Find an Equation of the Plane That Passes Through the Points (4, 1, 4), (5, 8, 6), and (4, 5, 1)

To find the equation of the plane that passes through the three given points, we can use the following approach:

1. Let the three points be A(4, 1, 4), B(5, 8, 6), and C(4, 5, 1).
2. We first find two vectors that lie in the plane. These can be formed using the given points:
• Vector AB: B – A = (5 – 4, 8 – 1, 6 – 4) = (1, 7, 2)
• Vector AC: C – A = (4 – 4, 5 – 1, 1 – 4) = (0, 4, -3)
3. Now, we find the normal vector to the plane by taking the cross product of these two vectors:

Normal Vector = AB × AC = | i  j  k  |
                     | 1  7  2 |
                     | 0  4 -3 |

4. Calculating the determinant gives:

i(7*-3 - 2*4) - j(1*-3 - 2*0) + k(1*4 - 0*7) = i(-21 - 8) - j(-3) + k(4)

5. This simplifies to:

Normal Vector = (-29, 3, 4)

6. Now, we can use the point-normal form of the plane equation, given by:

n_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0

7. Using point A(4, 1, 4) and the normal vector (-29, 3, 4):

-29(x - 4) + 3(y - 1) + 4(z - 4) = 0

8. This further simplifies to:

-29x + 116 + 3y - 3 + 4z - 16 = 0

9. Finally, we can rearrange this into the standard form:

-29x + 3y + 4z + 97 = 0

10. This is the equation of the plane that passes through the points (4, 1, 4), (5, 8, 6), and (4, 5, 1).