# Find an equation for the plane containing the points A(0,4,2) and B(-5,0,7)and perpendicular to the plane 2x+2y+4z=3

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need write the vector `bar(AB)` , that is parallel to the plane you need to find, such that:

`bar(AB) = <x_B - x_A,y_B - y_A,z_B - z_A>`

`bar(AB) = <-5-0,0-4,7-2> => bar(AB) = <-5,-4,2>`

You also need that the normal vector to the plane `2x+2y+4z=3` is parallel to the plane whose equation you need to find

`bar n = <2,2,4>`

Notice that evaluating the cross product of the vectors `bar(AB)` and `bar n` yields the normal vector to the plane you need to find, such that:

`bar n x bar(AB) = [(bar i, bar j, bar k),(2,2,4),(-5,-4,2)]`

`bar n x bar (AB) = 4 bar i - 8 bar k - 20 bar j + 10 bar k + 16 bar i - 4 bar j`

`bar n x bar (AB) = 20 bar i - 24 bar j + 2 bar k`

You may evaluate the equation of the plane using the normal vector `bar n_1 = <20,-24,2>` and a pointÂ `A(0,4,2)` contained by the plane, such that:

`20*(x - 0) + (-24)(y - 4) + 2(z - 2) = 0`

`20x - 24y + 2z + 96 - 4 = 0`

`20x - 24y + 2z + 92 = 0 => 10x - 12y + z + 46 = 0`

Hence, evaluating the equation of the plane, under the given conditions, yields `10x - 12y + z + 46 = 0` .