All subjects · Further Mathematics · WFM03 Further Pure Mathematics 3

Shortest distances between lines

Edexcel IAL Further Mathematics · WFM03 Further Pure Mathematics 3

Skew lines are neither parallel nor intersecting. The shortest distance between them is along their common perpendicular, whose direction is the vector product of their directions.

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Skew lines

For r = a + λb and r = c + μd, the common perpendicular has direction n = b × d. The shortest distance is the length of the projection of AC (from a point on one line to a point on the other) onto n:

d = |(c - a) · (b × d)| / |b × d|.

If this is 0, the lines intersect (provided they are not parallel).

Parallel lines and a point to a line

If b and d are parallel, b × d = 0 and the formula fails. The distance between parallel lines, or from a point P to the line r = a + λb, is

|(p - a) × b| / |b|,

because |AP × b| = |AP||b| sin θ and |AP| sin θ is the perpendicular distance.

Exam approach

1. Write down a, b, c, d clearly.

2. Find b × d and check it is non-zero.

3. Find c - a and the scalar product.

4. Divide by |b × d| and take the modulus.

You may also be asked to show lines intersect: equate components and solve for λ and μ, or show the shortest distance is 0.

Worked example

Find the shortest distance between r = (2i + k) + λ(i - j) and r = (j + 3k) + μ(j + 2k).

  1. b × d = (1, -1, 0) × (0, 1, 2) = (-2, -2, 1), length 3
  2. c - a = (0, 1, 3) - (2, 0, 1) = (-2, 1, 2)
  3. (c - a) · (b × d) = 4 - 2 + 2 = 4

Answer: Shortest distance = 4/3

Key idea: Skew lines: |(c - a) · (b × d)|/|b × d|. Point to line or parallel lines: |(p - a) × b|/|b|. Zero distance means the lines meet.

Check yourself

  1. Find the shortest distance between r = (i + 2k) + λ(i + 2j - k) and r = (j - k) + μ(2i + j + k), to 3 decimal places.
    Show answer

    0.577. b × d = (3, -3, -3), c - a = (-1, 1, -3), and the scalar product is -3 - 3 + 9 = 3. |b × d| = sqrt(27) = 3sqrt(3), so d = 3/(3sqrt(3)) = 0.577.

  2. Find the distance from P(3, 1, 2) to the line r = i + λ(i + 2j + 2k), to 3 decimal places.
    Show answer

    1.374. AP = (2, 1, 2). AP × b = (1×2 - 2×2, 2×1 - 2×2, 2×2 - 1×1) = (-2, -2, 3), length sqrt(17). Divide by |b| = 3: 1.374.

  3. The lines r = (i + 2j + ck) + λ(i + k) and r = j + μ(i + j + 2k) intersect, where c is a constant. Find c.
    Show answer

    2. b × d = (1, 0, 1) × (1, 1, 2) = (-1, -1, 1). c - a = (-1, -1, -c). For intersecting lines (c - a) · (b × d) = 1 + 1 - c = 0, so c = 2.

  4. The shortest distance between two non-parallel lines is 0. What does this mean?
    • The lines are parallel
    • The lines intersect
    • The lines are skew
    • The lines are perpendicular
    Show answer

    The lines intersect. A zero distance means some point is on both lines.

  5. In d = |(c - a) · (b × d)|/|b × d|, why divide by |b × d|?
    • To make the answer an integer
    • Because b × d is parallel to each line
    • To convert the answer from a volume to an area
    • To project c - a onto a unit vector along the common perpendicular
    Show answer

    To project c - a onto a unit vector along the common perpendicular. (c - a) · n/|n| is the component of c - a in the direction of n, which is the length of the common perpendicular.

  6. Find the distance between the parallel lines r = (i + 2j + 3k) + λ(i + j + k) and r = (2i + 4k) + μ(i + j + k), to 3 decimal places.
    Show answer

    2.449. c - a = (1, -2, 1). (c - a) × b = (-2×1 - 1×1, 1×1 - 1×1, 1×1 - (-2)×1) = (-3, 0, 3), length sqrt(18). Divide by sqrt(3): sqrt(6) = 2.449.

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Last updated 8 November 2026. Spotted a mistake? Tell us.