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Given two orthogonal (orthonomal also) vectors $(1,0,0)^T,(0,\frac{1}{\sqrt{5}},\frac{-2}{\sqrt{5}})^T$

I want to find a third vector $q_3$ such that they become all orthogonal to each other. Of course one should directly apply the gram-schimdt process. But by my books definition is as follows :

Suppose $(\mathbf{u}_1, \mathbf{u}_2, \dots \mathbf{u}_n)$ is a basis for a finite dimensional inner product space $V$. Let \begin{eqnarray*} \mathbf{v}_1 &=& \mathbf{u}_1,\\ \mathbf{v}_2 &=& \mathbf{u}_2 -\frac{\langle \mathbf{u}_2,\mathbf{v}_1\rangle}{\langle \mathbf{v}_1,\mathbf{v}_1\rangle}\mathbf{v}_1,\\ \mathbf{v}_3 &=& \mathbf{u}_3 -\frac{\langle \mathbf{u}_3,\mathbf{v}_1\rangle}{\langle \mathbf{v}_1,\mathbf{v}_1\rangle}\mathbf{v}_1-\frac{\langle \mathbf{u}_3,\mathbf{v}_2\rangle}{\langle \mathbf{v}_2,\mathbf{v}_2\rangle}\mathbf{v}_2,\\ \vdots\\ \mathbf{v}_n &=& \mathbf{u}_n -\frac{\langle \mathbf{u}_n,\mathbf{v}_1\rangle}{\langle \mathbf{v}_1,\mathbf{v}_1\rangle}\mathbf{v}_1-\frac{\langle \mathbf{u}_n,\mathbf{v}_2\rangle}{\langle \mathbf{v}_2,\mathbf{v}_2\rangle}\mathbf{v}_2-\cdots-\frac{\langle \mathbf{u}_n,\mathbf{v}_{n-1}\rangle}{\langle \mathbf{v}_{n-1},\mathbf{v}_{n-1}\rangle}\mathbf{v}_{n-1} \end{eqnarray*} Then $\{\mathbf{v}_1, \mathbf{v}_2,\dots,\mathbf{v}_n\}$ is an orthogonal basis for $V$ .

So a question popped out, my aim is to find $\mathbf{v}_3$, since I already know what my $\mathbf{v}_1,\mathbf{v}_2$ are, It suffices to just choose my $\mathbf{u}_3$, so when choosing my $\mathbf{u}_3$, should I just choose $\mathbf{u}_3$ to be any vector from a basis that forms $\mathbb{R}^3$ (avoiding $(1,0,0)^T$ in this case) or is there a systematic way to choose $\mathbf{u}_3$?

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  • $\begingroup$ The idea of the Gram-Schmidt process is that any basis can be turned into an orthogonal basis through this process and so it would be natural to choose a linearly independent vector $\textbf{u}_3$ at "random". However in Your case You might just set $\textbf{u}_3=\textbf{u}_1\times \textbf{u}_2$ that is already orthogonal $\endgroup$ – Peter Melech Sep 23 '18 at 10:03
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In such cases, you normally apply the Gram-Schmidt process when have got a triple of vectors and want to construct an orthonormal basis. Here you have only two vectors.

I would introduce a new vector $(a,b,c), a,b,c \in \mathbb{R}$, and start finding the conditions for which is linearly independent with the others, i.e. $$ \det \left( \begin{matrix} 1& 0 & 0\\ 0& 1/\sqrt{5}& -2/\sqrt{5} \\ a & b& c \end{matrix} \right) \neq 0. $$ As you can compute, this gives $c + 2b \neq 0$. Under this condition, you can apply the Gram-Schmidt process and find an orthonormal basis.

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