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Find an orthonormal basis of the plane $x_1 - 5x_2 - x_3 = 0$

I'm having trouble with this problem.

So I picked the vectors $u_1 = \begin{bmatrix}1\\0\\1\end{bmatrix}$ and $u_2 = \begin{bmatrix}5\\1\\0\end{bmatrix}$ then I found

norm = $\sqrt{1+0+1} = \sqrt{2}$

I also found that norm$_2 = \sqrt{5+1+0} = \sqrt{6}$ so then i thought my orthonomal basis would be

$\begin{bmatrix}\frac{1}{\sqrt{2}}\\0\\\frac{1}{\sqrt{2}}\end{bmatrix}, \begin{bmatrix}\frac{5}{\sqrt{6}}\\\frac{1}{\sqrt{6}}\\0\end{bmatrix}$

but it is saying it is incorrect

Majid
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Yusha
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  • Those two vectors aren't orthogonal. One of them isn't normalized. Furthermore, what is the question? – Aweygan Jul 28 '16 at 04:10

2 Answers2

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An $\color{red}{\textrm{ortho}}\color{blue}{\textrm{normal}}$ set of vectors is $\color{red}{\textrm{orthogonal}}$ and each vector in it is $\color{blue}{\textrm{normalized}}$.

Your set has neither of those properties.

To orthogonalize your set, use the Gram-Schmidt process.

To normalize a vector $\mathbf v$, you divide the vector by its norm: $\frac{\mathbf v}{\|\mathbf v\|}$, where the norm of the vector $\mathbf v = (v_1, \dots, v_n)$ is defined as $\|\mathbf v\|:= \sqrt{{v_1}^2 + \cdots + {v_n}^2}$.

Community
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For a plane it is easy to construct an orthonormal basis with clever adhoc method.

The equation given is $x_1-5x_2-x_3=0$. There is definitely a solution with $x_2=0$ with is $(1,0,1)$. This can be the first vector of an orthonormal basis. (We will normalize it later).

The second vector should also satisfy the given equation and further perpendicular to the first solution. This gives additional constrain $x_1+x_3=0$.

That is,solve 2 eqns in 3 unknowns. Solve this and get any solution vector. Now normalize both the solutions. Done.

P Vanchinathan
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  • Since we’re working in $\mathbb R^3$, we can also find a second basis vector via cross product: $(1,-5,-1)\times(1,0,1)$ and then, of course, normalize. – amd Jul 28 '16 at 18:52