In three dimensions two straight lines are either parallel, intersecting or skew. They are parallel if their direction vectors are multiples of each other. If not, set the two vector equations equal: solve two of the component equations for λ and μ, then check the third. If it holds, the lines intersect; if it fails, they are skew, which can only happen in 3D.
Classifying two lines
- Parallel? Check whether one direction vector is a multiple of the other.
- If not parallel, set r₁ = r₂ and write three component equations.
- Solve two of them for λ and μ.
- Substitute into the third: if it works, the lines intersect; if not, they are skew.
Acute angle between lines: cos θ = |d₁ · d₂| / (|d₁||d₂|).
Parallel lines
Two lines are parallel when one direction vector is a scalar multiple of the other. Only the directions matter; the points the lines pass through make no difference. If two parallel lines also share a point, they are the same line.
Do the lines intersect?
Set the two vector equations equal, using different parameters, such as λ for the first line and μ for the second. Matching the i, j and k components gives three equations in two unknowns. Solve two of them, then substitute into the third. If the third equation is satisfied, the lines meet; substitute λ into the first line to get the point of intersection.
Skew lines
Lines are skew if they are neither parallel nor intersecting. A full proof needs both parts: show the direction vectors are not multiples of each other, giving a reason, and show that the values of λ and μ from two equations do not satisfy the third. Missing either part loses marks.
Finding an unknown constant
If you are told the lines intersect, the three equations must be consistent. Solve the two equations that do not contain the unknown, then substitute into the third to find it. If the lines are to be parallel instead, match the ratios of the direction vector components.
The angle between two lines
Exam questions often go on to ask for the acute angle between intersecting lines. Use the direction vectors and the scalar product: cos θ = |d₁ · d₂| / (|d₁||d₂|). The modulus on top makes sure the angle is acute.
Worked examples
Example 1. Show that r = (i + 2k) + λ(2i + j + 3k) and r = (3i + j) + μ(4i + 2j + 6k) are parallel.
- Direction vectors: 2i + j + 3k and 4i + 2j + 6k.
- 4i + 2j + 6k = 2(2i + j + 3k).
Answer: One direction is a multiple of the other, so the lines are parallel.
Marks: M1 for comparing directions, A1 for the conclusion.
Example 2. Show that r = (2i + 3j + 2k) + λ(i + j) and r = (i + 5j + k) + μ(2i − j + k) intersect, and find the point.
- x: 2 + λ = 1 + 2μ. y: 3 + λ = 5 − μ. z: 2 = 1 + μ.
- From z: μ = 1. Then x: 2 + λ = 3, so λ = 1.
- Check y: 3 + 1 = 4 and 5 − 1 = 4, which agree.
- Point: (2, 3, 2) + 1(1, 1, 0).
Answer: They intersect at (3, 4, 2).
Marks: M1 for the three equations, M1 for solving and checking, A1 for the point.
Example 3. Prove that r = (j + 2k) + λ(i + 2j + k) and r = (i + 3k) + μ(2i + j − k) are skew.
- Not parallel: i + 2j + k and 2i + j − k are not multiples (the i ratio is 2 but the j ratio is ½).
- x: λ = 1 + 2μ. y: 1 + 2λ = μ.
- Solving: λ = −1, μ = −1.
- Check z: 2 + λ = 1, but 3 − μ = 4. These are not equal, so the lines do not meet.
Answer: Not parallel and not intersecting, so the lines are skew.
Marks: B1 for not parallel with a reason, M1 for solving two equations, M1 for the third failing, A1 for skew.
Example 4. The lines r = (i + j + k) + λ(2i + j + 3k) and r = (2i + 3j + pk) + μ(i − j + k) intersect. Find the constant p.
- x: 1 + 2λ = 2 + μ. y: 1 + λ = 3 − μ.
- Adding: 2 + 3λ = 5, so λ = 1, then μ = 1.
- z: 1 + 3λ = p + μ gives 4 = p + 1.
Answer: p = 3.
Marks: M1 for λ and μ, M1 for substituting into z, A1 for p = 3.
Example 5. Show that r = (−i + k) + λ(3i + j + 2k) and r = (i − j + 4k) + μ(i + 2j − k) intersect, find the point, and the acute angle between them.
- x: −1 + 3λ = 1 + μ. y: λ = −1 + 2μ. Solving gives λ = 1, μ = 1.
- Check z: 1 + 2(1) = 3 and 4 − 1 = 3, which agree. Point: (2, 1, 3).
- d₁ · d₂ = 3 + 2 − 2 = 3. |d₁| = √14 and |d₂| = √6.
- cos θ = 3 / √84, so θ = 70.9°.
Answer: They meet at (2, 1, 3), at an acute angle of 70.9°.
Marks: M1 for intersection, A1 for the point, M1 for cos θ, A1 for 70.9°.
Common mistakes
- Using the same parameter for both lines. Use λ for one line and μ for the other.
- Not checking the third equation. Two equations can always be solved; the third decides whether the lines meet.
- An incomplete skew proof. Show both that the lines are not parallel and that they do not intersect.
- Comparing positions instead of directions for parallel lines. Only the direction vectors decide whether lines are parallel.
- Getting an obtuse angle between lines. Use the modulus of the scalar product to give the acute angle.
How pairs of lines are examined
Intersecting and skew lines are Pure 4 content (WMA14) in Edexcel International A-Level; in Edexcel A-Level they belong to Further Maths rather than to single Maths (9MA0). Questions usually ask you to:
- show that two lines intersect and find the point of intersection
- prove that two lines are skew
- find an unknown constant given that two lines meet
- find the acute angle between two lines
Questions like these, with full mark schemes, are in the Pure 4 vectors exam questions. For more, try the vectors worksheet or the vectors revision engine.
This builds on vector equations of lines. The angle uses the scalar product.
Frequently asked questions
What are skew lines?
Lines in three dimensions that are not parallel and do not intersect.
How do you show two lines intersect?
Set their equations equal, solve two component equations for the parameters, and show that those values also satisfy the third equation.
Can lines be skew in two dimensions?
No. In two dimensions, any two lines that are not parallel must intersect.