Find an equation of a plane containing the three points (-1, -4, 2), (1, -6, 1), (1, -5, 3)?

Question

which the coefficient of x is -3.

Answer 1

Equation of the plane containing the three points (-1, -4, 2), (1, -6, 1), (1, -5, 3) will be, in the determinant form:


| x - 1 - 4 2 |
| |
| y 1 - 6 1 | = 0.
| |
| z 1 - 5 3 |

Answer 2

Find an equation of a plane containing the three points
P(-1, -4, 2); Q(1, -6, 1); and R(1, -5, 3).

Create two directional vectors in the plane.

u = PQ = = <1+1, -6+4, 1-2> = <2, -2, -1>
v = PR = = <1+1, -5+4, 3-2> = <2, -1, 1>

The normal vector n, to the plane is perpendicular to both of them. Take the cross product.

n = u X v = <2, -2, -1> X <2, -1, 1> = <-3, -4, 2>

With the normal vector to the plane and a point in the plane we can write the equation to the plane.
Let's choose Q(1, -6, 1).

-3(x - 1) - 4(y + 6) + 2(z - 1) = 0
-3x + 3 - 4y - 24 + 2z - 2 = 0
-3x - 4y + 2z - 23 = 0