All subjects · Further Mathematics · WFM03 Further Pure Mathematics 3

Equations of planes

Edexcel IAL Further Mathematics · WFM03 Further Pure Mathematics 3

A plane is fixed by one point and a normal direction, or by one point and two directions within it. F3 uses the scalar product form r · n = p, the parametric form r = a + sb + tc and the cartesian form ax + by + cz = d.

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Scalar product form

If A is on the plane and n is a normal vector, every point R satisfies (r - a) · n = 0, so r · n = a · n = p.

Writing r = (x, y, z) and n = (n1, n2, n3) gives the cartesian form n1x + n2y + n3z = p. So the coefficients of x, y, z are a normal vector.

If n is a unit vector, |p| is the perpendicular distance from the origin to the plane; in general that distance is |p|/|n|.

Parametric form

r = a + sb + tc, where A is a point on the plane and b and c are non-parallel vectors in the plane.

To convert to r · n = p: n = b × c, then p = a · n.

For a plane through three points A, B, C: use b = AB and c = AC.

Checking

Worked example

Find the cartesian equation of the plane through A(1, 0, 0), B(0, 2, 0) and C(0, 0, 3).

  1. AB = (-1, 2, 0), AC = (-1, 0, 3)
  2. n = AB × AC = (6, 3, 2)
  3. p = a · n = (1, 0, 0) · (6, 3, 2) = 6

Answer: 6x + 3y + 2z = 6

Key idea: r · n = p with p = a · n; cartesian n1x + n2y + n3z = p; r = a + sb + tc with n = b × c. Distance from O to the plane is |p|/|n|.

Check yourself

  1. A plane passes through (1, 2, 3) and has normal 2i - j + 4k. Its equation is r · (2i - j + 4k) = p. Find p.
    Show answer

    12. p = (1, 2, 3) · (2, -1, 4) = 2 - 2 + 12 = 12.

  2. Which is the cartesian form of r · (3i - 2j + k) = 5?
    • 3x + 2y + z = 5
    • x/3 - y/2 + z = 5
    • (x - 3) - (y + 2) + (z - 1) = 5
    • 3x - 2y + z = 5
    Show answer

    3x - 2y + z = 5. r · n = 3x - 2y + z.

  3. The plane through A(1, 0, 0), B(0, 2, 0) and C(0, 0, 3) has equation 6x + 3y + 2z = d. Find d.
    Show answer

    6. n = AB × AC = (6, 3, 2), and d = a · n = 6. Check B: 3 × 2 = 6.

  4. The plane r = (i + j) + s(i + 2k) + t(j - k) has cartesian equation 2x - y - z = d. Find d.
    Show answer

    1. n = (1, 0, 2) × (0, 1, -1) = (-2, 1, 1), so -2x + y + z = -1, which is 2x - y - z = 1.

  5. Find the perpendicular distance from the origin to the plane 2x - y + 2z = 9.
    Show answer

    3. |n| = sqrt(4 + 1 + 4) = 3, so the distance is 9/3 = 3.

  6. Which vector is normal to the plane 4x - y + 3z = 7?
    • 4i + j + 3k
    • 7i
    • 8i - 2j + 6k
    • i - 4j + 3k
    Show answer

    8i - 2j + 6k. The normal is (4, -1, 3), and any multiple of it, such as (8, -2, 6), is also normal.

  7. Which point lies on the plane r · (1, 2, -1) = 4?
    • (3, 1, 1)
    • (1, 2, -1)
    • (4, 0, 1)
    • (0, 2, 1)
    Show answer

    (3, 1, 1). 3 + 2 - 1 = 4.

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