Find the the minimal distance from the point $P=(17, -19, 0)$ to the plane $V$ in $\Bbb R^3$ spanned by the vectors $u_1 = (4, -4, -2)$ and $u_2 = (-4, 1, 1)$.
So, I tried to apply the Best Approximation Theorem where the $\mathrm{dist}(P,W) = \| P – \mathrm{proj}_w P \|$ where $W = \mathrm{span} \{u_1, u_2\}$. But I realize that to apply the Best Approximation theorem $u_1$ and $u_2$ must be orthogonal and they are not.
So, to find an orthogonal basis for $W$ I used The Gram-Schmidt Process where I let $v_1 = u_1 =(4, -4, -2)$ and I obtained $v_2 = (-14/9, -13/9, -2/9)$. Hence, $\{v_1, v_2\}$ is an orthogonal set.
After this I tried to apply the Best Approximation Theorem to find the minimal distance but I still do not get the correct answer.
I hope my question makes sense.
Best Answer
You have $W = \mathrm{span}\{ (4,-4,-2), (-4,1,1) \}$. First, orthonormalize the basis of $W$ to find $\{v_1, v_2\}$ using the Gram-Schmidt algorithm. After this, an easy way to compute the projection on $W$ would be this : compute the projections of $P$ on $v_1$ and $v_2$ as follows : $$ \mathrm{proj}_{v_1}(P) = (P \cdot v_1)v_1 $$ and $$ \mathrm{proj}_{v_2}(P) = (P \cdot v_2)v_2 $$ Now that you have this, $$ \mathrm{proj}_{W}(P) = \mathrm{proj}_{v_1}(P) + \mathrm{proj}_{v_2}(P). $$ Therefore you can compute the norm of $P - \mathrm{proj}_{W}(P)$ and get the distance from $P$ to $W$. 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 $u_1$ and $u_2$ to be orthogonal. What requires orthogonality is the technique used to compute the projections, because you want to project $P$ on $v_1$ and $v_2$ and then add the individual projections. This does not work when $v_1$ and $v_2$ are not orthogonal (make yourself a little drawing if you want to be convinced, it's quite obvious).
Hope that helps,