Let’s look at a basic, straightforward example

of finding the equation of a line in space. Here are the directions: find the equation

of the line through point 0, 2, negative 5 and in the direction of the vector 1, negative

1, 1. First of all, what’s the equation for a line in space? The general expression looks

like this: R, vector R, equals vector R0 plus T times vector V. Here vector R sub 0 corresponds

to the position vector pointing to the point you’re interested in. Here we want an equation

of a line through point 0, 2, negative 5 so, vector R zero is going to be the position

vector that goes through the origin and that point. Then, vector V is the direction vector

of your line. Really this is kind of analogous to point-slope form and when you’re drawing

the line on the two dimensional plane using point-slope form you’re given a point that

you want the line to pass through then, you’re also given the slope, which tells you the

direction that the line is going to go in that’s really happening here as well. You’re

given a point, this time is in three dimensional space, then we’re given a vector that tells

us in what direction we want the line to point. Really all we have to do here is plug in.

Out to the side here, let’s just write down what vector our 0 is. Once again, this is

going to be the vector that goes between points, the point I’m interested in, 0, 2, negative

5 and the origin. The components of that vector are just going to be 0, 2, negative 5. Then,

we’re given vector V explicitly that’s just 1, negative 1, 1. If I plug that into my expression

for the equation of a line it’s going to look like this: vector R0, so, 0, 2, negative 5

plus T, here T is a scalar quantity, T times vector V. We could leave this in this form

or we could combine these vectors. If I were to combine these vectors, it would like this:

the first component would be 0 plus 1T then, 2 minus 1T, then, negative 5 plus 1T or I

could write that as, T comma 2 minus T comma negative 5 plus T. This gives me the equation

of my line in space. Basically, T is my parameter here, if I were to let T vary for all values

of T between negative infinity and infinity this vector valued function would sweep out

the entire line. In this direction we’re asked to find the line through the points P0 at

point negative 3, 5, 2 and P1 at point 2, 0, 8. Once again, let’s just start by writing

down the equation of a line in space. That equation is given by vector R equals vector

R sub 0 plus T times vector V. Now what we’re actually given in this example is simply 2

points. In a more simple example you might be given 1 point, P0, then vector V, which

gives you the direction of the line but, here I’m just given 2 points. This is kind of analogous

to the situation just in 2 dimensions where you’re working in point-slope form and you’re

just given 2 points and you’re asked to find the equation of the line between them. In

that case, what you start by doing is actually finding the slope between the two lines. That’s

basically what we’re going to be doing here instead of finding the slope; we’re going

to be finding the direction vector, V, between these 2 points. Let’s start by then finding

vector V. Vector V is going to be the vector that goes between points P0 and P1. If I work

that out just using the points P1 and P0, the components of that vector are going to

be 2 minus negative 3, 0 minus 5 and then, 8 minus 2. If I work that out that’s going

to give me 5, negative 5 and 6 as the components of my vector V. From here I’m basically doing

what I do in any other equation of a line problem. I have my point P0 from that I can

write down R0, right? R0 is just the vector that goes between the origin and point P0

so, its’ components are just negative 3, 5, 2. Then, I just plug it into my formula because

I have R0 and I have vector V. Then, the equation of my line, vector R is going to be given

by vector R0, negative 3, 5, 2 plus T times vector V, which was 5, negative 5, 6. If I

actually go through and add these vectors I’m going to get negative 3 plus 5T then,

5 minus 5T then, 2 plus 6T. This gives me the equation of my line in space.