Final answer:
To find the equation of the plane that passes through points A(2,1,1), B(-1,-1,10) and C(1,3,-4), determine two vectors in the plane, calculate their cross product to get the normal vector, and use the normal vector and a point on the plane to write the plane's equation, which is 10x - 22y - 6z = -8.
Step-by-step explanation:
To find an equation of the plane that passes through the points A(2,1,1), B(-1,-1,10) and C(1,3,-4), we start by finding two vectors that lie in the plane by subtracting the coordinates of these points:
- Vector AB = B - A = (-1,-1,10) - (2,1,1) = (-3,-2,9)
- Vector AC = C - A = (1,3,-4) - (2,1,1) = (-1,2,-5)
Next, we find the normal vector to the plane by taking the cross product of AB and AC:
Normal Vector, N = AB x AC
= |i j k|
|-3 -2 9|
|-1 2 -5|
By computing the determinant, we get N = (10, -22, -6).
The general equation of a plane is given by ax + by + cz = d, where (a,b,c) is the normal vector to the plane. Using point A(2,1,1) and normal vector N(10,-22,-6), we substitute x, y and z from point A into the equation:
10(2) - 22(1) - 6(1) = d
Hence, d = 20 - 22 - 6 = -8.
The equation of the plane is:
10x - 22y - 6z = -8