Talk:Geometry:WAXS 3D

Check

We define:

${\begin{alignedat}{2}d^{\prime }&={\sqrt {x^{2}+d^{2}+z^{2}}}=\|\mathbf {v} _{1}\|\\(v_{2y})&=(d\cos \theta _{g}-z\sin \theta _{g})\\(v_{2y})^{2}&=(d\cos \theta _{g}-z\sin \theta _{g})^{2}\\&=d^{2}\cos ^{2}\theta _{g}-dz\cos \theta _{g}\sin \theta _{g}+z^{2}\sin ^{2}\theta _{g}\end{alignedat}}$

And solve:

${\begin{alignedat}{2}q^{2}&=[(q_{x})^{2}+(q_{y})^{2}+(q_{z})^{2}]\\\left({\frac {q}{k}}\right)^{2}d^{\prime 2}&={\begin{alignedat}{2}[&\left(x\cos \phi _{g}-\sin \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})\right)^{2}\\&+\left(x\sin \phi _{g}+\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})-d^{\prime }\right)^{2}\\&+\left(d\sin \theta _{g}+z\cos \theta _{g}\right)^{2}]\end{alignedat}}\\&={\begin{alignedat}{2}[&\left(x\cos \phi _{g}-\sin \phi _{g}(v_{2y})\right)^{2}\\&+\left(x\sin \phi _{g}+\cos \phi _{g}(v_{2y})-d^{\prime }\right)^{2}\\&+\left(d\sin \theta _{g}+z\cos \theta _{g}\right)^{2}]\end{alignedat}}\\&={\begin{alignedat}{2}[&x^{2}\cos ^{2}\phi _{g}-x\cos \phi _{g}\sin \phi _{g}(v_{2y})+\sin ^{2}\phi _{g}(v_{2y})^{2}\\&+x^{2}\sin ^{2}\phi _{g}+x\sin \phi _{g}\cos \phi _{g}(v_{2y})-d^{\prime }x\sin \phi _{g}\\&+x\sin \phi _{g}\cos \phi _{g}(v_{2y})+\cos ^{2}\phi _{g}(v_{2y})^{2}-d^{\prime }\cos \phi _{g}(v_{2y})\\&-d^{\prime }x\sin \phi _{g}-d^{\prime }\cos \phi _{g}(v_{2y})+d^{\prime 2}\\&+d^{2}\sin ^{2}\theta _{g}+dz\sin \theta _{g}\cos \theta _{g}+z^{2}\cos ^{2}\theta _{g}]\end{alignedat}}\\&=?\\&=?\\&=?\\&=2d^{\prime 2}-2d^{\prime }x\sin \phi _{g}+2d^{\prime }\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})\\&=2d^{\prime }\left(d^{\prime }-x\sin \phi _{g}+\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})\right)\\\left({\frac {q}{k}}\right)^{2}&=2\left(1-{\frac {x\sin \phi _{g}+\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})}{d^{\prime }}}\right)\end{alignedat}}$