#### Question

Find the vector equation of the plane passing through the points `hati +hatj-2hatk, hati+2hatj+hatk,2hati-hatj+hatk`. Hence find the cartesian equation of the plane.

#### Solution

`Let `

`bar(AB)=barb-bara=(hati+2hatj+hatk)-(hati+hatj-2hatk)=hatj+3hatk`

`bar(AC)=barc-bara=(2hati-hatj+hatk)-(hati+hatj-2hatk)=hati-2hatj+3hatk`

`bar(AB)xxbar(AC)=|[hati,hatj,hatk],[0,1,3],[1,-2,3]|`

`=hati(3+6)-hatj(0-3)+hatk(0-1)`

`=9hati+3hatj-hatk`

Then the equation of required plane is,

`barr.barn=bara.barn`

`barr.(9hati+3hatj-hatk)=(hati+hatj-2hatk).(9hati+3hatj-hatk)`

`barr.(9hati+3hatj-hatk)=9+3+2`

`barr.(9hati+3hatj-hatk)=14`

The cartesian equation of the plane is given by,

`(xhati+yhatj+zhatk).(9hati+3hatj-hatk)=14, `

9x+3y-z=14

The cartesian equation of the plane is 9x + 3y - z = 14.