In this section we need to take a look at the equation of a
line in . As we saw in the previous section the
equation does not describe a line in ,
instead it describes a plane. This doesn’t mean however that we can’t write
down an equation for a line in 3D space.
We’re just going to need a new
way of writing down the equation of a curve.
So, before we get into the equations of lines we first need
to briefly look at vector functions.
We’re going to take a more in depth look at vector functions later. At this point all that we need to worry about
is notational issues and how they can be used to give the equation of a curve.
The best way to get an idea of what a vector function is and
what its graph looks like is to look at an example. So, consider the following vector function.
A vector function is a function that takes one or more
variables, one in this case, and returns a vector. Note as well that a vector function can be a
function of two or more variables.
However, in those cases the graph may no longer be a curve in space.
The vector that the function gives can be a vector in
whatever dimension we need it to be. In
the example above it returns a vector in . When we get to the real subject of this
section, equations of lines, we’ll be using a vector function that returns a
vector in
Now, we want to determine the graph of the vector function
above. In order to find the graph of our
function we’ll think of the vector that the vector function returns as a
position vector for points on the graph.
Recall that a position vector, say ,
is a vector that starts at the origin and ends at the point .
So, to get the graph of a vector function all we need to do
is plug in some values of the variable and then plot the point that corresponds
to each position vector we get out of the function and play connect the
dots. Here are some evaluations for our
example.
So, each of these are position vectors representing points on
the graph of our vector function. The
points,
are all points that
lie on the graph of our vector function.
If we do some more evaluations and plot all the points we
get the following sketch.
In this sketch we’ve included the position vector (in gray
and dashed) for several evaluations as well as the t (above each point) we used for each evaluation. It looks like, in this case the graph of the
vector equation is in fact the line .
Here’s another quick example. Here is the graph of .
In this case we get an ellipse. It is important to not come away from this
section with the idea that vector functions only graph out lines. We’ll be looking at lines in this section,
but the graphs of vector functions do not have to be lines as the example above
shows.
We’ll leave this brief discussion of vector functions with
another way to think of the graph of a vector function. Imagine that a pencil/pen is attached to the
end of the position vector and as we increase the variable the resulting
position vector moves and as it moves the pencil/pen on the end sketches out
the curve for the vector function.
Okay, we now need to move into the actual topic of this
section. We want to write down the
equation of a line in and as suggested by the work above we will
need a vector function to do this. To
see how we’re going to do this let’s think about what we need to write down the
equation of a line in . In two dimensions we need the slope (m) and a point that was on the line in
order to write down the equation.
In that is still all that we need except in this
case the “slope” won’t be a simple number as it was in two dimensions. In this case we will need to acknowledge that
a line can have a three dimensional slope.
So, we need something that will allow us to describe a direction that is
potentially in three dimensions. We
already have a quantity that will do this for us. Vectors give directions and can be three
dimensional objects.
So, let’s start with the following information. Suppose that we know a point that is on the
line, ,
and that is some vector that is parallel to the
line. Note, in all likelihood, will not be on the line itself. We only need to be parallel to the line. Finally, let be any point on the line.
Now, since our “slope” is a vector let’s also represent the
two points on the line as vectors. We’ll
do this with position vectors. So, let and be the position vectors for P_{0} and P respectively. Also, for no
apparent reason, let’s define to be the vector with representation .
We now have the following sketch with all these points and
vectors on it.
Now, we’ve shown the parallel vector, ,
as a position vector but it doesn’t need to be a position vector. It can be anywhere, a position vector, on the
line or off the line, it just needs to be parallel to the line.
Next, notice that we can write as follows,
If you’re not sure about this go back and check out the
sketch for vector addition in the vector
arithmetic section. Now, notice that
the vectors and are parallel.
Therefore there is a
number, t, such that
We now have,
This is called the vector
form of the equation of a line. The
only part of this equation that is not known is the t. Notice that will be a vector that lies along the line and
it tells us how far from the original point that we should move. If t
is positive we move away from the original point in the direction of (right in our sketch) and if t
is negative we move away from the original point in the opposite direction of (left in our sketch). As t
varies over all possible values we will completely cover the line. The following sketch shows this dependence on
t of our sketch.
There are several other forms of the equation of a
line. To get the first alternate form
let’s start with the vector form and do a slight rewrite.
The only way for two vectors to be equal is for the
components to be equal. In other words,
This set of equations is called the parametric form of the equation of a line. Notice as well that this is really nothing
more than an extension of the parametric equations
we’ve seen previously. The only
difference is that we are now working in three dimensions instead of two
dimensions.
To get a point on the line all we do is pick a t and plug into either form of the
line. In the vector form of the line we
get a position vector for the point and in the parametric form we get the
actual coordinates of the point.
There is one more form of the line that we want to look
at. If we assume that a, b,
and c are all nonzero numbers we can
solve each of the equations in the parametric form of the line for t.
We can then set all of them equal to each other since t will be the same number in each. Doing this gives the following,
This is called the symmetric
equations of the line.
If one of a, b, or c does happen to be zero we can still write down the symmetric
equations. To see this let’s suppose
that . In this case t will not exist in the parametric equation for y and so we will only solve the
parametric equations for x and z for t. We then set those equal
and acknowledge the parametric equation for y
as follows,
Let’s take a look at an example.
Example 1 Write
down the equation of the line that passes through the points and . Write down all three forms of the equation
of the line.
Solution
To do this we need the vector that will be parallel to the line. This can be any vector as long as it’s
parallel to the line. In general, won’t lie on the line itself. However, in this case it will. All we need to do is let be the vector that starts at the second
point and ends at the first point.
Since these two points are on the line the vector between them will
also lie on the line and will hence be parallel to the line. So,
Note that the order of the points was chosen to reduce the
number of minus signs in the vector.
We could just have easily gone the other way.
Once we’ve got there really isn’t anything else to do. To use the vector form we’ll need a point
on the line. We’ve got two and so we
can use either one. We’ll use the
first point. Here is the vector form
of the line.
Once we have this equation the other two forms
follow. Here are the parametric
equations of the line.
Here is the symmetric form.

Example 2 Determine
if the line that passes through the point and is parallel to the line given by ,
and passes through the xzplane. If it does give
the coordinates of that point.
Solution
To answer this we will first need to write down the
equation of the line. We know a point
on the line and just need a parallel vector.
We know that the new line must be parallel to the line given by the
parametric equations in the problem statement. That means that any vector that is parallel
to the given line must also be parallel to the new line.
Now recall that in the parametric form of the line the
numbers multiplied by t are the
components of the vector that is parallel to the line. Therefore, the vector,
is parallel to the given line and so must also be parallel
to the new line.
The equation of new line is then,
If this line passes through the xzplane then we know that the ycoordinate of that point must be zero. So, let’s set the y component of the equation equal to zero and see if we can solve
for t. If we can, this will give the value of t for which the point will pass
through the xzplane.
So, the line does pass through the xzplane. To get the
complete coordinates of the point all we need to do is plug into any of the equations. We’ll use the vector form.
Recall that this vector is the position vector for the
point on the line and so the coordinates of the point where the line will
pass through the xzplane are .
