# Let's see how to know the intersection of two lines in vector calulation in A-Level Maths!

**Intersection of two lines **

When you have two lines in a plane [a two-dimensional space], either they’re parallel or they intersect.

If they’re parallel, they have the same gradient [which you can spot in Cartesian equations because they have the same value of m and in vector equations because they have the same direction vector.]

In three dimensions, there’s an additional possibility: that two lines which are notparallel do not intersect. Such a pair of lines are called skew.

In fact, if you chose two 3D lines at random, they’d probably be skew.

We need now to translate this into the algebra involving the equations of two lines in 3D.

## Example

Find if the following two lines intersect. If they do, then find the position vector of their intersection

We must first look at the two direction vectors and notice that they are not the same and one is not amultiple of the other. That means that the lines are not parallel, so we must proceed to investigate whether they intersect.

If they do intersect, then we should be able to find values of land m such that the two equations give the same position vector.

To explore this, we write out the three equations equating the x, y and z components:

3 + λ = 4 + 2m

2 + 2λ = 1 + m

4 + 5λ = 3 + 3m

Before we process these, it might be sensible to tidy them up:

λ – 2m = 1 ……………………..(i)

2λ – m = -1……………………..(ii)

5λ – 3m = -1 …………………..(iii)

Now we must stop and think. We have three equations in two unknowns. Any two of them should yield asolution for λ and m which might or might not work in the third.

If it does work, then those values satisfy all three equations and we have an intersection.

If it doesn’t work, then the two lines must be skew. We’ll choose the first two equations to solve.

(ii)x 2: 4λ – 2m = -2…………………………(iv)

(iv) – (i): 3λ = -3

λ = 1

(ii): 2 – m = -1

m = 3

So λ = 1, m = 3 is a solution satisfying the first two equations. We must now see if it works in (iii).

LHS = 5 – 9 = -4 which does not equal to

As this solution does not work in (iii), that means that there are no values of λ and m which work in all three equations. In other words, the two lines are skew.

If this pair of values did work in all three equations, that would show that there is an intersection and we could find its position vector either by putting this value of λ in the equation for *l*_{1} or by putting this value of min the equation for *l*_{2}.

**Example:**

Find where the lines meet:

Matching the x-values gives: 2 + λ = 3 - μ

Matching the y-values gives: 3 - 2λ = -4 + 3μ

Rearrange these to get:

λ + μ = 1

2λ + 3μ = 7

**These solve to give:** λ = -4, μ = 5

**Substituting these values into our line equations we get the point of intersection as being:**

x = -2, y = 11.

Therefore the lines meet at (-2, 11).

In 3-D two lines do not have to meet. If they do not meet then the lines are called '**skew**'.

Finding where two lines meet is the same as for 2-D, except there will be three simultaneous equations (in two variables) to solve. (A solution that does not work for all three equations means the lines do not meet and are therefore skew.)

Drafted by Eunice (Maths)