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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?

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1 Answer 1

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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?

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  • $\begingroup$ Yes, it does. Thanks! $\endgroup$
    – user775075
    Jul 21, 2020 at 18:34

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