Find the acute angle between a line and a plane

First separate the base point and direction:

$\begin{aligned}&\mathbf r\\ & =(1,2,-1)+\lambda(1,-1,1).\end{aligned}$

The line direction is $\mathbf m=(1,-1,1)$ and plane normal is $\mathbf n=(2,-1,1)$. Their dot product is $4$, and their magnitudes are $\sqrt3$ and $\sqrt6$.

Method 1: find the angle to the normal, then its complement. Write that angle as $\theta$:

$\begin{aligned}4&=\sqrt6\sqrt3\cos\theta,\\\theta&=\cos^{-1}\!\left(\frac4{\sqrt{18}}\right)\\&=19.4712206\ldots^\circ.\end{aligned}$

The source gives $19.471^\circ$ to 3 d.p. Hence the line–plane angle is $90^\circ-\theta=70.5287794\ldots^\circ$, or $70.5^\circ$ to 1 d.p.

Method 2: use the line–plane sine formula directly.

$\begin{aligned}\alpha&=\sin^{-1}\!\left(\frac{|\mathbf m\cdot\mathbf n|}{|\mathbf m||\mathbf n|}\right)\\&=\sin^{-1}\!\left(\frac4{\sqrt3\sqrt6}\right)\\&=70.5^\circ\quad\text{(to 1 d.p.).}\end{aligned}$

The source reuses $\theta$ for the line–plane angle in Method 2. Here $\alpha$ distinguishes that angle from Method 1’s angle to the normal; both printed methods and results are preserved. The line’s base point and the plane constant do not affect their angle.