find the plane that contains A(0,3,1); B(-1,2,3); C(4,1,1)?

Answer 1

Form the vectors AB and AC,
Cross them to find a vector normal to the plane.
Use one of the points on the plane to plug into the general equation for a plane. Then dot them and set it equal to zero.

*=0

Solve that and get your plane equation.

Answer 2

You need the equation? Probably the simplest way is to write the equation in general form, z = ax + by + c (works unless the plane is parallel to the z-axis, which this one isn't). Then use the given points to set up a system of three equations for three unknowns. For example, A(0,3,1) means that x=0, y=3, z=1 are coordinates of a point on the plane, and if those values are substituted in the equation of the plane it will balance. So:

1 = 0a + 3b + c
3 = -1a + 2b + c
1 = 4a + 1b + c

Solving, I get a = -2/3, b = -4/3, c = 5, so the equation is:

z = - 2x/3 - 4y/3 + 5.

Answer 3

Find the plane that contains A(0,3,1); B(-1,2,3); C(4,1,1).

We can use the three points to create two vectors that we will use to find the normal vector to the plane.

vector AB = <-1-0,2-3,3-1> = <-1,-1,2>
vector AC = <4-0,1-3,1-1> = <4,-2,0>

The cross product ot AB and AC will give us the normal vector of the plane.

AB X AC = 4i + 8j + 6k
Divide by 2 and select 2i + 4j + 3k.

Now use the normal vector and one of the points in the plane to get the equation of the plane. Let's use A(0,3,1).

Plane:
2(x - 0) + 4(y - 3) + 3(z - 1) = 0
2x + 4y - 12 + 3z - 3 = 0
2x + 4y + 3z -15 = 0

Answer 4

Define a plane through three points
The plane passing through three points , and can be determined by the following determinant equations:

(See wikipedia article "plane")