Final answer:
To find the equation of the plane passing through three points, we can use vectors. Find two vectors in the plane, then find the cross product of those vectors to get the normal vector. Finally, use one of the points and the normal vector to write the equation of the plane.
Step-by-step explanation:
To find the equation of the plane that passes through three points, (4, 1, 4), (5, -8, 6), and (-4, -5, 1), we can use the concept of vectors.
Step 1: Find two vectors in the plane. We can take the vectors from point (4, 1, 4) to (5, -8, 6) and from (4, 1, 4) to (-4, -5, 1). These two vectors are:
V1 = (5-4, -8-1, 6-4) = (1, -9, 2)
V2 = (-4-4, -5-1, 1-4) = (-8, -6, -3)
Step 2: Find the cross product of V1 and V2 to get the normal vector of the plane. The cross product is given by:
N = V1 x V2 = (1*-6 - (-9)*(-8), (1*(-3) - (-9)*(-8)), (1*(-6) - (-9)*(-3))) = (-54 - 72, (-3 - 72), (-6 + 27)) = (-126, -75, 21)
Step 3: Use one of the points and the normal vector to write the equation of the plane. Using the point (4, 1, 4) and the normal vector (-126, -75, 21), the equation of the plane is:
-126x - 75y + 21z = -126*4 - 75*1 + 21*4
Simplifying, we get:
-126x - 75y + 21z = -615