If I have a given point $A(x_a,y_a,z_a)$ in the horizontal plane $z=1400$ and I rotate the plane in such a way that it is perpendicular to a line that makes an angle of $60^{\circ}$ with $Oz$ axes and its projection makes $45^{\circ}$ and $45^{\circ}$ angles with $Ox$ and $Oy$, what will the new coordinates of the point be?
Here's what I managed to do so far:
I found the vector perpendicular to the plane $\frac{3\sqrt{2}}{4\sqrt{2}} i + \frac{3\sqrt{2}}{4\sqrt{2}} j + \frac{1}{2}k;$ the new plane contains the point $(x_a,y_a,z_a)$, so the new plane equation is: $\frac{3\sqrt{2}}{4\sqrt{2}}(x-x_a)+\frac{3\sqrt{2}}{4 \sqrt{2}} (y-y_a)+\frac{1}{2}(z-z_a)=0.$ I have one equation missing (since $x_{a,new}=x_a$).
Thanks