Talk:Geometry:WAXS 3D

Check

We define:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{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{alignat} }

And solve:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} q^2 & = [ (q_x)^2 + (q_y)^2 + (q_z)^2 ] \\ \left ( \frac{q}{k} \right )^2 d^{\prime 2} & = \begin{alignat}{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{alignat} \\ & = \begin{alignat}{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{alignat} \\ & = \begin{alignat}{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 + d z \sin \theta_g \cos \theta_g + z^2 \cos^2 \theta_g ] \end{alignat} \\ & = ? \\ & = ? \\ & = ? \\ & = 2 d^{\prime 2} - 2 d^{\prime} x \sin \phi_g + 2 d^{\prime} \cos \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ & = 2 d^{\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{alignat} }