Question

Let y =

4

-6

6

U1 =

-2

-5

1

U2 =

-2

2

6

Find the distance from y to the plane in R3 spanned by U1 and U2. Exact answer please.

Answer 1

You have W=span{(-2,−5,1),(−2,2,6)}. First, orthonormalize the basis of WW to find {U1,U2}{U1,U2} using the Gram-Schmidt algorithm. After this, an easy way to compute the projection on WW would be this : compute the projections of PP on U1U1 and U2U2 as follows :
projv1(P)=(P⋅U1)U1
projv1(P)=(P⋅U1)U1
and
projv2(P)=(P⋅U2)U2
projv2(P)=(P⋅U2)U2
Now that you have this,
projW(P)=projU1(P)+projU2(P).
projW(P)=projU1(P)+projU2(P).
Therefore you can compute the norm of P−projW(P)P−projW(P) and get the distance from PP to WW. I leave the number crunching to you. If the number crunching went wrong I would need to see the numbers to help.

Note that your so-called "best approximation theorem" doesn't need the vectors u1u1 and u2u2 to be orthogonal. What requires orthogonality is the technique used to compute the projections, because you want to project PP on U1U1 and U2U2 and then add the individual projections. This does not work when U1U1 and U2U2 are not orthogonal (make yourself a little drawing if you want to be convinced, it's quite obvious).

Hope that helps,

_________

Or we can also do in this way;

Employ double cross product u3 = u1 X (u1 X u2) in plane V to apply Best Approximation theorem.