Calculate the sine of an angle in coordenate geometry Given two lines $a_1x+b_1y+c_1 = 0$, $a_2x+b_2y+c_2 = 0$ that make an angle $\alpha$ at their intersection, show that
$$\sin\alpha = \frac{a_2b_1-a_1b_2}{\sqrt{a_1^2+b_1^2}\sqrt{a_2^2+b_2^2}}$$
So I'm stuck here, I'm thinking making a circle with the intersection point as it center and radius 1 and consider the intersections of the lines with the circle and from there calculating the sine, or making a rough right triangle and do the same as that. But I'm pretty sure there is a better and simpler way to solve this problem.
 A: Going two three dimensions the vector product of two vectors $u$ and $v$ has the property:
$$
\lvert u \times v \rvert 
= \lvert u \rvert \lvert v \rvert \sin \alpha
$$
If one considers vectors
$$
u = (a_1, b_1, 0)^t \quad v = (a_2, b_2, 0)^t
$$
then
$$
u \times v = (0, 0, a_1 b_2 - a_2 b_1)^t
$$
and finally
$$
\sin \alpha 
= \frac{\lvert a_1 b_2 - a_2 b_1 \rvert}{\sqrt{a_1^2+b_1^2}\sqrt{a_2^2+b_2^2}}
$$
A: Hint:
$$a_1x+b_1y+c_1 = 0$$
$$a_2x+b_2y+c_2 = 0$$
$$\implies m_1=-\frac{a_1}{b_1},m_2=-\frac{a_2}{b_2}$$
$$\tan\alpha=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$ 
$$\tan\alpha=\left|\frac{a_1b_2-a_2b_1}{a_1a_2+b_1b_2}\right|$$ 
Then 
$$\sin\alpha=\frac{\left|\frac{a_1b_2-a_2b_1}{a_1a_2+b_1b_2}\right|}{\sqrt{1+
\left(\frac{a_1b_2-a_2b_1}{a_1a_2+b_1b_2}\right)^2
}}=\frac{|a_1b_2-a_2b_1|}{\sqrt{(a_1a_2+b_1b_2)^2+(a_1b_2-a_1b_2)^2}}= \frac{|a_2b_1-a_1b_2|}{\sqrt{a_1^2+b_1^2}\sqrt{a_2^2+b_2^2}}$$
A: The angle between the lines is the same as the angle between the perpendicular lines, so it's the angle between the line through $(0,0)$ and $(a_1,b_1)$ and the line through $(0,0)$ and $(a_2,b_2)$.  Convert to polar coordinates: let
\begin{align*}
a_1 &= r_1\cos\theta_1 & a_2 &= r_2\cos\theta_2 \\
b_1 &= r_1\sin\theta_1 & b_2 &= r_2\sin\theta_2
\end{align*}
Then
$$ \sin\alpha
= \sin(\theta_1-\theta_2)
= \sin\theta_1\cos\theta_2 - \sin\theta_2\cos\theta_1
= \frac{b_1}{r_1}\cdot\frac{a_2}{r_2} - \frac{b_2}{r_2}\cdot\frac{a_1}{r_1}
$$
