Equations of lines and planes

Question

Hi guys I just have a few technicality and terminology questions about equations of lines and planes.

Just say we have a line represented by the vector equation:

r = r0 + tv, where t is an element of real numbers and r0, r, v are vectors.

I kindly request to know if it is correct to say that r is the line or treat r as a function of t. I read around but I just want a second opinion. I know r is a position vector but can I also say that it is the line as well?

What do the symmetric equations of the line tell me. I mean just say I have two lines and they are parallel, will the symmetric equations show me any evidence if this?

Answer 1

The line is not $r$ ( note that $r$ is really vector $\vec r$, that I suppose in $\mathbb{R}^3$) but the graph of the function $$ f=\mathbb{R} \to \mathbb{R}^3 $$ defined as: $$ \vec r= f(t)=\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix}+t\begin{pmatrix}x_v\\y_v\\z_v\end{pmatrix}=\vec r_0+t\vec v $$

so we can say , with a little abuse, that $\vec r$ is a point of the line.

Answer 2

When you write $r=r_0+tv$ you are writting the vectorial equation of the line. Actually, $r$ is a function of $t$ and is the vector position of any point of the line. For each value of $t$ you get a different point. Note that this equation written in coordinates is $$(x,y,z)=(x_0,y_0,z_0)+t(v_1,v_2,v_3).$$ This also can be written as

$$\begin{cases}x&=x_0+tv_1\\y&=y_0+tv_2\\z&=z_0+tv_3\end{cases}$$ And eliminating $t$ as $$\dfrac{x-x_0}{v_1}=\dfrac{y-y_0}{v_2}=\dfrac{z-z_0}{v_3}.$$ (Be careful with the meaning if some $v_i=0.$)

Now if you have two lines $r=r_0+tv$ and $s=s_0+tw,$ how can you know if they are parallel? You only need to compare $v$ and $w.$ If they are proportional (that is, $\exists k: w=kv$ then they are paralle (or the same line). In other case, they are not parallel. So, if $$\frac{w_1}{v_1}=\frac{w_2}{v_2}=\frac{w_3}{v_3}$$ they are parallel (be careful with the meaning if some $v_i=0$) and in other case they are not.