How to find an equation of the plane containing the points (0, 1, 1), (2, 1, 0), and (1, 1, 0)?

To find the equation of the plane containing the given points, we can use the point-normal form of a plane equation. First, we need to determine vectors that lie in the plane.

Let the points be:

• P1 = (0, 1, 1)
• P2 = (2, 1, 0)
• P3 = (1, 1, 0)

We can find two vectors in the plane by subtracting the coordinates of these points:

• Vector A (from P1 to P2): A = P2 – P1 = (2 – 0, 1 – 1, 0 – 1) = (2, 0, -1)
• Vector B (from P1 to P3): B = P3 – P1 = (1 – 0, 1 – 1, 0 – 1) = (1, 0, -1)

Next, we calculate the normal vector to the plane by taking the cross product of vectors A and B:

Let A = (2, 0, -1) and B = (1, 0, -1).

N = A × B = | i  j  k  |
            | 2  0 -1  |
            | 1  0 -1  |

N = i(0*(-1) - 0*(-1)) - j(2*(-1) - (-1)*1) + k(2*0 - 0*1)

N = 0i - j(-2 + 1) + 0k = 0i + j + 0k = (0, 1, 0).

The normal vector N = (0, 1, 0) indicates the direction perpendicular to the plane.

Now, we can use the normal vector and one of the points (for instance, P1) to write the equation of the plane:

The general form of a plane equation is:

Ax + By + Cz = D

Using the normal vector (A, B, C) = (0, 1, 0) and point P1 (0, 1, 1):

0(x) + 1(y) + 0(z) = D

To find D, substitute the coordinates of P1:

0(0) + 1(1) + 0(1) = D 
D = 1.

Thus, the equation of the plane is:

y = 1.

This indicates that all points on the plane have a constant y-coordinate of 1, which geometrically describes a horizontal plane that intersects the y-axis at 1.