I have eventually end up rotating a point in $\mathbb {R^3}$n .
Let r$=(-1,-1,-1)$ orient the line L it spans, and let $R: \mathbb{R^3} \rightarrow \mathbb{R^3}$ be the rotation of $\mathbb{R^3}$ about $L$ according to the right hand rule along the angle $\theta= \frac{\pi}{6}$.
$a)$ FInd u, the unit vector in the rirection of r.
This was simple. $||u||=\sqrt{3}$ so u$=\frac{1}{\sqrt{3}}(-1,-1,-1)$
$(b)$ Find an orthonormal basis $\text{{$\bf a$,$\bf c$}}$ of the orthogonal compliment of L.
How would I do this? Since I am in $\mathbb{R^3}$, I feel like I could avoid using the gram schmidt process and use the cross product somehow. What does it mean by "orient the line it spans" though? I feel like I wrote the equation of a plane $-x-x-z=0$ or essentially $x+y+z=0$. So Maybe I could find two other points and cross them with each other to get an orthonormal basis?
I am not completely sure about this.
$(c)$ Using the vectors a,c and u, create an ordered orthonormal basis $B=a,b,u$ of $\mathbb{R^3}$, where $b= \pm c$ is chosen so that the triple $(\bf a, \bf b,\bf c)$ has the right hand orientation of $\mathbb{R^3}$
Is this not the same as part b? I am lost on this part... If someone could explain this, I would greatly appreciate it. Thank you!
Suppose that instead of $(-1, -1, -1)$ the vector $r$ was $(0, 3, 0)$. Then the answers would be
(a) $(0, 1, 0)$, as you know.
(b) $a = (1, 0, 0), c = (0, 0, 1)$. Note, though, that $a, c, r$ is not a positively oriented basis.
(c) Since $a, c, r$ is not positively oriented, we know that $a, -c, r$ is positively oriented.
Now you need to rotate by $\pi/6$ in the $a,-c$ plane...but I leave that to you.
Oh...by the way, for part b, there are infinitely many answers. So a good approach is to try to find a vector $a$ with, say, the $z$ coordinate being zero; that reduces your choices a good deal. And then yes, you can find $c$ as a cross product of $a$ and $r$ (or $r$ and $a$, or do something else and then use gram-schmidt)