The acute angle formed by intersecting lines is equal to the acute angle formed by their normals

Question

Whilst reading my geometry notes I came across the statement given without proof:

The acute angle formed by intersecting lines is equal to the acute angle formed by their respective normals

I decided to attempt to prove this statement in $\Bbb R^2$ and got stuck. Suppose line $L_1$ is given by the direction vector $(1, m_1)$, and line $L_2$ by $(1,m_2)$, then comparing the two formulas for $\theta$, the angle between the two lines is:

$cos (\theta) = \frac {1 + m_1m_2} {\sqrt {1+ m_1^2} \sqrt{1+m_2^2} }$

Then this should be equal to $ \frac {1 + \frac {1}{m_1m_2}} {\sqrt {1+ \frac {1} {m_1^2}} \sqrt{1+\frac {1} {m_2^2}} }$ by taking the normal vectors to have negative reciprocal of the respective direction vectors in $L_1,L_2$, but I am not sure how to sure these are equal, or if I have made a mistake.

Furthermore how can I generalize this to $\Bbb R^n$?

Answer 1

There is no need for lengthy formulas. Just remember the meaning of "normal." The normal to a line is perpendicular to it. So if one line is at an angle $\theta_1$ from the $x$-axis, and the other line is at an angle $\theta_2$ from the $x$-axis, their normals are at angles $\theta_1 + \frac\pi2$ and $\theta_2 + \frac\pi2$ from the $x$-axis. The angle between those normals is the difference of those two angles, $$\left(\theta_1 + \frac\pi2\right) - \left(\theta_2 + \frac\pi2\right) = \theta_1 - \theta_2,$$ which is the same as the angle between the two lines.

Answer 2

The two formulae ARE equal - multiply the "normal" one by $1 = \frac{m_1 m_2}{m_1 m_2}$. You get

$$ \frac {1 + \frac {1}{m_1m_2}} {\sqrt {1+ \frac {1} {m_1^2}} \sqrt{1+\frac {1} {m_2^2}} } \frac{m_1 m_2}{m_1 m_2} = \frac {{m_1m_2} +1} {m_1 \sqrt {1+ \frac {1} {m_1^2}} m_2\sqrt{1+\frac {1} {m_2^2}} } = \frac {{m_1m_2} +1} {\sqrt {{m_1^2} +1} \sqrt{{m_2^2} +1} } $$

In $n$ dimensions, by orthonormal transformation you can always choose a coordinate system such that the direction vector is $(1, m_1, 0 , \cdots , 0) $ and likewise for the other vectors. Then, the same formulae hold that you already have.