I am stuck on a question that involves vectors in a plane.
Show that the planes $ax+az=c$ and $bx-by=d$, where $a,b,c,d\in \mathbb{R}, a,b \ne 0$, always form an angle of $\pi/3 = 60^\circ$.
The link to solution to this problem is here (I do not understand it): https://asxtronomical.tumblr.com/post/626215701033664512 I would appreciate it if someone can explain to me step-by-step how to get to the solution.
The main idea would be to note that the angle between planes is the same as the angle between their normal vectors, which are $(a,0,a)$ and $(b,-b,0)$.
Now for any vectors $\vec{x}, \vec{y}$, the key to the angle $\theta_{xy}$ between them is in their dot product, which can be computed coordinate-wise, but is also given by $$ \vec{x} \cdot \vec{y} = \left|\vec{x}\right| \left|\vec{y}\right| \cos\theta_{xy}. $$ You have $\vec{x} = (a,0,a), \vec{y} = (b,-b,0)$. Can you find $\theta_{xy}$?