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Find two vectors $\mathbf{v}$ and $\mathbf{w}$ that are perpendicular to $(1,0,1)$ and to each other.

It is not to hard to find the particular vectors, but how can i find all vectors, that fit to the problem?

Can you help me please to explain it step by step. Thank you

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    $\begingroup$ I don't know what you mean. If you can find $v$ and $w$, then the space of all vectors perpendicular to $(1,0,1)$ will just be $\operatorname{span}(v,w)$. But the question doesn't appear to ask for that. It only wants $v$ and $w$. Are you sure you can find those vectors? $\endgroup$
    – user137731
    Oct 16, 2015 at 13:37
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    $\begingroup$ Does this question help you? I see your edit but I don't know what you're looking for. The question only asks for two vectors $v$ and $w$ that fit some conditions. If you can find those then you're done. The space of all vectors orthogonal to $(1,0,1)$ is a plane and once you've found $v$ and $w$ that plane will just be $\operatorname{span}(v,w)$. So if you choose any vector in that plane as $v'$ then take the cross product of $v'$ and $(1,0,1)$ to get $w'$ which'll also be solutions.. $\endgroup$
    – user137731
    Oct 16, 2015 at 13:44
  • $\begingroup$ The questin was to find particular vector. But i want also to know is it possible to find all vectors. Thanks for the help:) $\endgroup$
    – Splash
    Oct 16, 2015 at 13:50
  • $\begingroup$ $(-1,0,1)$ and $(0,1,0)$. $\endgroup$
    – Disintegrating By Parts
    Oct 16, 2015 at 15:56

1 Answer 1

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Hint: There is a unique plane in $\mathbb R^3$ to which the given vector ${\mathbf n}=(1,0,1)$ is normal. Just choose any orthogonal basis for the plane.

One way to do this is to rotate $\mathbf n$ through an angle $\pi/2$ in any direction, then form the cross product of the result and the original vector to get a third vector orthogonal to both of the others.

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  • $\begingroup$ letting v= (x, y, z), v is orthogonal to (1, 0, 1) if and only if x+ z= 0. One such vector is (0, 1, 0). Any vector, (x, y, z) orthogonal to that must satisfy y= 0. (1, 0, -1) satisfies both so one answer to your question is "(0, 1, 0) and (1, 0, -1)". Since those are obviously independent, the two-dimensional space of all such vectors is a(0, 1, 0)+ b(1, 0, -1)= (b, a, -b) for all numbers a and b. $\endgroup$
    – user247327
    Oct 16, 2015 at 13:53

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