= Solution
Number the <simple roots> so that $\alpha_1$ is short and $\alpha_2$ is long. In a Euclidean plane take
$$
\alpha_1=(1,0),\qquad \alpha_2=(-3/2,\sqrt3/2).
$$
Their squared lengths are $1,3$, and their angle is $150^\circ$. The positive <roots of a root system> are
$$
\alpha_1,\quad\alpha_2,\quad\alpha_1+\alpha_2,\quad2\alpha_1+\alpha_2,\quad3\alpha_1+\alpha_2,\quad3\alpha_1+2\alpha_2;
$$
their negatives give the remaining six roots of the <G2 root system>. The three short positive <roots of a root system> are $\alpha_1$, $\alpha_1+\alpha_2$, and $2\alpha_1+\alpha_2$.
Solving $\langle\omega_j,\alpha_i^\vee\rangle=\delta_{ij}$ gives the <fundamental weights>
$$
\boxed{\omega_1=2\alpha_1+\alpha_2=(1/2,\sqrt3/2),\qquad \omega_2=3\alpha_1+2\alpha_2=(0,\sqrt3).}
$$
In this case both <fundamental weights> are themselves roots; the diagram marks their locations on the two hexagons.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-6-g2-roots.png]
{title=The twelve G2 roots with short-first simple roots and fundamental weights marked on the short and long hexagons}
Let $a=n_1$ and $b=n_2$. Relative to the <simple coroots>, the six positive <coroots>, in the root order above, have coordinates
$$
(1,0),\quad(0,1),\quad(1,3),\quad(2,3),\quad(1,1),\quad(1,2).
$$
For example, $\alpha_1+\alpha_2$ has squared length one, so its <coroot> is $2\alpha_1+2\alpha_2=\alpha_1^\vee+3\alpha_2^\vee$; the analogous length computation gives each other entry. Since $\rho=\omega_1+\omega_2$, the pairings of $\lambda+\rho$ with these <coroots> are $a+1$, $b+1$, $a+3b+4$, $2a+3b+5$, $a+b+2$, and $a+2b+3$. Their values at $a=b=0$ multiply to $1\cdot1\cdot4\cdot5\cdot2\cdot3=120$. The <Weyl dimension formula> therefore yields the <G2 dimension polynomial>
$$
\boxed{\dim L(a\omega_1+b\omega_2)=\frac{(a+1)(b+1)(a+b+2)(a+2b+3)(a+3b+4)(2a+3b+5)}{120}.}
$$
This holds for all nonnegative <integers> $a,b$. In particular it gives <dimensions> $7$ and $14$ for $L(\omega_1)$ and $L(\omega_2)$, and $27$ for $L(2\omega_1)$. Reversing the numbering of the <simple roots> would interchange $a,b$ in the <polynomial>, so the short-first convention matters.
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