How do you find the equation of a plane that passes through the points (2, 1, 2), (3, 8, 6), and (2, 3, 1)?

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To find the equation of a plane that passes through three points in three-dimensional space, we can follow a systematic approach. Let’s denote the three given points as:

  • Point A: (2, 1, 2)
  • Point B: (3, 8, 6)
  • Point C: (2, 3, 1)

The general equation of a plane can be expressed as:

A(x - x_0) + B(y - y_0) + C(z - z_0) = 0

where (x0, y0, z0) is a point on the plane, and (A, B, C) is a normal vector to the plane.

Step 1: Find Two Vectors in the Plane

First, we will create two vectors in the plane using the given points:

AB = B - A = (3 - 2, 8 - 1, 6 - 2) = (1, 7, 4)
AC = C - A = (2 - 2, 3 - 1, 1 - 2) = (0, 2, -1)

Step 2: Find the Cross Product

Next, we need to find the normal vector to the plane, which can be found using the cross product of vectors AB and AC:

N = AB × AC = |i   j   k|
                |1   7   4|
                |0   2  -1|

Calculating the determinant, we have:

N = i(7 * -1 - 4 * 2) - j(1 * -1 - 4 * 0) + k(1 * 2 - 7 * 0)
  = i(-7 - 8) - j(-1 - 0) + k(2 - 0)
  = -15i + j + 2k

Thus, the normal vector is:

N = (-15, 1, 2)

Step 3: Write the Equation of the Plane

Now, we can substitute the normal vector and the coordinates of one of the points (let’s use point A) into the plane equation:

-15(x - 2) + 1(y - 1) + 2(z - 2) = 0

Expanding this gives:

-15x + 30 + y - 1 + 2z - 4 = 0

Combining like terms:

-15x + y + 2z + 25 = 0

Step 4: Final Equation

Finally, we can rearrange this equation into the standard form:

15x - y - 2z = 25

So, the equation of the plane that passes through the points (2, 1, 2), (3, 8, 6), and (2, 3, 1) is:

15x - y - 2z = 25

This equation effectively describes the desired plane in 3D space.

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