Question

Find the Cartesian forms of the equations of the following planes.

\[\vec{r} = \left( 1 + s + t \right) \hat{i}  + \left( 2 - s + t \right) \hat{i}  + \left( 3 - 2s + 2t \right) \hat{k}\]

 

Answer 1

`\text{ The given equation of the plane is } `
\[ \vec{r} = \left( 1 + s + t \right) \hat{i} + \left( 2 - s + t \right) \hat{j} + \left( 3 - 2s + 2t \right) \hat{k}  \]
\[ \Rightarrow \vec{r} = \left( \hat{i} + 2 \hat{j}  + 3 \hat{k}  \right) + s \left( \hat{i}  - j - 2 \hat{k}  \right) + t \left( \hat{i}  + \hat{j} + 2 \hat{k} \right)\]
\[ \text{ We know that the equation }  \vec{r} = \vec{a} + s \vec{b} + t \vec{c} \text{ represents a plane passing through a point whose position vector is } \vec{a} \text{ and parallel to the vectors }  \vec{b} \text{ and }  \vec{c} .\]
\[\text{ Here } , \vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} ; \vec{b} = \hat{i} - j - 2 \hat{k}  ; \vec{c} = \hat{i} + \hat{j} + 2 \hat{k}  \]
\[\text{ Normal vector}, \vec{n} = \vec{b} \times \vec{c} \]
\[ = \begin{vmatrix}\hat{i}  & \hat{j}  & \hat{k}  \\ 1 & - 1 & - 2 \\ 1 & 1 & 2\end{vmatrix}\]
\[ = 0 \hat{i}  - 4 \hat{j}  + 2 \hat{k}  \]
\[ = - 4 \hat{j}  + 2 \hat{k}  \]
\[ \text{ The vector equation of the plane in scalar product form is } \]
\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]
\[ \Rightarrow \vec{r} . \left( - 4 \hat{j} + 2 \hat{k} \right) = \left( \hat{i} + 2 \hat{j}  + 3 \hat{k}  \right) . \left( - 4 \hat{j}+ 2 \hat{k}  \right)\]
\[ \Rightarrow \vec{r} . \left[ - 2 \left( 2 \hat{j} - \hat{k} \right) \right] = 0 - 8 + 6\]
\[ \Rightarrow \vec{r} . \left[ - 2 \left( 2 \hat{j} - \hat{k}  \right) \right] = - 2\]
\[ \Rightarrow \vec{r} . \left( 2 \hat{j}  - \hat{k}  \right) = 1\]
\[\text{ For Cartesian form, let us substitute }  \vec{r} = x \hat{i}  + y \hat{j} + z \hat{k}  \text{ here. Then, we get } \]
\[\left( x \hat{i} + y \hat{j}  + z \hat{k}  \right) . \left( 2 \hat{j}  - \hat{k}  \right) = 1\]
\[ \Rightarrow 2y - z = 1\]