4.04e - Intersecting, parallel or skew lines | OCR A-Level Further Mathematics A

Determine whether two lines meet, run parallel or are skew, and find their intersection when it exists. You will use separate parameters for the two lines, check all three coordinates in space, and distinguish separate parallel lines from two equations of the same line.

How lines can be arranged

Two roads cross on a map. Must they meet in space? A bridge gives a familiar counterexample. In coordinates, consider

l_1:\ (x,y,z)=(t,0,0),\qquad l_2:\ (x,y,z)=(0,s,2)

Their views from above cross at $(0,0)$ , but one line has height 0 and the other height 2. How can we prove that two lines really meet, rather than merely appear to cross?

You already know

A line has a starting point and a non-zero direction: $\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}$ . A point lies on the line only when one parameter value gives all its coordinates. Proportional direction vectors describe parallel directions.

[DIAGRAM: asset_name: Lines crossing in projection but not in space; asset_slug: h245_4_04e_skew_lines; caption: The two lines share an apparent crossing from above, but their heights differ; description: Three-dimensional coordinate diagram of the x-directed line (t,0,0) and the y-directed line (0,s,2), with the points (0,0,0) and (0,0,2) marked and joined by a dashed vertical guide. A side annotation explains that the lines are non-parallel and do not intersect.]
The two lines share an apparent crossing from above, but their heights differ

Intersecting, parallel and skew lines

Distinct lines are intersecting if they share a point. They are parallel if their direction vectors are proportional; distinct parallel lines share no point. They are skew if they are neither parallel nor intersecting. Coincident lines are two descriptions of the same line, so every point is shared.

Two distinct intersecting lines share exactly one point: if they shared two distinct points, both would be the line through those points. Skew lines occur only in three dimensions or more. In a single plane, non-parallel lines must meet. This is the extra possibility that a flat sketch can hide.

Here is a useful mathematical surprise: two directions can be at right angles while the lines themselves never meet. The two lines in the opening example have directions $(1,0,0)$ and $(0,1,0)$ , but their constant heights prevent an intersection. Direction and location answer different questions.

Key idea

An intersection requires the same point on both lines: every coordinate must agree. A crossing on a sketch establishes neither the coordinates nor the height.

Before solving, be clear about what each word asserts.

Finding a common point

Use a different parameter for each line, equate corresponding coordinates and solve; then verify that the proposed common point satisfies every coordinate equation.

The parameters are labels chosen independently. An intersection may occur at $\lambda=2$ on one line and $\mu=1$ on the other. Setting $\lambda=\mu$ would impose an extra condition with no geometric reason.

Intersection condition

For lines $\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}$ and $\mathbf{r}=\mathbf{b}+\mu\mathbf{e}$ , a common point exists precisely when real values of $\lambda,\mu$ satisfy

\mathbf{a}+\lambda\mathbf{d}=\mathbf{b}+\mu\mathbf{e}

in every component.
Not in the formula booklet: learn it.

Choose two component equations that give enough information to solve for the parameters. In three dimensions, the remaining equation is a separate consistency check.

Worked example 1: intersecting lines in a plane

Question. Determine the point of intersection of

l_1:\mathbf{r}=\begin{pmatrix}1\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\1\end{pmatrix}, \qquad l_2:\mathbf{r}=\begin{pmatrix}6\\2\end{pmatrix}+\mu\begin{pmatrix}-1\\2\end{pmatrix}

Plan. Equate both coordinates, solve for the two parameters and substitute into either line.

\begin{aligned} 1+2\lambda&=6-\mu && x\text{-coordinates}\\ 2+\lambda&=2+2\mu && y\text{-coordinates}\\ \lambda&=2\mu && \text{second equation}\\ 4\mu+\mu&=5 && \text{first equation}\\ \mu&=1,\quad\lambda=2 && \text{solve} \end{aligned}

Why this step? Each line has its own parameter, but their coordinates must match at a shared point.

Answer. The intersection is $(1+4,2+2)=(5,4)$ .

Check. On $l_2$ , $(6,2)+1(-1,2)=(5,4)$ as well.

Worked example 2 changes one feature: the points and directions gain a third component. The first two equations still give the same candidate, but it must now pass a third test.

Worked example 2: a third coordinate to check

Question. Determine the point of intersection of

l_1:\mathbf{r}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\3\end{pmatrix}, \qquad l_2:\mathbf{r}=\begin{pmatrix}6\\2\\1\end{pmatrix}+\mu\begin{pmatrix}-1\\2\\4\end{pmatrix}

Plan. Solve the $x$ - and $y$ -equations, check $z$ , then state the common point.

\begin{aligned} 2\lambda+\mu&=5 && x\text{-coordinates}\\ \lambda-2\mu&=0 && y\text{-coordinates}\\ \lambda&=2,\quad\mu=1 && \text{solve the pair}\\ -1+3(2)&=5=1+4(1) && z\text{-coordinates agree} \end{aligned}

Why this step? Matching $x$ and $y$ alone only establishes a crossing in projection. Matching $z$ establishes a shared point in space.

Answer. The lines intersect at $(5,4,5)$ .

Check. Substituting into both full vector equations gives $(5,4,5)$ , so all three coordinates agree.

The final check is part of determining the intersection, not an optional afterthought.

Watch out

Wrong: “ $\lambda=2,\mu=1$ , so the lines intersect,” after using only two coordinates. Right: substitute those values into the third equation and evaluate both sides before concluding.

OCR's “Determine” asks for justification: show the equations and consistency check, not just the output of a simultaneous-equation solver.

Parallel, coincident or skew

Check directions first: proportional directions call for a point-on-line check; non-proportional directions call for an intersection check.

Direction vectors	Point check	Conclusion
Proportional	One starting point lies on the other line	Coincident
Proportional	One starting point does not lie on the other line	Distinct parallel lines
Not proportional	All coordinate equations are consistent	Intersecting
Not proportional	Coordinate equations are inconsistent	Skew

For proportional directions, why is checking one point enough? If they share one point $P$ , both lines contain every real multiple of the same direction from $P$ , so they are the same line. If they share no starting point, their parallel directions cannot bring them together later. These classification tests are not in the formula booklet: learn them.

Worked example 3: inconsistent coordinates mean skew

Question. Determine whether the following lines intersect, are parallel or are skew.

l_1:\mathbf{r}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\3\end{pmatrix}, \qquad l_2:\mathbf{r}=\begin{pmatrix}5\\4\\6\end{pmatrix}+\mu\begin{pmatrix}-1\\2\\4\end{pmatrix}

Plan. Compare directions, then solve two coordinates and test the third.

\begin{aligned} (-1,2,4)&\ne k(2,1,3)\text{ for any }k && \text{not parallel}\\ 2\lambda+\mu&=4,\quad \lambda-2\mu=2 && x\text{ and }y\\ \lambda&=2,\quad\mu=0 && \text{solve}\\ z_1&=-1+3(2)=5 && \text{first line}\\ z_2&=6+4(0)=6 && \text{second line} \end{aligned}

Why this step? The only parameter pair that matches $x$ and $y$ fails to match $z$ , so there is no possible common point.

Answer. The lines are skew: they are not parallel and do not intersect.

Check. The two candidate points are $(5,4,5)$ and $(5,4,6)$ , displaying the mismatch explicitly.

Worked example 4 changes one feature: the second line's direction becomes a multiple of the first. The starting point stays the same. A point-on-line check now settles the classification more efficiently.

Worked example 4: parallel directions and different lines

Question. Classify

l_1:\mathbf{r}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\3\end{pmatrix}, \qquad l_2:\mathbf{r}=\begin{pmatrix}5\\4\\6\end{pmatrix}+\mu\begin{pmatrix}4\\2\\6\end{pmatrix}

Plan. Verify proportional directions and test whether $(5,4,6)$ lies on $l_1$ .

\begin{aligned} (4,2,6)&=2(2,1,3) && \text{parallel directions}\\ 1+2\lambda=5&\Rightarrow\lambda=2 && x\text{-coordinate}\\ 2+\lambda&=4 && y\text{ agrees}\\ -1+3\lambda&=5\ne6 && z\text{ fails} \end{aligned}

Why this step? Proportional directions alone allow either the same line or distinct parallel lines; the starting point distinguishes them.

Answer. The lines are distinct and parallel.

Check. The displacement between their starting points is $(4,2,7)$ , which is not a multiple of $(2,1,3)$ .

If only the second starting point's last coordinate were changed from 6 to 5, it would lie on $l_1$ at $\lambda=2$ . The directions would still be proportional, so the two equations would then describe coincident lines. An intersection problem that produces dependent equations is not automatically wrong: it may be describing this same-line case.

Watch out

Wrong: “No intersection, so skew.” Distinct parallel lines also have no intersection. Right: establish non-parallel directions as well before concluding that lines are skew.

The opening roads are skew: their directions are not proportional, and $z=0$ cannot also be $z=2$ . A complete coordinate check settles what a flat map cannot.

Quick Recap

Compare direction vectors first: proportional directions mean parallel or coincident lines.
Use different parameters and equate corresponding coordinates to find a common point.
In three dimensions, every coordinate equation must agree; two alone are insufficient.
Non-parallel lines with no intersection are skew. Parallel lines with one common point are coincident.

The opening lines cross only in projection: their different heights prove that they never meet.

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4.04e - Intersecting, parallel or skew lines

How lines can be arranged

Finding a common point

Parallel, coincident or skew

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