# If the angle between the line $x=y=cz$ and the plane $z=0$ is $45^\circ$, find all values of $c$.

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 $$x = 0$$

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 $$x = y =\pm \frac {\sqrt 2}{2} z$$