The area of parallelogram formed by the lines $$x \cos \alpha +y\sin\alpha = p,$$ $$x \cos\alpha +y\sin\alpha = q,$$ $$x\cos\beta + y\sin\beta = r$$ and $$x\cos\beta + y\sin\beta = s$$ for given values of $p,q,r$ and $s$ is least , if $(α-β) =?$
Find four points of intersection and then minimizing the area does not seem a good approach. Is there any smarter way to do it?
The first two lines are two parallel lines which make angle $\pi/2+\alpha$ with the positive X-axis and are at distances $|p-q|$ from each other.
The other two lines are two parallel lines which make angle $\pi/2+\beta$ with the positive X-axis and are at distance $|r-s|$.
The angle between the two pairs will obviously be $\alpha-\beta$
So if you draw a figure it will be clear that the area $$A=|p-q||r-s||\csc(\alpha-\beta)|$$
So $|\alpha-\beta|$ should be $\pi/2$