Find the Cartesian equation of the following planes:
(a) $\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=2$
(b) $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=1$
(c) $\vec{r} \cdot[(s-2 t) \hat{i}+(3-t) \hat{j}+(2 s+t) \hat{k}]=15$
(a) It is given that equation of the plane is
$\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=2$ ...(1)
For any arbitrary point $P(x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r}=x \hat{i}+y \hat{j}-z \hat{k}$
Substituting the value of $\vec{r}$ in equation (1), we obtain
$(x \hat{i}+y \hat{j}-z \hat{k}) \cdot(\hat{i}+\hat{j}-\hat{k})=2$
$\Rightarrow x+y-z=2$
This is the Cartesian equation of the plane.
(b) $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=1$ ...(1)
For any arbitrary point $P(x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r}=x \hat{i}+y \hat{j}-z \hat{k}$
Substituting the value of in equation (1), we obtain
$(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=1$
$\Rightarrow 2 x+3 y-4 z=1$
This is the Cartesian equation of the plane.
(c) $\vec{r} \cdot[(s-2 t) \hat{i}+(3-t) \hat{j}+(2 s+t) \hat{k}]=15$
For any arbitrary point $P(x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r}=x \hat{i}+y \hat{j}-z \hat{k}$
Substituting the value of in equation (1), we obtain
$(x \hat{i}+y \hat{j}-z \hat{k}) \cdot[(s-2 t) \hat{i}+(3-t) \hat{j}+(2 s+t) \hat{k}]=15$
$\Rightarrow(s-2 t) x+(3-t) y+(2 s+t) z=15$
This is the Cartesian equation of the given plane.