To find the equation of a plane given three points, we can use the vector method. The points given are:
- Point A: (0, 5, 5)
- Point B: (5, 0, 5)
- Point C: (5, 5, 0)
First, we can find two vectors that lie on the plane. We can create these vectors by subtracting the coordinates of the points:
- Vector AB = B – A = (5, 0, 5) – (0, 5, 5) = (5, -5, 0)
- Vector AC = C – A = (5, 5, 0) – (0, 5, 5) = (5, 0, -5)
Next, we will find the normal vector to the plane by taking the cross product of vectors AB and AC:
- AB x AC = |i j k|
- |5 -5 0|
- |5 0 -5|
Calculating this determinant gives:
i(-5* -5 - 0*0) - j(5*-5 - 0*5) + k(5*0 - (-5*5)) = i(25) - j(-25) + k(25) = (25, 25, 25)
So, the normal vector (N) to the plane is (25, 25, 25). This means that the coefficients of x, y, and z in the plane’s equation will be 25, 25, and 25, respectively.
The general equation of a plane can be expressed as:
A(x - x_0) + B(y - y_0) + C(z - z_0) = 0
Where (A, B, C) are the components of the normal vector and (x_0, y_0, z_0) is a point on the plane (we can use any of the three points). We’ll use point A (0, 5, 5):
25(x - 0) + 25(y - 5) + 25(z - 5) = 0
Simplifying this gives:
25x + 25y - 125 + 25z - 125 = 0
Or, we can divide through by 25 to simplify further:
x + y + z - 10 = 0
Thus, the equation of the plane passing through the points (0, 5, 5), (5, 0, 5), and (5, 5, 0) is:
x + y + z = 10