water speed of a river

Problem:

A boats speed is 1,70 m/s in still water.
It must cross a river with a width of 260 m.
The boats starting point is the origin on the xy-axsis (on the shore).
It has to dock 110 m to the right(in the positive x-direction) opposite of the starting point on the other shore(i.e. the point parallel to the starting point on the other side + 110 m).
The boat must sail in a 45° angle relative to the shore(x-axis) to arrive at that point.

What is the speed of the water current(water flows to the negative x-direction)?

What I have done:

It semms to be a pretty simple vector problem.
Just subtract the vector of the boat in moving water from the vector of the boat in still water(direct route) to get the vector of water flow.

I did this and got a nonzero y component of the water flow, which can't be true. How can it even be zero if only the sin(0°+180°*n)= 0 and the y components of the vectors aren't equal?

Thank you for your help

Let $v_b$ be the speed of boat in still water, $v_r$ is the speed of river. The speed of boat in $x$ direction (in still water at 45 degree angle): $v_x=v_b/\sqrt{2}$, the speed of boat in $y$ direction is $v_y=v_b/\sqrt{2}$. The river speed has negative $x$ direction as you correctly concluded. Thus, we need to subtract it from $v_x$. We have: $(v_b/\sqrt{2}-v_r)\cdot t=110$, $v_b/\sqrt{2} \cdot t=260$ (where $t$ is the time to cross the river). This will give us the equation to find $v_r$: $$\frac{v_b/\sqrt{2}}{v_b/\sqrt{2}-v_r}=\frac{260}{110}$$ Can you complete from here?