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I want ask which must be the dot product for vectors (1,1,0), (1,0,1), (0,1,1), so they can form a orthonormalbasis.

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Call the $i$th such vector $v_i$. Let $V$ denote the matrix satisfying $V_{ik}=(v_i)_k$. We seek an inner product $a,\,b\mapsto a_kM_{kl}b_l$, with implicit summation over repeated indices, so that $\delta_{ij}=V_{ik}M_{kl}V_{jl}$ i.e. $VMV^T=I$. So take $M=V^{-1}(V^T)^{-1}=(V^TV)^{-1}$. In this case $$V=\left(\begin{array}{ccc} 1 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 1 \end{array}\right),\,V^{T}V=\left(\begin{array}{ccc} 2 & 1 & 1\\ 1 & 2 & 1\\ 1 & 1 & 2 \end{array}\right),\,M=\frac{1}{4}\left(\begin{array}{ccc} 3 & -1 & -1\\ -1 & 3 & -1\\ -1 & -1 & 3 \end{array}\right).$$

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  • $\begingroup$ thank you, but i have to get a formula for dot product in this case.. $\endgroup$ – melanzana Mar 19 at 20:26
  • $\begingroup$ @melanzana You mean $(3(a_1b_1+a_2b_2+a_3b_3)-a_1b_2-a_2b_1-a_1b_3-a_3b_1-a_2b_3-a_3b_2)/4$? $\endgroup$ – J.G. Mar 19 at 20:39
  • $\begingroup$ yes, thank you very much!!! $\endgroup$ – melanzana Mar 20 at 18:15

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