Find the vector and cartesian equation of the plane passing through the point A(1, 1, -2), B(1,2,1) and C (2, -1,1).

Explanation

The position vectors `bara, barb, barc`, of the points A, B, C, w.r.t. the origin are
`bara=hati+hatj-2hatk, barb=hati+2hatj+hatk, barc=2hati-hatj+hatk`
the vector equation of the plane passing through the point `A(bara), B(barb)` and `C(barc)` is
`barr*(bar(AB)xxbar(AC))=bara*(bar(AB)xxbar(AC))" "...(1)`
where, `bar(AB) = barb = -bara = (hati + 2hatj + hatk) - (hati + hatj -2hatk)`
`= hatj + 3hatk`
and `bar(AC) = barc - bara = (2hati - hatj + hatk) - (hati + hatj - 2hatk)`
` = hati - 2hatj + 3hatk`
`therefore bar(AB)xxbar(AC)=|{:(hati,,hatj,,hatk),(0,,1,,3),(1,,"-2",,3):}|`
` = (3+6)hati -(0-3) ahtj + (0-1)hatk`
`= 9hati + 3hatj - hatk`
and `bara * (bar(AB) xx bar(AC)) = (hati + hatj - 2hatk) * (9hati + 3hatj - hatk)`
` = 1(9) + 1(3) + (-2)(-1)`
` = 9 + 3 + 2 = 14`
`therefore` from (1), the vector equation of the required plane is `barr*(9hati + 3hatj - hatk) =14.`
If `barr = xhati + yhatj + zhatk`, then the above equation becomes
`(xhati+yhatj+zhatk)*(9hati+3hatj-hatk)=14`
`therefore x(9) +y(3)+z(-1)=14`
`therefore 9x + 3y -z = 14`
This is the cartesian form of the equation of plane.