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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 $$q_2=v-langle v|q_1rangle q_1-langle v|q_3rangle q_3=(-1,-2,1)$$

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?

HINT

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

$$q_2=q_1times q_3$$

As an alternative by GS we have

$$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$

None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
$$ q_2=v-{langle v|q_1rangle over langle q_1|q_1rangle} q_1-{langle v|q_3rangle over langle q_3|q_3rangle} q_3 = ({1 over 2},-{1 over 2},0)$$.

$$q_2=v-frac{langle v|q_1rangle}{langle q_1|q_1rangle} q_1-frac{langle v|q_3rangle}{langle q_3|q_3rangle} q_3$$