Write in canonical form and find the angle between ell_1…
Write in canonical form and find the angle between $\ell_1:\ 2x-y-7=0,\ 2x-z+5=0$ and $\ell_2:\ 3x-2y+8=0,\ z=3x$.
Explanation of Method
Two lines given by planes, their canonical forms, and the angle between them.
Step 1: Canonical forms
\[ \vec d_1=\{2,-1,0\}\times\{2,0,-1\}=\{1,2,2\},\qquad x=0\Rightarrow M_1(0,-7,5) \]
\[\ell_1:\ \frac x1=\frac{y+7}{2}=\frac{z-5}{2}\]
\[ \vec d_2=\{3,-2,0\}\times\{3,0,-1\}=\{2,3,6\},\qquad x=0\Rightarrow M_2(0,4,0) \]
\[\ell_2:\ \frac x2=\frac{y-4}{3}=\frac z6\]
Step 2: Angle
\[ \cos\varphi=\frac{|2+6+12|}{3\cdot7}=\frac{20}{21} . \]
\[\varphi=\arccos\frac{20}{21}\approx17.75^\circ\approx17^\circ45'\]
Result
\[\boxed{\,\ell_1:\ \frac x1=\frac{y+7}{2}=\frac{z-5}{2}\,}\]
\[\boxed{\,\ell_2:\ \frac x2=\frac{y-4}{3}=\frac z6\,}\]
\[\boxed{\,\varphi=\arccos\frac{20}{21}\approx17.75^\circ\approx17^\circ45'\,}\]
Remark
More precisely $\varphi=17.7528^\circ=17^\circ45.2'$; the source rounds to $17^\circ46'$. Note also that the lines are skew, so the angle is the angle between their directions:
\[ \big[\overrightarrow{M_1M_2},\vec d_1,\vec d_2\big]=\begin{vmatrix}0&11&-5\\1&2&2\\2&3&6\end{vmatrix}=-17\ne0 . \]