Question

Find the Cartesian forms of the equations of the following planes. \[\vec{r} = \left( \hat{i}  - \hat{j}  \right) + s\left( - \hat{i}  + \hat{j}  + 2 \hat{k} \right) + t\left( \hat{i} + 2 \hat{j} + \hat{k}  \right)\]

Answer 1

\[\vec{r} = \left( \hat{i} - \hat{j} + 0 \hat{k} \right) + s \left( - \hat{i} + j + 2 \hat{k}  \right) + t \left( \hat{i} + 2 \hat{j} + \hat{k} \right)\]

\[ \text{ We know that the equation } \vec{r} = \vec{a} + s \vec{b} + t \vec{c} \text{ represents a plane passing through a point whose position vector is } \vec{a} \text{ and parallel to the vectors } \vec{b} \text{ and } \vec{c} .\]

\[\text{ Here } , \vec{a} = \hat{i}  - \hat{j}  + 0 \hat{k}  ; \vec{b} = - \hat{i}  + j + 2 \hat{k}  ; \vec{c} = \hat{i}  + 2 \hat{j}  + \hat{k}  \]

\[\text{ Normal vector,}  \vec{n} = \vec{b} \times \vec{c} \]

\[ = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ - 1 & 1 & 2 \\ 1 & 2 & 1\end{vmatrix}\]

\[ = - 3 \hat{i}  + 3 \hat{j}  - 3 \hat{k}  \]

\[\text{ The vector equation of the plane in scalar product form is } \]

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

\[ \Rightarrow \vec{r} . \left( - 3 \hat{i}  + 3 \hat{j}  - 3 \hat{k}  \right) = \left( \hat{i}  - \hat{j}  + 0 \hat{k}  \right) . \left( - 3 \hat{i} + 3 \hat{j}  - 3 \hat{k} \right)\]

\[ \Rightarrow \vec{r} . \left[ - 3 \left( \hat{i}  - \hat{j} + \hat{k}  \right) \right] = - 3 - 3 + 0\]

\[ \Rightarrow \vec{r} . \left[ - 3 \left( \hat{i}  - \hat{j} + \hat{k} \right) \right] = - 6\]

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

\[ \text{ For Cartesian form, let us substitute } \vec{r} = x \hat{i} + y \hat{j}  + z \hat{k} \text{ here. Then, we get } \]

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

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