All subjects · Mathematics · WMA14 (P4)

Intersecting, parallel and skew lines

In 3D, two lines can be parallel, can meet at a point, or can be skew: not parallel and never meeting. You decide which with simultaneous equations.

Parallel first

Compare the direction vectors. If one is a multiple of the other, the lines are parallel (or the same line).

Finding an intersection

Set the two equations equal: a + λb = c + μd. This gives three equations, one per component.

1. Solve two of them for λ and μ.

2. Check the third equation.

3. If it works, substitute λ into the first line to find the point of intersection.

Skew lines

If the lines are not parallel and the third equation fails, they do not meet: they are skew. Skew lines cannot happen in 2D, where non-parallel lines always meet.

Worked example

Show that r = (i + 2j + 3k) + λ(i - j + 2k) and r = (4i + 2j + 4k) + μ(2i + j - k) intersect, and find the point.

  1. x: 1 + λ = 4 + 2μ; y: 2 - λ = 2 + μ
  2. From y, μ = -λ. Then 1 + λ = 4 - 2λ, so λ = 1 and μ = -1.
  3. z: 3 + 2(1) = 5 and 4 - (-1) = 5, so the third equation holds.

Answer: They meet at (2, 1, 5)

Key idea: Parallel: direction vectors are multiples. Otherwise solve two component equations; if the third holds they intersect, if not they are skew.

Check yourself

  1. What does it mean for two lines to be skew?
    • They are parallel and do not intersect
    • They meet at right angles
    • They are the same line
    • They are not parallel and do not intersect
    Show answer

    They are not parallel and do not intersect. Skew lines go in different directions but never meet. This only happens in 3D.

  2. The lines r = (2i + k) + λ(i + 2j - k) and r = (i + 4j - 4k) + μ(i - j + 2k) intersect. Find the point of intersection.
    • (2, 0, 1)
    • (3, 2, 0)
    • (1, 4, -4)
    • (4, 4, -1)
    Show answer

    (3, 2, 0). x: 2 + λ = 1 + μ; y: 2λ = 4 - μ. So μ = 1 + λ and 2λ = 3 - λ, giving λ = 1, μ = 2. z: 1 - 1 = 0 and -4 + 4 = 0. The point is (3, 2, 0).

  3. What is the relationship between r = (j - k) + λ(2i + k) and r = (i - 2j + 3k) + μ(i + 3j - k)?
    • Intersecting
    • Parallel
    • The same line
    • Skew
    Show answer

    Skew. The directions are not multiples. x: 2λ = 1 + μ and y: 1 = -2 + 3μ give μ = 1, λ = 1. z: -1 + 1 = 0 but 3 - 1 = 2. The third equation fails, so they are skew.

  4. Line 1 has direction 2i - 4j + 6k and line 2 has direction -i + 2j - 3k. What can you conclude?
    • They are parallel or the same line, as 2i - 4j + 6k = -2(-i + 2j - 3k)
    • They must intersect
    • They must be skew
    • They are perpendicular
    Show answer

    They are parallel or the same line, as 2i - 4j + 6k = -2(-i + 2j - 3k). Direction vectors that are multiples give parallel lines. Check a point to see whether they are the same line.

  5. Why can two lines in a 2D plane never be skew?
    • Lines in 2D have no direction
    • 2D lines are always parallel
    • Two non-parallel lines in a plane always meet
    • Skew lines need a z-component of zero
    Show answer

    Two non-parallel lines in a plane always meet. In a plane, lines are either parallel or they cross.

  6. The lines r = i + λ(i + j + k) and r = (3i + ak) + μ(j + 2k) intersect. Find a.
    Show answer

    -2. x: 1 + λ = 3 gives λ = 2. y: λ = μ gives μ = 2. z: λ = a + 2μ gives 2 = a + 4, so a = -2.

  7. After solving two of the component equations for λ and μ, what must you do next?
    • Add λ and μ
    • Find the scalar product
    • Check that the values satisfy the third equation
    • Set λ = μ
    Show answer

    Check that the values satisfy the third equation. Only if all three equations hold do the lines meet.

Practise with instant feedback

Sources

Last updated 7 November 2026. Spotted a mistake? Tell us.