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Let $V = R^3$ with the standard inner product $u = (2,1,3)$ and $W = \{(x,y,z) : x + 3y - 2z = 0\}$

I came up with the basis $\{(-3,1,0), (2,0,1)\}$ but these are not orthogonal to each other. I'm not exactly sure how to approach this question, any help would be appreciated. Thanks

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There are many ways how to find an orthogonal projection.

You seem to want to use an orthogonal (or an orthonormal) basis of $W$ in some way.

If you already have a basis of $W$, you can get an orthogonal basis from it using Gram-Schmidt process.

Another way to do this. Let us choose $\vec b_1=(2,0,1)$ at the first vector basis. Now you want a find another vector which belongs to $W$ (i.e., it satisfies $x+3y-z=0$) and which is orthogonal to $\vec b_1$ (i.e., it satisfies $2x+z=0$). Can you find solution of these two equations? Can you use it to get an orthogonal basis of $W$?


Solution using a linear system. Here is another way to find an orthogonal projection. We are given a vector $\vec u=(2,1,3)$. And we want to express it as $\vec u=\vec u_1+\vec u_2$, where $\vec u_1 \in W$ and $\vec u_2=W^\bot$. We know bases of $W=[(-3,1,0),(2,0,1)]$ and of $W^\bot=[(1,3,-2)]$.

So we simply express the vector $\vec u$ as a linear combination $\underset{\in W}{\underbrace{c_1(-3,1,0)+c_2(2,0,1)}}+\underset{\in W^\bot}{\underbrace{c_3(1,3,-2)}}$.

To find $c_{1,2,3}$ it suffices to solve the system of equations $$ \left(\begin{array}{ccc|c} -3 & 2 & 1 & 2 \\ 1 & 0 & 3 & 1 \\ 0 & 1 &-2 & 3 \end{array}\right) $$ If you do so, you will find that the only solution is $c_1=\frac{17}{14}$, $c_2=\frac{20}7$, $c_3=-\frac1{14}$.
This gives you $\vec u_1=\underline{\underline{\frac1{14}(29,17,40)}}$ and $\vec u_2=\frac1{14}(-1,-3,2)$.

Projection to $W^\bot$. As mentioned in a comment, since $W^\bot$ is one-dimensional, it is easy to find projection to $W^\bot$. The vector $\vec a=\frac1{\sqrt{14}}(1,3,-2)$ is unit vector which spans $W^\bot$. The projection can be found as $$\vec u_2 = \vec u \vec a^T \vec a =\frac1{14} (2,1,3)\begin{pmatrix}1\\3\\-2\end{pmatrix}(1,3,-2)=-\frac1{14}(1,3,2).$$

Then the projection to the subspace $\vec u_1$ can be computed as $\vec u_1=\vec u-\vec u_2$.

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  • $\begingroup$ Thank you :) Exactly the kind of detail I was looking for. $\endgroup$ – ಠ_ಠ Feb 22 '16 at 17:04
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    $\begingroup$ @jia Note that you don’t always need to solve the full system of equations. It’s sometimes easy to find the projection $\vec u_2$ of $\vec u$ onto $W^\perp$, after which you can find $\vec u_1$ by subtraction. In this case, $W^\perp$ is one-dimensional, and the orthogonal projection of $\vec u$ onto that subspace is $\frac{\vec u\cdot\vec n}{\vec n\cdot\vec n}\vec n$, where $\vec n=(1,3,-2)$, the normal to the plane $W$. $\endgroup$ – amd Feb 22 '16 at 19:07

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