Question

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

(ii) $\overrightarrow{r}=\left(1+s+t\right)\hat{i}+\left(2-s+t\right)\hat{i}+\left(3-2s+2t\right)\hat{k}$

Answer 1

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

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