can someone help me understand the vector equation of a line segment ????

Question

chrisplusian · Answer

My textbook says:
"The line segment from $$r _{0} to r _{1}$$ is given by the vector equation:
$$r(t)=(1-t)r _{0}+tr _{1}          $$
where$$0\le t \le1$$

chrisplusian · Answer

Why does t have to be restricted that way?

chrisplusian · Answer

By the way "r" , "r sub 1", and "r sub 0" are all vectors and t is a scalar value.

anonymous · Answer

$t$ is restricted that way because the segment only is defined between $\vec{r_0}$ and $\vec{r_1}$ can you please specify here (for the sake of the argument) that these are not just vectors but position vectors? Or does your text book talk about vectors?

So for the line segment you have:
$$\Large \vec{r(0)}=\vec{r_0} $$
and the second restriction (bound) is at t=1

$$\Large \vec{r(1)}=\vec{r_1} $$

chrisplusian · Answer

I am reading back through give me one second

chrisplusian · Answer

These vectors are said to be "the proof of the equation of a line in three dimensions". When I look at the geometric proof of the line it uses "rsub 0" and a scalar times a direction vector = "r" so it would seem that they are position vectors. (By position vectors we do mean vectors with their initial points at the origin right?)

anonymous · Answer

Exactly, that is what a position vector is, it's position (unlike as for a direction vector) is fixed at the origin. A direction vector can be anywhere on the xy-plane, or the xyz-euclidean room, in Rome or in Paris :-) 

I understand the formula above, however this is the parametric/vector equation of a line I use:

$$\Large \vec{r_x}=\vec{0P} +t\vec{v}  $$
0P is a position vector here, read as "Vector from 0 (Origin) to P", and v is a direction vector, you need both to describe a line. Just imagine how you'd draw a line on your paper, first you need to place your pen somewhere, as soon as you did that, you have fixed a point. But now you only have a point, that's not a line yet, you also have to decide in which direction you want to draw your line, this is given by your direction vector. 

We can let $t$ vary from various scales to create a segment.

chrisplusian · Answer

Yes thinking about it that way the $$r _{0}$$ is a position vector. By the way how do you put the arrow above the letter? I have tried that and so far have been unsuccessful.

anonymous · Answer

Let me give you another insight which might help you to understand the formula a bit better, let me give you two points $P_1$ and $P_2$ as in the language of vector geometry, I have just given you two position vectors (in any dimensions). If you want a vector expression from the line segment that describes the line from $P_1$ to $P_2$ you compute it as follows.

Select $\vec{0P_1}$ as your initial position vector, compute your direction vector $\vec{v}=\vec{0P_2}-\vec{0P1}$ and plug it back into the formula:
$$\Large \vec{r_x}=\vec{0P_1}+t\vec{v}=\vec{0P_1}+t(\vec{0P_2}-\vec{0P1}) $$ and reorder it to get:
$$\Large  \vec{r_x}=(1-t)\vec{0P_1}+t\vec{0P_2}$$

There is only a difference in notation between the two, phyisicians usually use the r notation to remind us of the radius of a circle that starts at the origin, mathematicians use the 0.

The Command to $\LaTeX$ vectors is \vec{v}

chrisplusian · Answer

I don't get it thanks though