## (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.

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or you might like to jump to equations of lines in 3 dimensions in terms of x, y and z