1, 2, 5) and B(-1, 3, 1) } ,\]

\[ \vec{n} = \vec{AB} = \vec{OB} - \vec{OA} =\left( - \text{ i} + \text{ 3 } \hat{j} + \hat{k} \right) - \left( \hat{i} + \text{ 2 }\hat{j} + \text{ 5 }\hat{k} \right) = - \text{  2} \hat{i}  + \hat{j} - \text{ 4 }\hat{k} \]

\[\text{ We know that the vector equation of the plane passing through a point }  \vec{a} \text{ and normal to } \vec{n} \text{ is }\]

\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]

\[\text{ Substituting } \vec{a} = \hat{i} - \hat{j} + \hat{k}  \text{ and }  \vec{n} = - \text{  2 } \hat{i}+ \hat{j} - \text{  4 } \hat{k} , \text { we get }\]

\[ \vec{r} . \left( - \text{ 2 } \hat{i}] + \hat{j} - 4 \hat{k} \right) = \left( \hat{i} - \hat{j} + \hat{k} \right) . \left( - \text{ 2 }\hat{i} + \hat{j}  - 4 \hat{k} \right)\]

\[ \Rightarrow \vec{r} . \left( \text{ - 2 } \hat{i} + \hat{j}  - 4 \hat{k} \right) = \text{ - 2 - 1 - 4 }\]

\[ \Rightarrow \vec{r} . \left[ - \left( \text{ 2  }\hat{i} - \hat{j} + 4 \hat{k} \right) \right] = - 7\]

\[ \Rightarrow \vec{r} . \left( \text{ 2 } \hat{i} - \hat{j} + 4 \hat{k} \right) = 7\]

\[\text{ For Cartesian form, we need to substitute } \vec{r} = x \hat{i} + \text{ y } \hat{j} + z \hat{k}  \text{ in the vector equation } .\]

\[\text{ Then, we get } \]

\[\left( \text{ x }\hat{i} + \text{ y }\hat{j} + z \hat{k} \right) . \left(  \text{ 2 }\hat{i} - \hat{j}  + 4 \hat{k} \right) = 7\]

\[ \Rightarrow 2x - y + 4z = 7\]