Question

Find the vector equation of the plane passing through the points $A(1,0,1),B(1,-1,1)$ and $C(4,-3,2)$.

Answer 1

$\overrightarrow{AB}=(1-1)\hat{i}+(-1-0)\hat{j}+(1-1)\hat{k}=-\hat{j}$

$\overrightarrow{BC}=(4-1)\hat{i}+(-3+1)\hat{j}+(2-1)\hat{k}=3\hat{i}-2\hat{j}+\hat{k}$

Let $\overrightarrow{n}$ be the normal of the plane and $\overrightarrow{r}$ be a point on the plane.

As $\overrightarrow{AB}$ and $\overrightarrow{BC}$ lie in the plane,

$\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{BC}$

$\therefore \overrightarrow{n}=-\hat{j}\times (3\hat{i}-2\hat{j}+\hat{k})$

$\therefore \overrightarrow{n}=-\hat{i}+3\hat{k}$

Now, equation of plane with normal $\overrightarrow{n}$ and passing through point $A$ is

$\overrightarrow{r}\cdot \overrightarrow{n}=\overrightarrow{A}\cdot \overrightarrow{n}$

$\therefore \overrightarrow{r}\cdot \overrightarrow{n}=(\hat{i}+\hat{k})\cdot (-\hat{i}+3\hat{k})$

$\therefore \overrightarrow{r}\cdot \overrightarrow{n}=2$

This is the required vector plane equation.