Expression for cosine/sine of the angle of intersection of two lines Question: consider two lines in the plane described in the primitive form
$y = mx+b$, $y=ux+B$.
It is assumed that $m$ is not equal to $u$ so that the lines intersect somewhere. Find an expression for the cosine of the angle that these lines make on intersection.
What exactly is this question asking?
 A: 
The angle formed by a line of the form $y=mx+b$ and the $x$-axis is $\arctan m$. 
The angle formed by a line of the form $y=ux+B$ and the $x$-axis is $\arctan u$. 
Hence the angle formed by the two lines is given by $\arctan m -\arctan u$.
Since $\arctan m -\arctan u = \arctan \frac{m-u}{1+mu}$, and  $\cos (\arctan x) = \frac{1}{\sqrt{1+x^2}}$, we have $\cos( \arctan m -\arctan u ) = \frac{1+mu}{\sqrt{1+m^2} \sqrt{1+u^2}}$.
Alternatively: The lines intersect at some point $(x,y)$ and another point on each line is given by $(x+1,y+m)$, $(x+1,y+u)$ respectively. These form two vectors $(1,m)$ and $(1,u)$, and the inner product gives the desired answer directly as $\cos \theta = \frac{\langle (1,m), (1,u) \rangle}{\|(1,m)\| \|(1,u)\|} = \frac{1+mu}{\sqrt{1+m^2} \sqrt{1+u^2}}$.
A: By geometry, if we translate these lines so that they intersect the origin, angles will be preserved (by isometry). Now, consider these lines as scalar multiples of $(1,m)$ and $(1,v)$. To find the angle between them, we consider the dot product $a \space.b=\mid a \mid \mid b \mid cos(\theta)$. So $(1,m).(1,v)=1+mv$. And, $\mid (1,m) \mid \mid (1,v) \mid=\sqrt{(1+m^2)(1+v^2)}$. Therefore, $cos(\theta)=\frac{1+mv}{\sqrt{(1+m^2)(1+v^2)}}$.
