# Is this a linear algebra problem?

I'm solving this question, but I don't know what it is.

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Consider a plane $x - y + z = 0$, and a point $b = (1, 2, 0)^T$

a) Find a basis of this plane, and a basis of orthogonal subspace to this plane.

b) Find the closest point $\hat{b}$ on the plane to b.

c) What is the error between b and $\hat{b}$?

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Is this a linear algebra problem? What's a basis of orthogonal subspace and closest point $\hat{b}$? What's the error between b and $\hat{b}$??

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«Is this a linear algebra problem?» is a new angle :) – Mariano Suárez-Alvarez Oct 26 '10 at 5:04
sorry, I can't understand.. – Brian Oct 26 '10 at 5:05
What Mariano means is that it's completely obvious that this is a linear algebra problem! – Hans Lundmark Oct 26 '10 at 6:26
In what context did you come across this? It looks like a typical exercise in a linear algebra course. If you are indeed taking such a course, your textbook should explain in detail what the word "basis" means, and how to solve this problem. And then you should also have access to teachers that you can ask; this sort of thing is easier to explain face to face than in writing. By the way, the word "error" here simply means the distance between the points $b$ and $\hat{b}$. – Hans Lundmark Oct 26 '10 at 6:31

Yes, this is a linear algebra problem. Looks like a homework.

Hint: You are working in $\mathbb{R}^3$. Your plane is given by 1 equation, which strongly suggest that it has dimension 2. Hence you should try to find 2 vectors linear independant and you will have your first basis.

Then, 2 vector spaces are orthogonal if any vector of the first one is orthogonal to any vector of the second one (if $v_1 = (x_1, y_1, z_1)^T$ and $v_2 = (x_2, y_2, z_2)^T$, then they are orthogonal if and only if $x_1x_2 + y_1y_2 + z_1z_2 = 0$). When we are talking about the orthogonal subspace in this context, you are supposed to find one with maximal dimension. For example $V = {(0,0,0)^T}$ is always orthogonal to any other subspace, therefore try to find one with dimension > 0. Hint: the dimension of a subspace + dimension of it's orthogonal complement should equals the dimension of the space you are working in.

When you have solved a), then we might help you to solve b) and c).

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(a) To find a basis for the plane $x-y+z = 0$, you could solve this equation in terms of $x$ and $z$: $y= x+z$. Then the set of vectors of your plane could be described as:

$$V = \left\{ (x, x+z, z) \ \vert \ x,z \in \mathbb{R} \right\} \ .$$

From this description it's easy to find a basis for your plane $V$: it will have two vectors, say $u,v \in \mathbb{R}^3$:

$$V = [u,v] \ .$$

Then, the orthonormal subspace to your plane is a straight line, generated by the cross product of $u$ and $v$

$$V^\bot = [u \times v] \ .$$

(b) The closest point $\hat{b}$ to $b$ in the plane $V$ is the orthogonal projection of $b$ onto $V$. In this case, you don't really need the formulae you'll see in Wikipedia, because it's just the intersection of $V$ with the straight line with direction $u\times v$ that passes through $b$. That is, you have to solve the system of linear equations

\begin{align} x - y +z &= 0 \\ (x,y,z) &= (1,2,0) + \lambda u\times v \ . \end{align}

(c) The "error" between $b$ and $\hat{b}$ is the same as the distance from $b$ to $\hat{b}$; that is, $\| b - \hat{b} \|$.

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