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$\mathcal{B}=\{v_1,v_2,v_3\}$ is an orthonormal basis, find all the vectors $v\in \mathbb{R}^3$ such that $\langle v, v_1+v_2-v_3\rangle=0$.

I've done the following: I've expanded $$\langle v, v_1+v_2-v_3\rangle=0$$

using the linearity property

$$\langle v,v_1\rangle+\langle v,v_2\rangle+\langle v,-v_3\rangle=0$$

I know that there are the vectors $v_{n\in\{1,2,3\}}=\{(0,0,1),(0,1,0),(1,0,0)\}$, but I guess that there are other vectors that suffice the conditions, the given basis is required to be orthonormal, then it should consist of unit vectors such that $|v_n|=1$. How do I find all the other vectors such that $|v_n|=1$?

From here I have no clue on how to proceed, I know that perhaps there may be a way to do it via parametrization. I've done some probing and I'm thinking that there are a lot of values that could do it, but I'm unable to think of a way to do it.

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  • $\begingroup$ I think the question asks you to find a basis for the orthogonal complement of $\operatorname{span}(v_1+v_2-v_3)$, i.e. you need to find a basis of the subspace that consists of all vectors orthogonal to $v_1+v_2-v_3$. $\endgroup$ – user1551 May 8 '14 at 19:58
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I'm not exactly sure why you're looking for other vectors with norm $1$; there are infinitely many such vectors, and there are a lot of choices for an orthonormal basis of $\mathbb{R}^3$ (in fact, unless the inner product is explicitly specified, it's not guaranteed that $(1, 0, 0)$ is a unit vector with respect to $\langle \cdot, \cdot \rangle$).


Here's a way to proceed in a fairly simple manner: Since $\mathcal{B}$ is a basis of $\mathbb{R}^3$, write $$v = c_1 v_1 + c_2 v_2 + c_3 v_3$$

Convince yourself, after using the linearity of $\langle \cdot, \cdot \rangle$ and orthonormality of the basis that

$$\langle v, v_1 + v_2 - v_3\rangle = 0 \iff c_1 + c_2 - c_3 = 0$$

To see this, use linearity in both of the two coordinates, and pull the constants $c_i$ outside of the inner product.

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  • $\begingroup$ How does this address the OP's question "How do I find all the other vectors such that $|v_n|=1$?" $\endgroup$ – Fly by Night May 8 '14 at 20:03
  • $\begingroup$ @FlybyNight I've edited my answer to hopefully address that; I don't see immediately how it's useful to try finding vectors with norm $1$ here. $\endgroup$ – user61527 May 8 '14 at 20:08
  • $\begingroup$ I don't see how it's useful either. What is useful is to address the OP :) $\endgroup$ – Fly by Night May 8 '14 at 20:20
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    $\begingroup$ @IgäriaMnagarka No, it means that every vector in the basis has length $1$, in the sense that $$\sqrt{\langle v_i, v_i \rangle} = 1$$ $\endgroup$ – user61527 May 10 '14 at 19:43
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    $\begingroup$ Yes, which is why orthogonality is important here. So we have $\langle v_i, v_i \rangle = 1$ and $\langle v_i, v_j\rangle = 0$ if $i \ne j$. $\endgroup$ – user61527 May 10 '14 at 19:52

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