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Find an equation of the plane that passes through the points A(2,1,1) B(-1,-1,10) C(1.3.-4).

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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

User Peter Farmer
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