3.3 Theorem. Assume that every $3$-dimensional quadratic space over $K$ is isotropic. Let $\phi$ be a regular $n$-dimensional quadratic space. Then $$ \phi \cong \langle \delta, 1, \dotsc, 1 \rangle \qquad \text{where $\delta = \det(\phi)$}. $$ In particular, two regular quadratic spaces are isometric if and only if they have equal dimension and determinant. Moreover, $$ \hat{W}(K) \cong \mathbb{Z} \times K^\bullet / K^{\bullet 2} \quad\text{and}\quad W(K) \cong Q(K). $$
Proof: Since $\langle \alpha, \beta, -1 \rangle$ is isotropic this space splits off a hyperbolic plane and thus $\langle \alpha, \beta, -1 \rangle \cong \langle \gamma, 1, -1 \rangle$. By the cancellation law $\langle \alpha, \beta \rangle \cong \langle \gamma, 1 \rangle$. Comparing determinants we see that for any $2$-dimensional form $\langle \alpha, \beta \rangle \cong \langle \alpha \beta, 1 \rangle$. Using this and proceeding inductively we get $$ \langle \alpha_1, \dotsc, \alpha_n \rangle \cong \langle \alpha_1 \dotso \alpha_n, 1, \dotsc, 1 \rangle. $$ This proves the first assertion. The second is an immediate consequence. We leave it to the reader to verify that \begin{align*} (\dim, \det) &\colon \hat{W}(K) \to \mathbb{Z} \times K^\bullet / K^{\bullet 2} \\ (e,d) &\colon W(K) \to Q(K) \end{align*}
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This is excerpt from W. Scharlau: Quadratic and Hermitian Forms, page 38.
For defination of $e, d$ and other maps use page 35.
First of all I don't understand why they use specific three dimensional space as $\langle \alpha, \beta , -1 \rangle$ . Now since we know the above space is isotropic we can split it into hyperbolic plane. Also after that by cancellation law we get $\langle \alpha, \beta \rangle \cong \langle \gamma, 1\rangle$. After that how do we know from looking at the determinant that $\langle \alpha, \beta \rangle \cong \langle \alpha \beta, 1 \rangle$? After that how do we get isomorphisms by the maps $(\dim,\det)$ and $(e,d)$?