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