First, for each line we create a parallel vector by choosing two points and forming the vector associated to the directed line segment between them. For
\(\ell_1 \text{,}\) we’ll use intercepts
\(P(-2/3,0) \) and
\(Q(0,-4/3) \text{.}\) The directed line segment between them is
\(\abray{PQ} = (2/3,-4/3) \text{,}\) but let’s clear fractions and use
\(\uvec{v} = (2,-4) \text{.}\) For
\(\ell_2 \text{,}\) the
\(y \)-intercept is obviously
\(R(0,-1) \text{,}\) and by guess-and-check we find a second point
\(S(4,4) \text{.}\) Then a parallel vector is
\(\uvec{w} = \abray{RS} = (4,5) \text{.}\)
Now we can determine one of the angles between the lines by calculating the angle between our two vectors.
\begin{align*}
\cos\theta \amp = \frac{\udotprod{v}{w}}{\unorm{v} \unorm{w}} \\
\amp = \frac{-12}{\sqrt{20} \sqrt{41}} \\
\amp = - \frac{6}{\sqrt{5} \sqrt{41}}
\end{align*}
\begin{equation*}
\theta = \inv{\cos}\left( - \frac{6}{\sqrt{5} \sqrt{41}} \right) \approx 2.00
\end{equation*}
This is an obtuse angle since it is greater than \(\pi / 2 \approx 1.57 \text{.}\) We could compute the complementary acute angle \(\theta' \) by \(\theta' = \pi - \theta \text{,}\) but if we aren’t careful we may introduce round-off error. We can just as easily repeat our calculation with \(\uvec{v} \) replaced by \(- \uvec{v} \text{.}\) Since \(\dotprod{(- \uvec{v})}{\uvec{w}} = -(\udotprod{v}{w}) \) and \(\norm{-\uvec{v}} = \unorm{v} \text{,}\) we have
\begin{equation*}
\theta' = \inv{\cos}\left( \frac{6}{\sqrt{5} \sqrt{41}} \right) \approx 1.14 \text{.}
\end{equation*}
