Question

Compute the distance from the point x = (-5,5,2) to the subspace W of R³ consisting of all vectors orthogonal to the vector (1, 3, -2).

Answer 1

Let (p,q,r) be an arbitrary vector orthogonal to the vector (1, 3, -2).Then (p,q,r).(1,3,-2) = 0 or, p+3q-rz = 0 or, p = -3q+2r.

Then (p,q,r) = (-3q+2r,q,r) = q(-3,1,0)+r(2,0,1). It means that (p,q,r) is a linear combination of 2 linearly independent vectors (-3,1,0), (2,0,1). Hence W = span {(-3,1,0), (2,0,1)}.

Now, let u = (-3,1,0) and v = (2,0,1).

We have proj_u (x) = [(x.u)/(u.u)]u = [(15+5+0)/(9+1+0)]u = 2u = 2(-3,1,0)= (-6,2,0) and proj_v(x) = [(x.v)/(v.v)]v =[(-10+0+2)/(4+0+1)]v = -(8/5) (2,0,1) = (-16/5,0,-8/5).

Then proj_W(x) = proj_u (x)+ proj_v (x) = (-6,2,0)+ (-16/5,0,-8/5) = (-46/5, 2, -8/5) = y (say)

Since proj_W(x) is closest to x in W, hence the distance from the point X= (-5,5,2) to the subspace W of R³ , hence the required distance is || x-y|| = || ( -5,5,2)-(-46/5, 2, -8/5)|| =||( 21/5, 3,18/5)|| = √(441/25+ 9+324/25) = √(990/25) = (3/5)√110.