Angles between straight lines

Two secant straight lines $r$ and $s$ determine four equal angles two by two; this is due to the fact that they are opposite angles in virtue of the apex. The smallest of the angles $\alpha$ and $\beta$ is defined as the angle between the straight lines $r$ and $s$.

In case of the drawing, the angle between the straight lines $r$ and $s$ it would be $\widehat{rx}=b$.

A way of determining the above mentioned angle is from the scalar product of the director vectors of the straight lines $r$ and $s$. Let $\overrightarrow{u}$ and $\overrightarrow{v}$ be director vectors of the straight lines $r$ and $s$ respectively.

The scalar product of the vectors $\overrightarrow{u}$ and $\overrightarrow {v}$ is:$$\overrightarrow{u}\cdot \overrightarrow{v}=|\overrightarrow{u}||\overrightarrow{v}|\cos \widehat{(\overrightarrow{u}, \overrightarrow{v})}$$Now, let's observe that by taking a vector director of $r$ and one of $s$, the angle formed by the above mentioned vectors coincides with the angle between both straight lines, if it is acute, or with its supplementary if it is obtuse:

Therefore, the cosine of the angle between two straight lines will coincide, except for the sign, with that of the angle that its director vectors form, and therefore we have that:$$\cos \widehat{(r,s)}=|\cos \widehat{(\overrightarrow{u}, \overrightarrow{v})}|$$This last step is because $$\cos (a) = - \cos (180 - a)$$ This way, if we isolate in the formula of the scalar product,$$\cos \widehat{(r,s)}=|\cos \widehat{(\overrightarrow{u},\overrightarrow{v})}|=\displaystyle \frac{|\overrightarrow{u} \cdot \overrightarrow{v}|}{|\overrightarrow{u}||\overrightarrow{v}|}$$Note that: The scalar product between two vectors $\overrightarrow{u}=(u_1,u_2)$ and $\overrightarrow{v}=(v_1,v_2)$ is defined as $$\overrightarrow{u} \cdot \overrightarrow{v}=u_1 \cdot v_1+u_2\cdot v_2$$Therefore, if we remember that the expression of the module of a vector is $$\displaystyle |\overrightarrow{v}|=\sqrt{v_1^2+v_2^2}$$We have that in coordinates the expression of the cosine of the angle between two straight lines is:$$\cos \widehat{(r,s)}=|\cos \widehat{(\overrightarrow{u},\overrightarrow{v})}|=\displaystyle \frac{|\overrightarrow{u} \cdot \overrightarrow{v}|}{|\overrightarrow{u}||\overrightarrow{v}|}=\frac{|u_1\cdot v_1+u_2\cdot v_2|}{\sqrt{u_1^2+u_2^2}\sqrt{v_1^2+v_2^2}}$$

Determine the angle formed by the straight lines $r$ and $s$, which equations are, respectively, $3x - 2y - 1 = 0$ and $-x + 2y - 3 = 0$.

Let $\overrightarrow{u}= (2, 3)$ and $\overrightarrow{v} = (2, 1)$ be director vectors of the straight lines $r$ and $s$ respectively.

Then, applying the previous formula we have $$\cos \widehat{(r,s)}=|\cos \widehat{(\overrightarrow{u},\overrightarrow{v})}|=\displaystyle \frac{|\overrightarrow{u} \cdot \overrightarrow{v}|}{|\overrightarrow{u}||\overrightarrow{v}|}=\frac{|u_1\cdot v_1+u_2\cdot v_2|}{\sqrt{u_1^2+u_2^2}\sqrt{v_1^2+v_2^2}}=$$ $$=\displaystyle\frac{|2 \cdot 2+ 3\cdot 1|}{\sqrt{2^2+3^2}\sqrt{2^2+1^2}}=\frac{7}{\sqrt{65}}$$ Therefore, if we take the calculator we have $$\widehat{rs}=\arccos(\cos (\widehat{rs}))=\arccos \Big(\displaystyle \frac{7}{\sqrt{65}}\Big)=29.7^\circ$$