Why is $\cos\theta$ used in three-dimensional geometry in place of $\tan\theta$ or $\sin\theta$?

Question

While going through the introduction of 3D geometry, I'm finding the discussion starts with $\cos\theta$ called "direction cosines".

My query is, in 2D geometry, we find simpler and clear to understand equations of a line as $y= mx+ c$, so

1. Why all of a sudden we introduce $\cos\theta$ and also take help of a parallel vector which is not a part of the line?

2. Why do we depend on a vector introduction? Can not the equation of a line be derived using its own points?

3. Why can not it be used with $\tan\theta$ or $\sin\theta$ notations?

4. Why is it that the equation $ax+by+cz+ d=0$, which looks like the equation of a straight line, is the equation of a plane?

5. Due to all this 3D happens to be inaccessible to understand in the introduction itself. Why it becomes so complex?

6. Can 2D geometry be done using cosine or sine functions? or is $\tan\theta$ fixed for it, and $\cos\theta$ is fixed for 3D geometry?

Thanks.

Answer 1

I count six questions there. Taking #4, the reason $ax + by + cz = d$ is a plane is because we are in three dimensions. Each linear equation is a constraint, and reduces the dimension of the space by one. So when you are in two dimensions, one equation reduces it to a line. When you are in three dimensions, one equation reduces it to a plane. In higher dimensions one linear equation restricts to an $n-1$ dimensional subspace, and is called a "hyperplane".

If you start with the equation of a line in two dimensions, say $y = 2x + 3$, if you move to three dimensions and write the same equation, that equation has no restriction on z. So if you take the line in the x-y plane in 3D and move it directly up and down to different values of z, you will get a plane parallel to the z axis. Tilted planes require the more general $ax + by + cz = d$.

Answer 2

There are numerous formulations for lines in 2D and 3D. Some have direct analogies and some do not.

Here is a sampling of different ways to define a line in both 2D and 3D.

$$\begin{array}{r|l|l} \text{formulation} & \text{2D} & \text{3D}\\ \hline \text{ray from point} & \begin{pmatrix}x\\ y \end{pmatrix}=\begin{pmatrix}r_{x}\\ r_{y} \end{pmatrix}+t\,\begin{pmatrix}e_{x}\\ e_{y} \end{pmatrix} & \begin{pmatrix}x\\ y\\ z \end{pmatrix}=\begin{pmatrix}r_{x}\\ r_{y}\\ r_{z} \end{pmatrix}+t\,\begin{pmatrix}e_{x}\\ e_{y}\\ e_{z} \end{pmatrix}\\ \text{between two points} & \begin{pmatrix}x\\ y \end{pmatrix}=(1-t)\begin{pmatrix}a_{x}\\ a_{y} \end{pmatrix}+t\,\begin{pmatrix}b_{x}\\ b_{y} \end{pmatrix} & \begin{pmatrix}x\\ y\\ z \end{pmatrix}=(1-t)\begin{pmatrix}a_{x}\\ a_{y}\\ a_{z} \end{pmatrix}+t\,\begin{pmatrix}b_{x}\\ b_{y}\\ b_{z} \end{pmatrix}\\ \text{implicit} & e_{x}\,(y-r_{y})+e_{y}\,(r_{x}-x)=0 & \begin{pmatrix}e_{y}\,(z-r_{z})+e_{z}\,(r_{y}-y)=0\\ e_{x}\,(z-r_{z})+e_{z}\,(r_{x}-x)=0\\ e_{x}\,(y-r_{y})+e_{y}\,(r_{x}-x)=0 \end{pmatrix}\\ \text{Plücker} & \begin{pmatrix}m_{z}\end{pmatrix}=\begin{pmatrix}x\\ y \end{pmatrix}\times\begin{pmatrix}e_{x}\\ e_{y} \end{pmatrix} & \begin{pmatrix}m_{x}\\ m_{y}\\ m_{z} \end{pmatrix}=\begin{pmatrix}x\\ y\\ z \end{pmatrix}\times\begin{pmatrix}e_{x}\\ e_{y}\\ e_{z} \end{pmatrix}\\ \text{equation} & a\,x+b\,y+c=0 & \dfrac{x-r_{x}}{e_{x}}=\dfrac{y-r_{y}}{e_{y}}=\dfrac{z-r_{z}}{e_{z}}=t\\ \text{slope-intercept} & y=m\,x+b & ?\\ \text{angle }\varphi\text{-offset }d & x\,\sin\varphi-y\cos\varphi=d & \text{N/A}\\ \text{projective meet} & \begin{pmatrix}x\\ y\\ 1 \end{pmatrix}\cdot\begin{pmatrix}n_{y}\\ n_{x}\\ -d \end{pmatrix}=0 & \begin{aligned}\begin{pmatrix}e_{x}\\ e_{y}\\ e_{z} \end{pmatrix} & =\begin{pmatrix}n_{x}\\ n_{y}\\ n_{z} \end{pmatrix}\times\begin{pmatrix}n_{x}\\ n_{y}\\ n_{z} \end{pmatrix}\\ \begin{pmatrix}m_{x}\\ m_{y}\\ m_{z} \end{pmatrix} & =d_{1}\begin{pmatrix}n_{x}\\ n_{y}\\ n_{z} \end{pmatrix}-d_{2}\begin{pmatrix}n_{x}\\ n_{y}\\ n_{z} \end{pmatrix} \end{aligned} \end{array}$$

In the above list ($x$,$y$ ..) denotes a point on the line, ($r_x$,$r_y$..) denotes a known point, ($e_x$,$e_y$ ..) denotes line direction vector, ($n_x$,$n_y$ ..) denotes perpendicular direction, $t$ is either a free parameter or a constant, ($a$,$b$,$c$) are equation coefficients, $\varphi$ is the direction angle from horizontal of a 2D line and $d$ the offset of the line from the origin. Oh ,and ($m_x$,$m_y$ ..) are parameters that describe the moment of the line about the origin. And ($a_x$,$a_y$..) ($b_x$,$b_y$..) are the coordinates of two points.

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I am not sure where you see a $\cos \varphi$ to define a line, but my guess is that you saw that in the context of the dot product $\cdot$

$$ \| a \cdot b \| = \| a \| \| b \| \cos \varphi $$

This particular identity isn't of much use for defining lines. The question needs more details to be answered fully.