(17) What does the equation 3x + 4y = 12 give us?
It's a straight line, isn't it? The kind of straight line which has
become very familiar through years of school.
But what it actually describes depends on how many dimensions we are
working in.
First, we look at 3x + 4y = 12 in two dimensions.
This is a straight line, and 3x + 4y = 12 is what is called its
cartesian equation.
All points (x,y) which fit this equation will lie on this line.
For example, (0,3), (4,0), (-4,6) and (-2,9/2) all lie on it.
How would we write its vector equation?
As we saw in finding the vector equation of
a line, we need to know two things.
First, we need the position vector of a known point on the line.
In my drawing below, I've chosen the point A with coordinates (-2,9/2).
The position vector of A from O is
given by a = -2i + 9/2j.
We also need a direction vector to tell us the direction the line is going
in.
Now, we can see from the drawing that the gradient of this line is -3/4.
(Or we can say 3x + 4y = 12 so 4y = 12 - 3x so y = 3 - 3/4x so the
gradient is -3/4.)
So the direction of this line is given by the
vector b = 4i - 3j.
Therefore we can write down the vector equation of the line as
r = a + tb
or r = -2i + 9/2j + t(4i - 3j).
Different values of t give us the position vectors of different points on
the line. Here's some examples.
t=0 gives r = -2i + 9/2j which corresponds
to the point A (-2,9/2).
t=1 gives r = 2i + 3/2j which corresponds to the
point (2,3/2).
t = 3/2 gives r = 4i which corresponds to the point (4,0).
t = 1/2 gives r = 3j which corresponds to the point (0,3).
It's important to realise that you have lots of choices for the way in
which you write the vector equation of a line.
The drawings below show two more ways in which we could have written
the vector equation of the line 3x + 4y = 12.
This drawing gives us r = 3j + t(4i - 3j).
Now, putting t = 0 gives us r = 3j corresponding to (0,3).
Putting t = -1/2 gives us r = -2i + 9/2j corresponding
to (-2,9/2).
Or, we could have chosen a and b like this.
Now we have r = 4i + t(-4i + 3j).
This time, I've chosen
the direction vector so it points up the line instead of down it. Since the
straight line stretches infinitely either way, this works equally well.
I could also have chosen -8i + 6j or
2i - 3/2j for example, if I'd wanted.
Now, t = 0 gives r = 4i corresponding to (4,0).
What value of t gives you r = 2i + 3/2j?
Putting t = 1/2 will give this.
Now, what does the equation 3x + 4y = 12 give
in 3 dimensions?
Is it still the same line?
No, it can't be, because, although the same pairs of x and y values fit this
equation, we've now got to consider z values as well. For each pair of values
which work for x and y, z can take any value and this equation will still
be satisfied.
So, for example, each of the points (0,3,0), (0,3,20), (0,3,-2), (0,3,100)
lie on
3x + 4y = 12.
What we've got here is a plane, with the z-axis running parallel to it.
Here is a drawing of part of it.
You can see that each of the points I've given above will lie on this plane.
In fact, they all lie on the straight line in which the plane
3x + 4y = 12 cuts the vertical plane of x = 0.
Similarly, the points (4,0,12), (4,0,-50), (4,0,70) etc all lie on the
straight line where the plane
3x + 4y = 12 cuts the horizontal plane
of y = 0.
We can find out a bit more about the equation 3x + 4y = 12 by using the
results from equations of planes in the
form Ax + By + Cz = D.
These tell us that 3i + 4j is a normal vector to our plane.
This vector is 5 units in length, so n = 1/5(3i + 4j)
is a unit normal vector to our plane.
We know that the point with position vector a = 3j lies in
the plane.
Using the vector equation of the plane
in the form n.r = n.a = D gives us
D = 1/5(3i + 4j).(3j) = 12/5.
The perpendicular distance from any point on the z-axis to the
plane 3x + 4y = 12 is 12/5 units.
So, if 3x + 4y = 12 describes a plane, what does the equation
of a line in 3 dimensions look like?
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