Answer:
The equation of the plane is
8(x - 5) - 7(y - 4) - 5(z - 5) = 0
8x - 7y - 5z + 13 = 0
Explanation:
Given 3 points, P(x₁, y₁, z₁), Q(x₂, y₂, z₂), and R(x₃, y₃, z₃).
We can calculate the equation of the plane through those points as
a(x - x₀) + b(y - y₀) + c(z - z₀) = 0, where (x₀, y₀, z₀) are the coordinates of any one of the points P, Q, or R, and <a,b,c> is a vector perpendicular to the plane.
The vector perpendicular to the plane is obtained by writing vector PQ and PR and taking the cross or vector product.
For this question,
P = (5, 4, 5)
Q = (-5, -1, -4)
R = (-2, 1, -2)
PQ = (-5, -1, -4) - (5, 4, 5) = (-10, -5, -9)
= (-10î - 5ĵ - 9ķ)
PR = (-2, 1, -2) - (5, 4, 5) = (-7, -3, -7)
= (-7î - 3ĵ - 7ķ)
PQ × PR is then
| î ĵ ķ |
|-10 -5 -9|
|-7 -3 -7|
= î [(-5×-7) - (-9×-3)] - ĵ [(-10×-7) - (-9×-7)] + ķ [(-10×-3) - (-7×-5)]
= î (35 - 27) - ĵ (70 - 63) + ķ (30 - 35)
= 8î - 7ĵ - 5ķ
Hence, (a, b, c) = (8, -7, -5)
And using point P as (x₀, y₀, z₀) = (5, 4, 5)
The equation of the plane is
a(x - x₀) + b(y - y₀) + c(z - z₀) = 0
8(x - 5) - 7(y - 4) - 5(z - 5) = 0
8x - 40 - 7y + 28 - 5z + 25 = 0
8x - 7y - 5z = 40 - 28 - 25 = -13
8x - 7y - 5z + 13 = 0
Hope this Helps!!!