# Find an equation for the plane containing the line L:(x-4)/(-4) =(z+3)/2; y = -3 and the point Q(-3,2,-2).

sciencesolve | Certified Educator

You need to write the parametric form of the given line, such that:

`l: x = 4 - 4t`

`y = -3 + 0*t`

`z = -3 + 2t`

You need to consider the directional vector of the line using the coefficient of parameter t, such that:

`bar l = <-4,0,2>`

You may find the coordinates of other point P in the plane considering the parameter t = 0, such that:

`P = (4,-3,-3)`

You may evaluate the directional vector `bar (PQ)` such that:

`bar(PQ) = (x_P - x_Q) bar i + (y_P - y_Q) bar j + (z_P - z_Q) bar k`

`bar(PQ) = (4 + 3) bar i + (-3 + 2) bar j + (-3 - 2) bar k`

`bar(PQ) = 7 bar i - bar j - 5 bar k`

You may evaluate the normal vector `bar n` to the plane containing the line and the point Q such that:

`bar n = bar l x bar (PQ)`

`bar n = [(bar i, bar j, bar k),(-4,0,2),(7,-1,-5)]`

`bar n = 4 bar k + 14 bar j + 2 bar i - 20 bar j`

`bar n = 2 bar i - 6 bar j + 4 bar k`

`bar n = <2,-6,4>`

Dividing by 2 yields:

`bar n = <1,-3,2>`

You may write the equation of the plane such that:

`P: (x - x_Q) - 3(y - y_Q) + 2(z - z_Q) = 0`

`P: (x + 3) - 3(y - 2) + 2(z + 2) = 0`

Hence, evaluating the equation of the plane, under the given conditions, yields `P: (x + 3) - 3(y - 2) + 2(z + 2) = 0` .