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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.
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