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

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

• 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? – user137731 Oct 16 '15 at 13:37
• 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.. – user137731 Oct 16 '15 at 13:44
• The questin was to find particular vector. But i want also to know is it possible to find all vectors. Thanks for the help:) – Daniel Yefimov Oct 16 '15 at 13:50
• $(-1,0,1)$ and $(0,1,0)$. – DisintegratingByParts Oct 16 '15 at 15:56

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.