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 $α$ and $β$ is defined as the angle between the straight lines $r$ and $s$ .

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

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 $\vec{u}$ and $\vec{v}$ be director vectors of the straight lines $r$ and $s$ respectively.

The scalar product of the vectors $\vec{u}$ and $\vec{v}$ is: $\vec{u} \cdot \vec{v} = | \vec{u} | | \vec{v} | \cos \hat{(\vec{u}, \vec{v})}$ 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:

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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: $\cos \hat{(r, s)} = | \cos \hat{(\vec{u}, \vec{v})} |$ This last step is because $\cos (a) = - \cos (180 - a)$ This way, if we isolate in the formula of the scalar product, $\cos \hat{(r, s)} = | \cos \hat{(\vec{u}, \vec{v})} | = \frac{| \vec{u} \cdot \vec{v} |}{| \vec{u} | | \vec{v} |}$ Note that: The scalar product between two vectors $\vec{u} = (u_{1}, u_{2})$ and $\vec{v} = (v_{1}, v_{2})$ is defined as $\vec{u} \cdot \vec{v} = u_{1} \cdot v_{1} + u_{2} \cdot v_{2}$ Therefore, if we remember that the expression of the module of a vector is $| \vec{v} | = \sqrt{v_{1}^{2} + v_{2}^{2}}$ We have that in coordinates the expression of the cosine of the angle between two straight lines is: $\cos \hat{(r, s)} = | \cos \hat{(\vec{u}, \vec{v})} | = \frac{| \vec{u} \cdot \vec{v} |}{| \vec{u} | | \vec{v} |} = \frac{| u_{1} \cdot v_{1} + u_{2} \cdot v_{2} |}{\sqrt{u_{1}^{2} + u_{2}^{2}} \sqrt{v_{1}^{2} + v_{2}^{2}}}$

Example

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

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

Then, applying the previous formula we have $\cos \hat{(r, s)} = | \cos \hat{(\vec{u}, \vec{v})} | = \frac{| \vec{u} \cdot \vec{v} |}{| \vec{u} | | \vec{v} |} = \frac{| u_{1} \cdot v_{1} + u_{2} \cdot v_{2} |}{\sqrt{u_{1}^{2} + u_{2}^{2}} \sqrt{v_{1}^{2} + v_{2}^{2}}} =$ $= \frac{| 2 \cdot 2 + 3 \cdot 1 |}{\sqrt{2^{2} + 3^{2}} \sqrt{2^{2} + 1^{2}}} = \frac{7}{\sqrt{65}}$ Therefore, if we take the calculator we have $\hat{r s} = \arccos (\cos (\hat{r s})) = \arccos (\frac{7}{\sqrt{65}}) = {29.7}^{\circ}$