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Consider the vectors $q_1=(1,1,1)$ and $q_3=(1,1,-2)$. I need to find a third vector $q_2$ such that $\{q_1,q_2,q_3\}$ is a arthogonal basis for $\mathbb{R}^3$.

My problem is the following: I did take $v=(1,0,0)$ and I did verify that $\{q_1,q_3,v\}$ is a basis for $\mathbb{R}^3$. Then I did take $$q_2=v-\langle v|q_1\rangle q_1-\langle v|q_3\rangle q_3=(-1,-2,1)$$

And, by Gram-Schmidt process, $q_2$ must be orthogonal to $q_1$ and $q_3$. But, as we can see, it does not happen. So, where is my mistake?

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HINT

Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by

$$q_2=q_1\times q_3$$

As an alternative by GS we have

$$q_2=v-\langle v|\hat q_1\rangle \hat q_1-\langle v| \hat q_3\rangle \hat q_3=(1,0,0)-\frac13(1,1,1)-\frac16(1,1,-2)=\left(\frac12,-\frac12,0\right)$$

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  • $\begingroup$ I don't know why I don't use cross product before, It is a simple way to solve the problem. $\endgroup$ – Gödel Nov 22 '18 at 17:10
  • $\begingroup$ @Gödel Yes indeed it is the faster way in that case if we can't see that by inspection. $\endgroup$ – gimusi Nov 22 '18 at 17:18
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None of $q_1$ and $q_2$ are normalized. Hence, the formula would be $$ q_2=v-{\langle v|q_1\rangle \over \langle q_1|q_1\rangle} q_1-{\langle v|q_3\rangle \over \langle q_3|q_3\rangle} q_3 = ({1 \over 2},-{1 \over 2},0)$$.

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$$q_2=v-\frac{\langle v|q_1\rangle}{\langle q_1|q_1\rangle} q_1-\frac{\langle v|q_3\rangle}{\langle q_3|q_3\rangle} q_3$$

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