# Find the minimal distance from the point $P$ to the plane $V$ spanned by $u_1$ and $u_2$

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.

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Um, $u_1$ and $u_2$ don't need to be orthogonal. The distance between a point and a plane is always the norm of the difference between the point and its projection on the plane because the projection is the closest point in the plane to $P$. The way that the plane is defined doesn't change things at all, i.e. whether the basis chosen to define $W$ is orthonormal or not is irrelevant. –  Patrick Da Silva Nov 29 '12 at 23:35

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,

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Okay. I understand now. Thank You for the explanation. –  Diana Martinez Nov 29 '12 at 23:55

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

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That's just a computationally convenient (maybe...) way to orthogonalize the set \{u_1, u_2\}$into$\{u_1, u_3\}\$, but to be honest I'd rather use Gram-Schmidt ; only one dot product involved. –  Patrick Da Silva Nov 30 '12 at 4:44