The line $\vec r=2\hat i-\hat j+3 \hat k+\lambda (\hat i+\hat j + \sqrt{2}\hat k)$ is defined to make angles $\alpha, \beta$ and $\gamma$ with the xy, yz, and the zx planes respectively.

I simplified the directional vector of the line to $2(\frac {\hat i}{2}+\frac{\hat j}{2} + \frac{\hat k}{\sqrt2})$, and thought of the angle the line makes with each coordinate plane as the complement of the angle it makes with the coordinate axis normal to it. That implies that the directional cosines give you the sines of the required angles. With this, I concluded that $$\alpha=45°, \beta= 30° , \gamma=30°$$ But apparently I've got $\alpha$ and $\gamma$ mixed up. Where did I mess up here?

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    $\begingroup$ The angles with the $xz$- and $yz$-planes have to be the same. I suspect that the person who wrote the problem screwed up. By the way, your title says AXES, not PLANES. If you switch to axes, that solves the problem. $\endgroup$ Apr 24, 2021 at 2:48
  • $\begingroup$ @TedShifrin: good catch with the title, but how does that solve it? Is the book-answer still wrong? $\endgroup$
    – harry
    Apr 24, 2021 at 3:01
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    $\begingroup$ If you switch to angles with the coordinate axes, then the 45° one is in the right slot, but the 30°:should be 60°. The problem is a mess. Don’t stress over it. $\endgroup$ Apr 24, 2021 at 3:08
  • $\begingroup$ @TedShifrin: yes, I get that those are the angles with the axes. My initial title was wrong- it is the line-plane that the question asks. Either way, it seems like the questions's wrong, so I'll leave it at that. Thanks! $\endgroup$
    – harry
    Apr 24, 2021 at 3:21

1 Answer 1


In general, the angle between a line with direction $\mathbf v$ and a plane with normal $\mathbf n$ is $$\theta=\arcsin\left|\frac{\mathbf v \cdot\mathbf n}{\left|\mathbf v\right| \,\left|\mathbf n\right|} \right|.$$

Since the $xy-,$ $yz-, zx-$ planes have respective normals $\begin{pmatrix}0\\0\\1\end{pmatrix},\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix},$ as pointed out by Ted, your own answers are indeed correct.

P.S. The above formula has a similar derivation as yesterday's formula for the angle between two planes.

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    $\begingroup$ Haha, it's getting to when there's a 'yesterday's derivation', isn't it? Great answer as always. $\endgroup$
    – harry
    Apr 24, 2021 at 6:49
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    $\begingroup$ @HarryHolmes Hehe, actually, technically there isn’t a “yesterday’s derivation”, only a “yesterday’s formula”. I’ve omitted today’s derivation too, partly as it’s easy (and instructive) for you to derive (or look up). $\endgroup$
    – ryang
    Apr 24, 2021 at 7:03
  • $\begingroup$ Hi. Maybe you can help me with this: math.stackexchange.com/questions/4227486/… $\endgroup$
    – Gabriel
    Aug 19, 2021 at 15:36

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