$SO(3)$ contains no maximal tori In Linear Algebraic Groups I (Stanford, Winter 2010) (http://virtualmath1.stanford.edu/~conrad/252Page/handouts/alggroups.pdf ) course notes, there appears the theorem that if $G$ is a linear algebraic smooth connected group over $k$, then all $k$ split torus are conjugate.
In remark 4.2, They write about this not being true if we considered nonsplit toruses, claiming that this term may not be even well defined! They claim that in $SO(3)/\operatorname{Spec}(\mathbb{R})$, there is no maximal tori (while there is a maximal split one).
I don't understand how there is no maximal one, don't dimension considerations contradict this?
Exact quotes:

Theorem 4.2.1. If $G$ is any smooth connected linear algebraic $k$-group, all maximal $k$-split tori $T \subset G$ are $G(k)$-conjugate.


Remark 4.2.2. ... The theorem is false if “maximal $k$-tori” were to replace “maximal $k$-split tori”. And note that it is trivial (by dimension reasons) that maximal $k$-split tori always exist; in contrast, there may be no $k$-split maximal $k$-tori...

I assumed the latter maximal was a typo and they meant there may be no maximal $k$-tori.
 A: I think you have severely misread the remark.
Let us start by clarifying what maximal torus means, so that it is indeed obvious that maximal tori exist by a dimension argument.

Definition: Let $G$ be a reductive group over a field $K$. Then, a torus $T$ is maximal if it is not properly contained in a larger
torus.

This definition could, a priori, lead to some strangeness because for instance, one might have to maximal tori which don’t have the same dimension.
Thankfully, one has the following result.

Theorem (Grothendieck): Let $G$ be a reductive group over a field $K$, and let $T$ be a maximal torus of $G$. Then, $T_{\overline{K}}$ is a maximal torus of $G_{\overline{K}}$.

The reason this is significant is the other theorem you mention (note I am not presenting these results in the optimal pedagogical order).

Fact(Grothendieck): Let $G$ be a reductive group over a field $K$, then all maximal (with respect to inclusion) $K$-split tori are conjugate.

In particular, if $K$ is algebraically closed then all maximal tori are conjugate. Thus, if $T_1$ and $T_2$ are maximal tori for $G$ over $K$ (not necessarily algebraically closed) then $(T_1)_{\overline{K}}$ and $(T_2)_{\overline{K}}$ are maximal tori of $G_{\overline{K}}$ by the Theorem and thus are conjugate by the fact, and thus have the same dimension.
Thus, in fact, we see that a torus $T$ of $G$ is maximal if and only if it has maximal dimension (amongst tori) and so, in particular, it’s clear that maximal tori must always exist.
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Now, what is true, and remarked upon in Brian’s notes, is that rarely are all maximal tori of a group $G$ conjugate if $K$ is not algebraically closed. For instance, if $K=\mathbb{R}$ this happens if and only if $G/Z(G)$ is anisitropic (i.e. that $G/Z(G)$ contains no split tori, or equivalently that $(G/Z(G))(\mathbb{R})$ is compact).
Example: If one thinks about $\mathrm{GL}_{n,K}$, the general linear group over $K$, then its maximal tori are all of the form $\mathrm{Res}_{E/K}\mathbb{G}_{m,E}$ where $E/K$ is an etale algebra of degree $n$. Evidently these cannot, in general, be isomorphic let alone conjugate. For instance, if $n=2$ and $K=\mathbb{R}$ the conjugacy classes of maximal tori in $\mathrm{GL}_2$ are $2$, one corresponding to $\mathrm{Res}_{\mathbb{C}/\mathbb{R}}\mathbb{G}_{m,\mathbb{C}}$ (a torus of split rank $1$ and dimension $2$) and $\mathrm{Res}_{\mathbb{R}\times\mathbb{R}/\mathbb{R}}\mathbb{G}_{m,\mathbb{R}\times\mathbb{R}}\cong \mathbb{G}_{m,\mathbb{R}}^2$ (a torus of split rank $2$, and dimension $2$).
Random fact: The conjugacy classes of maximal tori in $G$ over $K$ can be classified in terms of group cohomology. Namely, there is a natural bijection of pointed sets between the maximal tori of $G$ and
$$\ker(H^1(K,N_G(T))\to H^1(K,G))$$
where $T$ is any given maximal torus. You can use this to verify my claims in the previous example.
So, in closing every reductive group has a maximal torus, and both notions coincide. The maximal tori of $\mathrm{SO}(n)$ are described in general, on Wikipedia (NB: since $\mathrm{SO}(n)$ is anisotropic, its maximal tori are precisely the Zariski closures of the maximal tori of the compact Lie group $\mathrm{SO}(n)(\mathbb{R})$).
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