Question

Find the vector equation of the plane through the point (2,0,-1) and perpendicular to the line joining the two points(1,2,3) and (3,-1,6).

Answer 1

Answer:

hope it helps you see the attachment for further information

Answer 2

The required equation of the plane is  2x-3y+3z=1  

Step-by-step explanation:

Given that the plane passing through the point (2,0,-1) and perpendicular to the line joining the two points (1,2,3) and (3,-1,6)

To find the vector equation of the plane :

The vector equation of plane passing through the point $\overrightarrow{a}$ and normal to $\overrightarrow{n}$ is  

$(\overrightarrow{r}-\overrightarrow{a})=0$

$\overrightarrow{r}=\overrightarrow{a}$  

Taking $\overrightarrow{n}$ on both sides we have

$\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}$

Here $\overrightarrow{a}=2i+0j+(-1)k$

∴  $\overrightarrow{a}=2i-k$

Hence the vector equation of the plane is  

$\overrightarrow{r}.(2i-3j+3k)=(2i-k).(2i-3j+3k)$ ( the points (3i,-1j,6k)-(1i,2j,3k)=2i-3j+3k )

$\overrightarrow{r}.(2i-3j+3k)=2(2)+0(-3)+(-1)(3)$

$=4-3$

$\overrightarrow{r}.(2i-3j+3k)=1$ is the equation of the plane

The cartesian equation is  

(xi+yj+zk).(2i-3j+3k)=1

x(2)+y(-3)+z(3)=1

2x-3y+3z=1 is the required equation of the plane