All subjects · Further Mathematics · WFM03 Further Pure Mathematics 3

Points, lines and planes

Edexcel IAL Further Mathematics · WFM03 Further Pure Mathematics 3

Once you can write a plane as ax + by + cz = d, many problems become short: how far a point is from the plane, where a line meets it, and the foot of the perpendicular from a point.

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Distance from a point to a plane

The perpendicular distance from (x1, y1, z1) to ax + by + cz = d is

|ax1 + by1 + cz1 - d| / sqrt(a2 + b2 + c2).

This is in the formula booklet. In vector form: |a · n - p|/|n| for the point A and plane r · n = p. The distance between two parallel planes r · n = p1 and r · n = p2 is |p1 - p2|/|n|.

Where a line meets a plane

Write the general point of the line in terms of λ, substitute into the plane's cartesian equation, solve for λ and substitute back.

If you get 0 = 0 the line lies in the plane; if you get a contradiction like 0 = 5 the line is parallel to the plane. Both happen when b · n = 0.

Foot of the perpendicular

The line through P perpendicular to the plane is r = p + λn. Where it meets the plane is the foot of the perpendicular; reflecting P uses the same λ doubled.

Angles between a line and a plane, or between two planes, are not listed in the F3 specification. If you meet one, it is a scalar product (P4) skill: sin θ = |b · n|/(|b||n|) for a line and a plane.

Worked example

Find where the line r = (i + 2k) + λ(i + j - k) meets the plane x + 2y + z = 9.

  1. General point (1 + λ, λ, 2 - λ)
  2. (1 + λ) + 2λ + (2 - λ) = 9, so 3 + 2λ = 9 and λ = 3
  3. Substitute λ = 3

Answer: (4, 3, -1)

Key idea: Point to plane: |ax1 + by1 + cz1 - d|/sqrt(a2 + b2 + c2). Line meets plane: substitute the general point. b · n = 0 means parallel to the plane or in it.

Check yourself

  1. Find the perpendicular distance from (3, 1, -2) to the plane 2x - y + 2z = 5. Give your answer as a fraction in its simplest form.
    Show answer

    4/3. |2(3) - 1 + 2(-2) - 5|/sqrt(4 + 1 + 4) = |-4|/3 = 4/3.

  2. The line r = (i + 2k) + λ(i + j - k) meets the plane x + 2y + z = 9 at P. Find the z-coordinate of P.
    Show answer

    -1. λ = 3 gives P = (4, 3, -1).

  3. The line through (5, 4, 3) perpendicular to the plane x + y + z = 3 is r = (5, 4, 3) + λ(1, 1, 1). Find λ at the foot of the perpendicular.
    Show answer

    -3. (5 + λ) + (4 + λ) + (3 + λ) = 3, so 12 + 3λ = 3 and λ = -3. The foot is (2, 1, 0).

  4. Find the distance from (5, 4, 3) to the plane x + y + z = 3, to 3 decimal places.
    Show answer

    5.196. |5 + 4 + 3 - 3|/sqrt(3) = 9/sqrt(3) = 3sqrt(3) = 5.196.

  5. A line with direction b and a plane with normal n satisfy b · n = 0. What can you conclude?
    • The line is perpendicular to the plane
    • The line meets the plane at exactly one point
    • The line passes through the origin
    • The line is parallel to the plane or lies in it
    Show answer

    The line is parallel to the plane or lies in it. b · n = 0 means the line is perpendicular to the normal, so it runs along the plane's direction.

  6. How is the line r = (i + j) + λ(i - j) related to the plane x + y + 2z = 5?
    • It lies in the plane
    • It is parallel to the plane and does not meet it
    • It meets the plane at exactly one point
    • It is perpendicular to the plane
    Show answer

    It is parallel to the plane and does not meet it. b · n = 1 - 1 + 0 = 0, so the line is parallel to the plane or in it. The point (1, 1, 0) gives 1 + 1 + 0 = 2, not 5, so the line is not in the plane.

  7. Find the distance between the parallel planes 2x - y + 2z = 3 and 2x - y + 2z = 12.
    Show answer

    3. |12 - 3|/sqrt(4 + 1 + 4) = 9/3 = 3.

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