Algebraic groups of relative rank 1 Let $K$ be a field, $V$ an $n-$dimensional $K$-vector space and $q: V \to K$ a quadratic form of Witt index $r$. Let $G:=SO(q)$ denote the special orthogonal group associated to $q$. 
Then $G$ is an algebraic $K$-group of $K-$rank $r$. 
If $q'$ is isometric to $q$, clearly $G':=SO(q')$ is isomorphic to $G$. But this is not necessary for $G$ to be isomorphic to $G'$ (take $n=1$).
My question is: How can I produce a lot of nonisomorphic $SO(q)$'s? 
Here "non-isomorphic" should be interpreted either as abstract groups or as algebraic groups.
I'm particularly interested in the case when the Witt rank is 1.  
 A: Assume $n \ge 5$ to avoid interference with exceptional isomorphisms among small Dynkin diagrams.  The case of odd $n$ should be easier, since in that case (with $n \ge 5$) there are no nontrivial diagram automorphisms and the group is adjoint (so it is its own automorphism scheme).  The problem is naturally viewed in terms of studying degree-1 Galois cohomology for automorphism schemes of non-degenerate quadratic spaces (i.e., orthogonal groups, and special orthogonal groups) and automorphism schemes of these automorphism schemes (e.g., automorphism scheme of ${\rm{SO}}(q)$, which we just saw is ${\rm{SO}}(q)$ itself via conjugation when $n$ is odd and $n \ge 5$).  So you should look in Serre's book on Galois cohomology.  You make no mention of specific $K$ of interest, so below I just make some general remarks. 
Over arithmetically interesting fields such as non-archimedean local fields and global fields (with some possible complications when there is a real place) one can use the result that degree-1 Galois cohomology of a connected and simply connected semisimple group always vanishes (apart from when there is a real place, in which case use Hasse principle instead) to get a handle on the degree-1 cohomology of a connected semisimple group in terms of the degree-2 cohomology of the (finite multiplicative) covering group for its simply connected central cover (spin group in case of special orthogonal groups). That brings in Brauer groups, which one understands for such fields.  Connectedness issues are also likely to intervene, since ${\rm{SO}}(q)$ is just the identity component of the automorphism scheme ${\rm{O}}(q)$ of $(V,q)$.  Serre's book has a detailed discussion of the relationship between $K$-forms of "tensor structures" (such as $(V,q)$) and cohomology of the associated automorphism group scheme, and over interesting fields where one has cohomological vanishing results (especially in the simply connected case) it is possible to use this viewpoint to make interesting conclusions.
A: Fleshing out the Galois cohomology approach suggested by Brian Conrad leads to a clean answer for all $n$, for all fields $K$ of characteristic not $2$, and for all nondegenerate quadratic forms of rank $n$ over $K$.  The answer is exactly what moonface claimed:

Given quadratic forms $q$ and $q'$, the algebraic groups $\operatorname{SO}(q)$ and $\operatorname{SO}(q')$ are isomorphic if and only if $q$ and $q'$ are similar, i.e., $q$ is equivalent to $\lambda q'$ for some $\lambda \in K^\times$.

The key observation is that the homomorphism from $\operatorname{O}_n$ to the automorphism group scheme of $\operatorname{SO}_n$ giving the conjugation action is surjective for all $n$, and the kernel is $\lbrace \pm 1 \rbrace$ for $n>2$.  Then for $n>2$, one has the exact sequence of pointed sets
$$ H^1(K,\lbrace \pm 1 \rbrace) \to H^1(K,\operatorname{O}_n) \to H^1(K,\operatorname{\bf Aut} \operatorname{SO}_n).$$
The first term is $K^\times/K^{\times 2}$, the second term is the set of equivalence classes of nondegenerate rank $n$ quadratic forms, and the third term is the set of isomorphism classes of $K$-forms of $\operatorname{SO}_n$.  The sequence (and its twists - remember that we are dealing with nonabelian cohomology) shows that two quadratic forms give rise to the same $K$-form of $\operatorname{SO}_n$ if and only if they are similar.
If $n=2$, a similar argument applies, though one can also see everything explicitly: every rank $2$ quadratic form is similar to $x^2-dy^2$ for some $d \in K^\times$, and the corresponding $\operatorname{SO}(q)$ is the "Pell equation torus" $x^2-dy^2=1$; both depend just on the image of $d$ in $K^\times/K^{\times 2}$.  I leave the cases $n=1$ and $n=0$ to those who like to think about such things.
More details: When $n$ is odd, we have $\operatorname{O}_n = \lbrace \pm 1 \rbrace \cdot \operatorname{SO}_n$, and all automorphisms of $\operatorname{SO}_n$ are inner.
When $n=2m$ for some $m \ge 2$, we have $-1 \in \operatorname{SO}_n$, so conjugation by an element of $\operatorname{O}_n$ outside $\operatorname{SO}_n$ gives an outer automorphism of $\operatorname{SO}_n$; correspondingly, the $D_m$ Dynkin diagram (interpreted appropriately for small $m$) has an involution.
Why does the rotation of the $D_4$ Dynkin diagram not give an extra outer automorphism when $n=8$?  The covering group between the simply connected form $\operatorname{Spin}_8$ and the adjoint form $\operatorname{PSO}_8$ is $(\mathbb{Z}/2\mathbb{Z})^2$, so there are three intermediate covers, one of which is $\operatorname{SO}_8$, but the rotation permutes these three.
(Thanks to my colleague David Vogan for discussing this with me.)
