Finding the angles a line makes with the coordinate planes

Question

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?

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.