91Ó°ÊÓ

The solution to the given problem involves the contraction of length in the horizontal and vertical direction for both frames. These lengths of both frames are compared for the solution.

The formula to calculate the component of length is given as:

$$\begin{gathered} L_{{}_{x^{\text{'}}}}=L_{0}cos\mathrm{θ} \\ {L}_{y^{\text{'}}}=L_{0}sin\mathrm{θ} \end{gathered}$$

Here, $L_0$ is the length of object and $\mathrm{θ}$ is the angle between frames.

The component of the length in frame S is given as:

$$\begin{gathered} L_x=L_0\cos {\mathrm{θ}}_{\sqrt{1-\frac{v^2}{c^2}}} \\ {L}_y=L_{0}sin\mathrm{θ} \end{gathered}$$

The x-component of length contract along direction of object while the y-component remains constant.

The angle of the rod between both frames is given as:

$$\begin{gathered} \tan \mathrm{Î\pm }=\frac{L_y}{L_x} \\ \tan \mathrm{Î\pm }=\frac{L_0\sin \mathrm{θ}}{L_0\cos {\mathrm{θ}}_{\sqrt{1}}\sqrt{1-\frac{v^2}{c^2}}} \\ \tan \mathrm{Î\pm }=\mathrm{γ}\tan \mathrm{θ} \end{gathered}$$

Here, $\mathrm{γ}$ is the Lorentz factor.

The above expression shows that y-component of object in both frames remains same and x-component only contracted. The origins of both frames at constant distance along y-direction.

Therefore, the origin of both frames constant along y-direction as they cross each other.