# Find the vector and Cartesian equation of the planes

Question:

Find the vector and Cartesian equation of the planes

(a) that passes through the point $(1,0,-2)$ and the normal to the plane is $\hat{i}+\hat{j}-\hat{k}$.

(b) that passes through the point $(1,4,6)$ and the normal vector to the plane is $\hat{i}-2 \hat{j}+\hat{k}$.

Solution:

(a) The position vector of point $(1,0,-2)$ is $\vec{a}=\hat{i}-2 \hat{k}$

The normal vector $\vec{N}$ perpendicular to the plane is $\vec{N}=\hat{i}+\hat{j}-\hat{k}$

The vector equation of the plane is given by, $(\vec{r}-\vec{a}) \cdot \vec{N}=0$

$\Rightarrow[\vec{r}-(\hat{i}-2 \hat{k})] \cdot(\hat{i}+\hat{j}-\hat{k})=0$

$\vec{r}$ is the position vector of any point $P(x, y, z)$ in the plane.

$\therefore \vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$

Therefore, equation (1) becomes

$[(x \hat{i}+y \hat{j}+z \hat{k})-(\hat{i}-2 \hat{k})] \cdot(\hat{i}+\hat{j}-\hat{k})=0$

$\Rightarrow[(x-1) \hat{i}+y \hat{j}+(z+2) \hat{k}] \cdot(\hat{i}+\hat{j}-\hat{k})=0$

$\Rightarrow(x-1)+y-(z+2)=0$

$\Rightarrow x+y-z-3=0$

$\Rightarrow x+y-z=3$

This is the Cartesian equation of the required plane.

(b) The position vector of the point $(1,4,6)$ is $\vec{a}=\hat{i}+4 \hat{j}+6 \hat{k}$

The normal vector $\vec{N}$ perpendicular to the plane is $\vec{N}=\hat{i}-2 \hat{j}+\hat{k}$

The vector equation of the plane is given by, $(\vec{r}-\vec{a}) \cdot \vec{N}=0$

$\Rightarrow[\vec{r}-(\hat{i}+4 \hat{j}+6 \hat{k})] \cdot(\hat{i}-2 \hat{j}+\hat{k})=0$

$\vec{r}$ is the position vector of any point $P(x, y, z)$ in the plane.

$\therefore \vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$

Therefore, equation (1) becomes

$[(x \hat{i}+y \hat{j}+z \hat{k})-(\hat{i}+4 \hat{j}+6 \hat{k})] \cdot(\hat{i}-2 \hat{j}+\hat{k})=0$

$\Rightarrow[(x-1) \hat{i}+(y-4) \hat{j}+(z-6) \hat{k}] \cdot(\hat{i}-2 \hat{j}+\hat{k})=0$

$\Rightarrow(x-1)-2(y-4)+(z-6)=0$

$\Rightarrow x-2 y+z+1=0$

This is the Cartesian equation of the required plane.