How do you find the equation of a plane that passes through the points (0, 0, 0), (9, 9, 0), and (9, 9, 9)?

To find the equation of a plane that passes through three points, we first need to determine the points in a three-dimensional space. In this case, we have the points:

• P1 (0, 0, 0)
• P2 (9, 9, 0)
• P3 (9, 9, 9)

We can represent these points as vectors:

• P1 = (0, 0, 0)
• P2 = (9, 9, 0)
• P3 = (9, 9, 9)

Next, we need to find two vectors that lie on the plane. We can obtain these vectors by subtracting the coordinates of the points:

• Vector A = P2 – P1 = (9-0, 9-0, 0-0) = (9, 9, 0)
• Vector B = P3 – P1 = (9-0, 9-0, 9-0) = (9, 9, 9)

Now, we can find the normal vector to the plane by taking the cross product of vectors A and B:

Normal vector N = A × B

Calculating the cross product:

N = |i    j    k|
         |9    9    0|
         |9    9    9|

Expanding this, we get:

N = i(9 * 9 - 0 * 9) - j(9 * 9 - 0 * 9) + k(9 * 9 - 9 * 9)
N = i(81) - j(81) + k(0)
N = (81, -81, 0)

Thus, our normal vector is N = (81, -81, 0). Now, the general equation of a plane can be defined as:

Ax + By + Cz = D

Where (A, B, C) are the components of the normal vector, and D is a constant that we can calculate using one of the points. Using point P1 (0, 0, 0):

81(0) - 81(0) + 0(0) = D
D = 0

Therefore, the equation of the plane is:

81x - 81y + 0z = 0
or simply 81x - 81y = 0
which can be further simplified to x - y = 0

This means that the equation of the plane passing through the points (0, 0, 0), (9, 9, 0), and (9, 9, 9) is:

x – y = 0