(11) Finding the vector equation of a line
Using vectors gives us a very neat way of writing down an equation
which gives the position vector of any point on a given straight line.
This method works equally well in two or three dimensions.
Suppose we 've got a straight line like the one I show in the drawing
below. (You have to imagine that it extends infinitely far in either
direction.)
In order to write down the vector equation of this line, we need to know
two things.
- We have to know the position vector of some point which lies on the
line, like a on my diagram.
- We have to know a vector which gives the direction of the line,
like b in my diagram. This is called a
direction vector.
Then the position vector r of any general point P on the
line is given by the equation
r = a + tb
where t tells us how much of b we need to take in order to
get from A to P. (t = 3 for the particular P I have shown in my drawing.)
DANGER Notice that
writing r = ta + b would give you a completely
different line!
Here's the drawing again to make it easier to refer to.
It's important to realise that there are many possible ways of writing
the vector equation of any given line. So, in my example above, any
point A on the line would work equally well provided we knew its position
vector, and any vector lying parallel to b would work
equally well as a direction vector. (For example, I could have used 3b
or - b.)
(If you are visiting from finding vector equations
of planes, you can return here.)
Here are some examples of equations of particular lines so we can look in
more detail at how they actually work.
Suppose that Line A has the
equation r = i + 3k + t(2i + j +
k)
so that the a in my drawing is i + 3k and
my b is 2i + j + k.
Different values of t give the position vectors of different points on
this line.
Putting t = 0 gives r = i + 3k.
Putting t = 1 gives r = 3i + j + 4k.
Putting t = -1 gives r = -i - j + 2k.
Suppose there are two more lines B and C so that now we have
line A with
equation r = i + 3k + t(2i + j +
k)
line B with
equation r = i + 3k + s(i + 4j -
k)
and line C with equation
r = i + j + k + u(4i + 2j +
2k).
The letters s and u work in exactly the same way for their lines
as t does for line A.
Both the lines B and C have special relationships with line A. Can you
spot what they are?
Using the a and b of my drawing to refer to, we see
that lines A and B both have a = i + 3k but they
have different direction vectors. Therefore they cut each other at
point A with r = i + 3k.
Line C has the same direction as line A, (its direction vector is just
scaled up by a factor of 2), so either lines A and C are parallel or
they are really the same line.
Putting t = 0 for line A gives r = i + 3k but there
is no value we can give to u in line C which would
make r = i + 3k so therefore A and C are distinct
parallel lines.
Now we'll consider these two lines.
Line D has
equation r = i - j + 4k + s(i - j
+ k).
Line E has equation r = 2i + 4j + 7k +
t(2i + j + 3k).
They are not parallel since their direction vectors aren't parallel but do
they cut each other? (If not, they are what are called
skew lines.)
If they cut each other then the point P where they cut must lie on both
lines. We'll call its position vector p.
The point P can only exist if there are values of s and t so that
p = i - j + 4k + s(i - j
+ k) = 2i + 4j + 7k +
t(2i + j + 3k).
For this equation to have a solution, the components
in the i, j and k
directions must each seperately be equal. This would mean that
i + si = 2i + 2ti so giving 1 + s = 2 + 2t
-j - sj = 4j + tj so giving -1 - s = 4 + t
4k + sk = 7k + 3tk so giving 4 + s = 7 + 3t.
Is this possible?
Adding (1) and (2) gives 0 = 6 + 3t so t = -2 and 1 + s = 2 - 4 so s = -3.
THE LINES ONLY MEET if these values of s
and t also fit equation (3).
Putting s = -3 and t = -2 in 4 + s = 7 + 3t gives LHS = 4 - 3 =1 and
RHS = 7 - 6 =1 so the 3 equations are consistent (that is, there is a
solution which fits all 3 of them) and the lines do cut
each other.
Putting s = -3 in line D's equation gives the position vector of this
point of intersection as
p = i - j + 4k - 3(i - j
+ k) = - 2i + 2j + k.
You will see that putting t = -2 in line E's equation gives exactly the
same result.
It is possible to use a similar method as we've used here to find the
equations of lines to give us vector equations for planes. This is
what the next section is about.
Next
or you might like to jump to equations of lines
in 3 dimensions in terms of x, y and z
or back to the previous section
or return to the vector homepage