Welcome to a lesson on parametric equations
of a line in R three.
A line that is parallel to vector v with components a, b, c
that passes through the point x one, y one, z one,
is given as x equals x sub one plus a, t,
y equals y sub one plus b, t,
and z equals z sub one plus c, t.
So the components of the directional vector a, b, c
are the coefficients for t.
And the point on the line x sub one, y sub one and z sub one
are the constants in each of the three equations.
Vector v is called the directional vector for the line,
and a, b and c are sometimes called the directional numbers.
Let's take a look at a line in space
so we can get a better idea of what we're talking about.
Here we have a graph of a line
on the x, y, z coordinate system,
and if we rotate this system, you can get a better idea
of that line passing through space.
So when we talk about the parametric equations
of a line in space, we're talking about a line
that looks something like this.
Now there's one more thing we need to talk about
before we take a look at some examples.
Sometimes when graphing a line in space,
using parametric equations, you may be asked to
determine the orientation of a line.
And this refers to the direction the points on the line
would be plotted as t increases.
So here's the graph of the line defined by
x equals two t, y equals one plus t,
and z equals negative three plus two t.
We can see that as t increases,
all three of these coordinates will also increase,
because the coefficients of all the t terms are positive.
More specifically, as t increases, so does z,
and you can think of z as the height of the line.
So as t increases, this line is moving upward,
so we know the orientation of this line would be
from this direction to the left,
or in this direction here.
There's one more form of linear equation we should mention
for R three.
The parameter t can be eliminated to obtain what's called
the symmetric equation of a line in R three.
So this form of a line in space
is a symmetric equation of a line.
To obtain this, what we can do is solve each equation for t,
and then set them all equal to each other.
Let's go and take a look at an example.
Here we will determine the directional vector of the line
and two points on the line.
Remember the directional vector for the line
is determined by the coefficients of the t terms.
So the directional vector would have components
negative four,
positive three,
and positive one.
Now let's try to find two points on the line.
We can find the first point, we'll call it point P,
by the constants of each of these equations.
So the x coordinate would be positive three,
the y coordinate would have to be zero,
and the z coordinate would be negative five.
But they ask for two points on the line,
so to determine one more point on the line,
we'll have to select a value for t,
and sub it into each of these equations.
So for the second point, let's call it Q,
if we let t equal zero, we're going to come up with
the same point that we just found,
so let's go ahead and let t equal one.
So if t is equal to one, we have three minus four,
that would be negative one for the x coordinate,
positive three for the y coordinate,
and then for z we'd have negative five plus one,
and that would give us negative four.
Now remember there's an infinite
number of points on a line.
These are just two points that
are pretty convenient to find.
Let's go and take a look at the graph of this line.
Here it is in red.
And, again, if we rotate this coordinate system,
we get a better idea of the line's orientation in space.
Notice the line is below the x, y plane,
until t is greater than positive five.
So the orientation of this line would be from left to right
as t increases.
Let's go and take a look at another example.
Here we're given to points on a line,
and we want to determine the parametric equation of the line
and then also determine the symmetric equation of the line.
So looking at the parametric equations of a line,
we can let either of these points represent
x sub one, y sub one and z sub one.
But remember, to find the values of a, b and c
we have to find the directional vector of this line.
Now we can do that by finding the vector in component form
that would pass through these two points.
So we'll call our directional vector vector V.
For the exit point, it will be two minus three,
the y coordinate will be positive three minus four,
and the z component will be negative five minus seven.
So our directional vector, which will give us a, b and c,
will be negative five, negative one and negative 12.
And, again, we'll go ahead and use the first point,
three, four seven
to represent x sub one, y sub one and z sub one.
But we could use any other point on the line,
if we wanted to.
SO the parametric equations of the line passing through
these two points will be x equals three
minus five t,
y will equal four
minus one t or minus t,
and z will equal
seven
minus 12 t.
And the last part of this question
asks us to find the symmetric equation of the line.
Let's go ahead and do that on this screen.
Now we could go ahead and just use
the form of this equation here,
by doing I show the connection between this form
and the parametric equations.
So our parametric equations were
x equals three minus five t,
y equals four minus t,
and z equals seven minus 12 t.
So what we can do is solve each of these equations for t
and then set them equal to each other.
To solve this equation for t, we would subtract
three on both sides and then divide by negative five.
For this equation, we would subtract four
and then divide by negative one.
And here we would subtract seven and divide by negative 12.
Now that we have all these equations equal to t,
we can set the right sides equal to each other,
which will give us the symmetric equation of this line.
So we'd have x minus three divided by negative five
equals y minus four divided by negative one
equals z minus seven divided by negative 12
and now we have the symmetrical equation of this line.
I hope you found this video helpful.
Thank you for watching.