Answer:161x + 14y - 7z = -434
Step-by-step explanation:
To find a plane that passes through the given points, we can use the equation of a plane in 3D space. The equation of a plane can be written in the form:
Ax + By + Cz = D
Where (A, B, C) is the normal vector to the plane, and (x, y, z) are the coordinates of a point on the plane.
Let's use the point (-2, -4, 8) as the reference point to start.
Step 1: Calculate the Normal Vector
To find the normal vector (A, B, C), we can use the cross product of two vectors formed by pairs of the given points. Let's use the points (-2, -4, 8) and (-2, 3, -6) to calculate the normal vector:
Vector 1 = (-2, 3, -6) - (-2, -4, 8) = (0, 7, -14)
Vector 2 = (-3, 8, 7) - (-2, -4, 8) = (-1, 12, -1)
Normal Vector = Vector 1 × Vector 2
Using the cross-product formula:
(A, B, C) = (i, j, k)
A = det | 7 -14 |
| 12 -1 |
A = (7 * -1) - (-14 * 12) = -7 + 168 = 161
B = -det | 0 -14 |
| -1 -1 |
B = -((-1) * (-14)) - (0 * (-1)) = 14
C = det | 0 7 |
| -1 12 |
C = (-1 * 7) - (12 * 0) = -7
So, the normal vector is (161, 14, -7).
Step 2: Write the Equation of the Plane
Now that we have the normal vector, we can use the point (-2, -4, 8) to write the equation of the plane:
161x + 14y - 7z = D
Substitute the coordinates of the reference point to find D:
161(-2) + 14(-4) - 7(8) = D
-322 - 56 - 56 = D
D = -434
The equation of the plane passing through the given points is:
161x + 14y - 7z = -434