Find the Cartesian equation of the following planes:

Question:

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) $\vec{r} \cdot[(s-2 t) \hat{i}+(3-t) \hat{j}+(2 s+t) \hat{k}]=15$

Solution:

(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.

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