Find the Cartesian equation of the following planes:
(a) \vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=2
(b) \vec{r} \cdot(2 \widehat{i}+3 \widehat{j}-4 \widehat{k})=1
Find the Cartesian equation of the following planes:
(a) \vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=2
(b) \vec{r} \cdot(2 \widehat{i}+3 \widehat{j}-4 \widehat{k})=1

Solution:

(a) It is given that,
The equation of the plane.
Let \overrightarrow{\mathrm{r}} be the position vector of \mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z}) is given by \vec{r}=x \hat{i}+y \hat{j}+z \hat{k}
So,
\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=2
On substituting the value of \overrightarrow{\mathrm{r}}, we obtain
(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}-\hat{k})=2
As a result, the Cartesian equation is
x+y-z=2

(b) \vec{r} \cdot(2 \widehat{i}+3 \widehat{j}-4 \widehat{k})=1

Solution:
Let \vec{r} be the position vector of P(x, y, z) is given by
\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}
Therefore,
\overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{4} \mathrm{k})=1
On substituting the value of \overrightarrow{\mathrm{r}}, we obtain
(x \hat{i}+\hat{y j}+z \hat{k}) \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=1
As a result, the Cartesian equation is
2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}=1