Let be $A:V\to V$ a linear map and $\langle\cdot,\cdot\rangle$ a scalar product on $V.$
We say that $A$ is orthogonal when it preserves the scalar product, i.e.: $\langle Au,Av\rangle=\langle u,v\rangle,\ \forall u,v\in V.$
For the orthogonality of $A$ it is not only necessary but even sufficient that $A$ preserves the associated quadratic form, i.e.: $\langle Au,Au\rangle=\langle u,u\rangle,\ \forall u\in V.$
In order to prove the sufficiency, we can use the Polarization Identity: $\langle u,v\rangle=\frac{1}{4}\left(\langle u+v,u+v\rangle-\langle u-v,u-v\rangle\right)\tag{P.I.},\ \forall u,v\in V.$ Infact, let us assume that $A$ preserves the quadratic form associated to $\langle\cdot,\cdot\rangle,$ i.e.: $\langle Au,Au\rangle=\langle u,u\rangle,\ \forall u\in V,\tag{*}$ then, for any $u,v\in V,$ by the polarization identity we get $\langle Au,Av\rangle\stackrel{P.I.}{=}\frac{1}{4}\left(\langle A(u+v),A(u+v)\rangle-\langle A(u-v),A(u-v)\rangle\right)\\\stackrel{*}=\frac{1}{4}\left(\langle u+v,u+v\rangle-\langle u-v,u-v\rangle\right)\stackrel{P.I.}{=}\langle u,v\rangle,$ that is $A$ preserves the scalar product $\langle\cdot,\cdot\rangle,$ and we have done.