Question

Find the equation of the plane passing through the point $(1,-2,1)$ and perpendicular to the line joining the points $A(3,2,1)$ and $B(1,4,2)$.

Answer 1

Vector equation of plane is $(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0$

Plane passes through $(1,-2,1)$

$\Rightarrow \overrightarrow{a}=i-2j+k$

Plane is perpendiculat to line joing $(3,2,1)$ and $(1,4,2)$

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

So the vector equation of the plane is

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

Cartesian equation of plane is

$\{xi+yj+zk-(i-2j+k\left)\right\}.(-i+2j+k)=0\left\{\right(x-1)i+(y+2)j+(z-1\left)k\right\}.(-i+2j+k)=0-(x-1)+2(y+2)+(z-1)=0-x+2y+z+4=0x-2y-z-4=0$