Equations of lines in 3D

Three forms

A line through the point A (position vector a) with direction b:

• Vector form: r = a + λb
• Vector product form: (r - a) × b = 0, or r × b = a × b
• Cartesian form: (x - a1)/b1 = (y - a2)/b2 = (z - a3)/b3 (= λ)

If a component of b is zero, say b3 = 0, write the cartesian form as (x - a1)/b1 = (y - a2)/b2, z = a3.

Why (r - a) × b = 0 works

r - a is the vector from A to a general point R. R lies on the line exactly when AR is parallel to b, which is when (r - a) × b = 0.

To convert r × b = c back to vector form, find one point on the line (try setting one coordinate to 0) and keep b as the direction.

Using the forms

• To test whether a point lies on the line, find λ from one component and check it in the others.
• Two lines are parallel if their direction vectors are multiples of each other.
• A line through A and B has direction b = AB.

Worked example

Write r = (i + 2j - k) + λ(3i - j + 2k) in cartesian form and in the form r × b = c.

1. x = 1 + 3λ, y = 2 - λ, z = -1 + 2λ, so λ = (x - 1)/3 = (y - 2)/(-1) = (z + 1)/2
2. a × b = (2×2 - (-1)(-1), (-1)(3) - 1×2, 1×(-1) - 2×3) = (3, -5, -7)

Answer: (x - 1)/3 = (y - 2)/(-1) = (z + 1)/2; r × (3i - j + 2k) = 3i - 5j - 7k

Check yourself

1. Which is the cartesian form of r = (i + 2j - k) + λ(3i - j + 2k)?
   • (x - 3)/1 = (y + 1)/2 = (z - 2)/(-1)
   • (x + 1)/3 = (y + 2)/(-1) = (z - 1)/2
   • (x - 1)/3 = (y - 2)/(-1) = (z + 1)/2
   • x/3 = y/(-1) = z/2
   Show answer

   (x - 1)/3 = (y - 2)/(-1) = (z + 1)/2. Subtract the point's coordinates on top and divide by the direction components.

2. What does (r - a) × b = 0 say about the point R with position vector r?
   • The vector from A to R is parallel to b
   • The vector from A to R is perpendicular to b
   • R is the point A
   • R is at distance |b| from A
   Show answer

   The vector from A to R is parallel to b. A vector product is zero exactly when the vectors are parallel.

3. A line has equation (x - 2)/1 = (y + 1)/3 = (z - 4)/(-2). Find z at the point where x = 5.
   Show answer

   -2. λ = (5 - 2)/1 = 3, so z = 4 + 3(-2) = -2.

4. Which point lies on the line (r - (i + 2k)) × (2i + j + k) = 0?
   • (3, 1, 2)
   • (5, 2, 4)
   • (2, 1, 1)
   • (4, 2, 3)
   Show answer

   (5, 2, 4). (5, 2, 4) - (1, 0, 2) = (4, 2, 2) = 2(2, 1, 1), which is parallel to b.

5. The line r = (i + 2j + 3k) + λ(i + j) is written as r × b = c with b = i + j. Find c.
   • 3i - 3j + k
   • -3i + 3j + k
   • i + 2j + 3k
   • -3i + 3j - k
   Show answer

   -3i + 3j - k. c = a × b = (2×0 - 3×1, 3×1 - 1×0, 1×1 - 2×1) = (-3, 3, -1).

6. The line through A(1, 2, 3) and B(4, 0, 5) is r = a + λ(b - a). Find λ at the point (10, -4, 9).
   Show answer

   3. b - a = (3, -2, 2). From x: 1 + 3λ = 10 gives λ = 3; check y = 2 - 6 = -4 and z = 3 + 6 = 9.

7. Which line is parallel to r = λ(2i - j + 3k)?
   • r = (2i - j + 3k) + μ(i + j + k)
   • r = μ(3i - j + 2k)
   • r = (i + j) + μ(2i + j + 3k)
   • r = (i + k) + μ(-4i + 2j - 6k)
   Show answer

   r = (i + k) + μ(-4i + 2j - 6k). -4i + 2j - 6k = -2(2i - j + 3k), so the directions are parallel.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (issue 3), Unit FP3

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