S.O.S. Mathematics CyberBoard

Consider the following set of three-dimensional points
D = {(0,2,1), (0,4,3), (1,4,5), (1,8,6), (1,8,10), (4,8,14), (5,9,13)}

(a) Find the least squares estimate for z taken as a function of x and y.
(b) Find the least squares estimate for x taken as a function of y and z.
(c) Find the best fit plane, in term of principal component analysis.


My answers:

(a) Finding the best fit plane for z taken as a function of x and y
A + 0B + 2C = 1
A + 0B + 4C = 3
A + 1B + 4C = 5
A + 1B + 8C = 6
A + 1B + 8C = 10
A + 4B + 8C = 14
A + 5B + 9C = 13

LeastSq(A, B, C) = { -0.331, 1.385, 0.877 }
New points that lie on projection plane:
x: 0.000 0.000 1.000 1.000 1.000 4.000 5.000
y: 2.000 4.000 4.000 8.000 8.000 8.000 9.000
z: 1.423 3.176 4.561 8.068 8.068 12.222 14.483

(b) Finding the best fit plane for x taken as a function of y and z
A + 2B + 1C = 0
A + 4B + 3C = 0
A + 4B + 5C = 1
A + 8B + 6C = 1
A + 8B + 10C = 1
A + 8B + 14C = 4
A + 9B + 13C = 5

LeastSq(A, B, C) = { 1.714, 6.143, 7.429 }
New points that lie on projection plane:
x: -0.356 0.156 1.040 0.737 2.505 4.274 3.645
y: 2.000 4.000 4.000 8.000 8.000 8.000 9.000
z: 1.000 3.000 5.000 6.000 10.000 14.000 13.000

(c) Finding the variances of plane(a) and plane(b).
Avg(D) = (1.714, 6.143, 7.429)
Var(Proj(Avg, Plane(a))): 29.175
Var(Proj(Avg, Plane(b))): 30.515

Plane(a) has smaller variance, therefore Plane(a) is a better fit plane.


Are my answers correct at all???