If the angle between the line $x=y=cz$ and the plane $z=0$ is $45^\circ$, find all values of $c$.
Before actually doing the calculations I thought I would get $c=1$ since the angle between $x=y=z$ and the $xy$-plane is $45^\circ$.
However, using the facts I know I got a different result:
Denote the asked angle by $\alpha$. Since the direction vector of the line is $(1,1,1/c)$ and the normal vector of the plane is $(0,0,1)$, we have $$\sin\alpha= \frac{(1,1,\frac{1}{c})(0,0,1)}{\|(1,1,\frac{1}{c})\|\|(0,0,1)\|}=\frac{\frac{1}{c}}{\sqrt{2+\frac{1}{c^2}}}=\frac{1}{\sqrt 2}$$ Squaring both sides, we get $$2c^2+1=2 \implies c=\pm \frac{1}{\sqrt 2}$$
Which way is correct?
If $x = z$, and $y = 0$ the line makes a $45^\circ$ angle with the plane. Same for $y = z$ and $x = 0$
This might explain your intuition.
In fact the set of all lines though the origin that form a $45^\circ$ angle with the plane would be the cone. $x^2 + y^2 = z^2$
Restricting $x = y$ then $2x^2 = z^2$
or $x = y =\pm \frac {\sqrt 2}{2} z$
Does this help?