Suppose that $(V, \mathbb{R}, +, \cdot)$ satisfies all the axioms of a vector space but (not necessarily) the associativity of the addition. Suppose that an inner product is defined on $V$ (usual axioms make sense if associativity is not assumed).
After that, assume that one can prove that associativity holds for orthogonal vectors, i.e., that $$ u_1 + (u_2 + u_3) = (u_1 + u_2) + u_3 $$ whenever $u_1, u_2, u_3 \in V$ and $u_i \cdot u_j = 0$ for $i \neq j$. Can we conclude that $V$ is a vector space, i.e., that associativity is guaranteed on all vectors?
Why such a question? When one wants to convince graphically that associativity holds for Euclidean vectors in space, the picture is messy for vectors in general position. But the picture can convince anyone as soon as one can consider only orthogonal vectors.
Once you give yourself the inner product, the associativity of addition is an immediate result by computing $$ \lVert((v_1+v_2)+v_3)-(v_1+(v_2+v_3))\rVert^2= \langle ((v_1+v_2)+v_3)-(v_1+(v_2+v_3)), ((v_1+v_2)+v_3)-(v_1+(v_2+v_3))\rangle $$ and peel the parentheses \begin{align*} RHS&=\langle (v_1+v_2)+v_3, ((v_1+v_2)+v_3)-(v_1+(v_2+v_3))\rangle\\ &\quad-\langle v_1+(v_2+v_3),((v_1+v_2)+v_3)-(v_1+(v_2+v_3))\rangle\\ &=[\langle v_1+v_2, ((v_1+v_2)+v_3)-(v_1+(v_2+v_3))\rangle+\langle v_3, ((v_1+v_2)+v_3)-(v_1+(v_2+v_3))\rangle]\\ &\quad-[\langle v_1,((v_1+v_2)+v_3)-(v_1+(v_2+v_3))\rangle+\langle v_2+v_3, ((v_1+v_2)+v_3)-(v_1+(v_2+v_3))\rangle]\\ &=\left[\sum_{i=1}^3\langle v_i, ((v_1+v_2)+v_3)-(v_1+(v_2+v_3))\rangle\right]\\ &\quad-\left[\sum_{i=1}^3\langle v_i,((v_1+v_2)+v_3)-(v_1+(v_2+v_3))\rangle\right]\\ &=0. \end{align*} So $((v_1+v_2)+v_3)-(v_1+(v_2+v_3))=0$, i.e., $(v_1+v_2)+v_3=v_1+(v_2+v_3)$.