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

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jigsfix | Student, Undergraduate | eNoter

Posted February 3, 2012 at 10:03 PM via web

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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted February 3, 2012 at 10:36 PM (Answer #1)

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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` .

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