I have a frame $s'$ moving with velocity $v_1$ along the x-direction with respect to a frame $s$. A frame $s''$ is moving with velocity $v_2$ along the y-direction with respect to $s'$.
I want to find the Lorentz transformation matrix from $s$ to $s''$, which I have below, just from multiplying the matrices with the corresponding boosts.
Note: I'm working in c=1 units. The first matrix is the boost from $s'$ to $s''$ in y, and the second is the boost from $s$ to $s'$ in x.
$$\left[\begin{array}{cccc}{\gamma_{2}} & {0} & {-v_{2} \gamma_{2}} & {0} \\ {0} & {1} & {0} & {0} \\ {-v_{2} \gamma_{2}} & {0} & {\gamma_{2}} & {0} \\ {0} & {0} & {0} & {1}\end{array}\right]\left[\begin{array}{cccc}{\gamma_{1}} & {-v_{1} \gamma_{1}} & {0} & {0} \\ {-v_{1} \gamma_{1}} & {\gamma_{1}} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {1}\end{array}\right] = \left[\begin{array}{cccc}{\gamma_{1} \gamma_{2}} & {-v_{1} \gamma_{1} \gamma_{2}} & {-v_{2} \gamma_{2}} & {0} \\ {-v_{1} \gamma_{1}} & {\gamma_{1}} & {0} & {0} \\ {-v_{2} \gamma_{2} \gamma_{1}} & {v_{1} v_{2} \gamma_{1} \gamma_{2}} & {\gamma_{2}} & {0} \\ {0} & {0} & {0} & {1}\end{array}\right]$$
Now let's say that we are working with a "boat problem", so that frame $s$ is a person on a riverbank, $s'$ is the rest frame of water moving with $v_1$ along the x-direction, and $s''$ is the rest frame of a boat traveling with $v_2$ perpendicular to the water's motion. I want to find the angle ($\phi = arctan(y/x)$) that the boat makes with respect to the shore. I know to use my final matrix, and I know that the second and third rows respectively correspond to y and x, but I am not sure how to find the equations for y and x to plug into my angle formula.
