Vtyjkyuk

3.3 Theorem.
Assume that every $3$-dimensional quadratic space over $K$ is isotropic.
Let $phi$ be a regular $n$-dimensional quadratic space.
Then
$$
phi
cong langle delta, 1, dotsc, 1 rangle
qquad
textwhere $delta = det(phi)$.
$$
In particular, two regular quadratic spaces are isometric if and only if they have equal dimension and determinant.
Moreover,
$$
hatW(K)
cong mathbbZ times K^bullet / K^bullet 2
quadtextandquad
W(K)
cong Q(K).
$$

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
beginalign*
(dim, det)
&colon hatW(K)
to mathbbZ times K^bullet / K^bullet 2
\
(e,d)
&colon W(K)
to Q(K)
endalign*

(Original image here.)

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 $langle alpha, beta rangle cong langle gamma, 1rangle$. 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 $(e,d)$?