vectors problem

Question

Find the distance between the point (10,100,1000) and the plane [x,y,z]=[1,1,1]+s[5,−5,3]+t[−1,1,3]

Answer 1

x = 1 + 5s - t; y = 1 - 5s + t; z = 1 + 3s + 3t;

Square of distance: f(s, t) = d² = (5s - t - 9)² + (t - 5s - 99)² + (3s + 3t - 999)²;

To get minimum: f_s = 10(5s - t - 9) - 10(t - 5s - 99) + 6(3s + 3t - 999) = 0;

f_t = - 2(5s - t - 9) + 2(t - 5s - 99) + 6(3s + 3t - 999) = 0

118s - 2t = 5094

-2s + 22t = 6174

1296s = 62208; s = 48; t = (118·48 - 5094)/2 = 285

min d = [(5·48 - 285 - 9)² + (285 - 5·48 - 99)² + (3·48 + 3·285 - 999)²]^1/2 = 5832^1/2 = 76.37