Vector Equations of Lines

Question

I'm taking multivariable calculus right now and I'm a bit confused on the vector equation of a line.

My book says that the line passing through the point $(x_0,y_0,z_0)$ and parallel to the vector $v=\langle v_1,v_2,v_3 \rangle$ can be expressed as$$\langle x,y,z \rangle = \langle x_0,y_0,z_0 \rangle + t\langle v_1,v_2,v_3\rangle$$ My confusion with this is how it represents a line. For any t, this function returns a vector with the components of a point on the line. But a line is a set of points not a set of vectors, so how does this represent a line?

Answer 1

They are called parametric equations of a line. You say simply that the line through the point $(x_0,y_0,z_0)$ (for $t=0$) and it consists of all points "proportional" to $(v_1,v_2,v_3)$. In other words we have the following representation of the line $r$

$r$ : $\lbrace(x_0,y_0,z_0)\rbrace \cup \mathrm{span}(v)$

with $v=(v_1,v_2,v_3) \neq \overline{o}$, and $\mathrm{span}(v)=\lbrace tv : t \in \mathbb{R} \rbrace$. At this point, as suggested in the comments of your question, you can have the rapresentation on the coordinate axes, using the canonical basis of $\mathbb{R}^3$ which is $\lbrace i,j,k \rbrace = \lbrace e_1,e_2,e_3 \rbrace$.

Answer 2

A mixed expression where there is some doubts about what is what (between points and vectors) can always be transformed into an homogenous expression. In the present case, rewrite:

$$\langle x,y,z \rangle = \langle x_0,y_0,z_0 \rangle + t\langle v_1,v_2,v_3\rangle$$

under the form:

$$\langle x,y,z \rangle - \langle x_0,y_0,z_0 \rangle = t\langle v_1,v_2,v_3\rangle$$

In other words: $\vec{M_0M}=t\vec{u}$

which is homogeneous (vector = vector).