Find the equation of a plane in three points P1=(1,0,1), P2=(0,1,1), P3=(1,1,0)

### 1 Answer | Add Yours

You should remember that if the points `P_1,P_2,P_3` are in a plane, hence the vectors `bar (P_1P_2)` and `bar (P_1P_3)` are in the plane containing these points.

Evaluating the cross product of the vectors `bar (P_1P_2)` and `bar (P_1P_3)` yields an orthogonal vector `bar N` to the plane containing the vectors `bar (P_1P_2)` and `bar (P_1P_3).`

Hence `bar (P_1P_2)Xbar (P_1P_3) = ` `[[i,j,k],[x_(P_2)-x_(P_1),y_(P_2)-y_(P_1),z_(P_2) - z_(P_1)],[x_(P_3)-x_(P_1),y_(P_3)-y_(P_1),z_(P_3) - z_(P_1)]]`

`bar (P_1P_2)Xbar (P_1P_3) = [[i,j,k],[0-1,1-0,1-1],[1-1,1-0,0-1]]` = `[[i,j,k],[-1,1,0],[0,1,-1]]`

Evaluating the determinant yields orthogonal vector: `bar N = -bar i - bar j - bar k`

You need to find the equation of the plane containing the points `P_1,P_2,P_3 ` considering the condition that `bar N` is orthogonal to vector formed by `P_1` and any point `P` in the plane such that:

`bar N*bar (P_1P) = 0` ( the dot product is zero if vectors are othogonal)

`(-bar i - bar j - bar k)((x-1)bar i + ybar j + (z-1)bar k)=0`

`-(x - 1) - y - (z - 1) = 0`

Opening the brackets yields:

`-x + 1 - y - z + 1 = 0 =gt x + y + z - 2 = 0`

**Hence, the equation of plane containing the points `P_1,P_2,P_3` is `x + y + z - 2 = 0` .**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes