Equation of a Line in Space, Examples

Equation of a Line in Space, Examples


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.

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