Final answer:
The equation of the plane passing through the points (2, −1, 5), (0, 0, 6), and (6, 7, −1) is found using vector operations to determine a normal vector to the plane and then using the point-normal form of a plane's equation. The resulting equation of the plane is 12x - 8y - 18z + 4 = 0.
Step-by-step explanation:
To find the equation of a plane that passes through three points, we can use the fact that the vector normal to the plane is orthogonal to any vector lying on the plane. First, we find two vectors that lie on the plane by subtracting the coordinates of the given points:
- Vector A: (0, 0, 6) - (2, -1, 5) = (-2, 1, 1)
- Vector B: (6, 7, -1) - (2, -1, 5) = (4, 8, -6)
We then find the cross product of these two vectors to get the normal vector, N, to the plane:
N = A × B = ({ i, j, k },
-2, 1, 1,
4, 8, -6)
= (1(8) + 1(4), -(-2(-6) + 1(4)), -2(8) - 1(-2))
= (12, -8, -18)
The equation of the plane in point-normal form is (x - P) · N = 0, where P is a point on the plane, and N is the normal vector. Using the point (2, -1, 5) and the normal vector (12, -8, -18), the equation of the plane becomes:
12(x - 2) - 8(y + 1) - 18(z - 5) = 0
Simplifying, the plane equation is:
12x - 8y - 18z + 4 = 0