Find the foot of a perpendicular from a point to a line
Choose a point on the line whose separation from the given point has zero scalar product with the line direction.
Section 3.7 · Vectors
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.7. Paper 1 coordinate geometry and algebra are assumed. Work in two or three dimensions as stated. The general section theorem is not included; midpoint geometry is required. The cross product, shortest distance between skew lines and their common perpendicular are not required. Scalar products include angles between lines and the foot of a perpendicular from a point to a line, including 3D solids. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Write vector line equations and calculate scalar products.
- Find a displacement magnitude and interpret segment parameter bounds.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For d = (2, 1, 0)ᵀ, what is d ⋅ d?
What you will learn
- Find a perpendicular foot using H = a + td and (p − H) ⋅ d = 0.
- Check the foot's line membership and perpendicularity, and distinguish a full-line foot from a nearest segment point.
Perpendicularity determines the line parameter
Let a line have position r = a + td with non-zero direction d, and let P have position p. A foot H on the line has position h = a + td. The displacement from H to P is p − a − td. Perpendicularity requires (p − a − td) ⋅ d = 0. Rearranging gives t = [(p − a) ⋅ d]/(d ⋅ d). The positive denominator d ⋅ d = |d|² makes the foot unique for a non-zero line direction.
For a = (1, 1, 0)ᵀ, d = (2, 1, 0)ᵀ and P = (2, 4, 0), p − a = (1, 3, 0)ᵀ. Its dot product with d is 5 and d ⋅ d is 5, so t = 1 and H = (3, 2, 0). The displacement H→P is (−1, 2, 0)ᵀ, whose dot product with d is −2 + 2 = 0. The point-to-line distance is |p − h| = √5.
Check the geometry and any bounds
There are two separate checks: h must satisfy the vector line equation, and the separation vector must be perpendicular to its direction. A point off the line can have a perpendicular-looking displacement but cannot be the foot. For any other point h + sd on the full line, the squared distance from P is |p − h|² + s²|d|² because the perpendicular cross term is zero. This establishes that H is the nearest full-line point, without using a cross product.
A foot on an infinite line can lie outside a finite segment. For segment A = (0, 0, 0) to B = (2, 0, 0) and P = (3, 1, 0), the full-line foot is H = (3, 0, 0), corresponding to t = 3/2 in A + t(B − A). It lies outside 0 ≤ t ≤ 1. The nearest segment point is endpoint B, at distance √2, whereas the full-line distance is 1. Check bounds before interpreting the result as the distance to a finite object.
Pause and explain
A calculated foot has t = 3/2 for a segment defined by 0 ≤ t ≤ 1. What follows?
Worked example
Find the foot from P = (2, 4, 0) to r = (1, 1, 0)ᵀ + t(2, 1, 0)ᵀ and the point-to-line distance.
Show the worked solution
- Use t = [(1, 3, 0) ⋅ (2, 1, 0)]/[(2, 1, 0) ⋅ (2, 1, 0)] = 5/5 = 1.
- The foot is H = (3, 2, 0); H→P = (−1, 2, 0)ᵀ has zero dot product with the line direction.
- The distance is √(1 + 4) = √5; H also satisfies the line equation at t = 1.
Answer H = (3, 2, 0); point-to-line distance √5.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find the foot from P = (4, 3, 0) to r = t(3, 4, 0)ᵀ and its distance from P.
Give me a hint
The anchor is the origin; use p ⋅ d divided by d ⋅ d.
Compare my reasoning
- The parameter is (12 + 12)/(9 + 16) = 24/25.
- H = (72/25, 96/25, 0), so H→P = (28/25, −21/25, 0)ᵀ and its dot product with d is zero.
- The distance is √(28² + 21²)/25 = 35/25 = 7/5.
H = (72/25, 96/25, 0); distance 7/5.
Look for these in your work
- Use the squared direction magnitude in the denominator.
- Verify the foot and compute the separation magnitude exactly.
Find the foot from P = (4, 1, 3) to r = (1, 0, 2)ᵀ + t(1, 2, 2)ᵀ, and its distance from P.
Give me a hint
Subtract the anchor first, then apply the perpendicularity equation.
Compare my reasoning
- p − a = (3, 1, 1)ᵀ gives numerator 7 and denominator 9, so t = 7/9.
- H = (16/9, 14/9, 32/9); H→P = (20/9, −5/9, −5/9)ᵀ has zero dot product with (1, 2, 2)ᵀ.
- The distance is √450/9 = 5√2/3; substitute t = 7/9 back into the full line equation to check H.
H = (16/9, 14/9, 32/9); distance 5√2/3.
Look for these in your work
- Include the non-origin anchor in the displacement.
- Check all three foot coordinates and the exact perpendicular distance.
A finite beam runs from A = (0, 0, 0) to B = (2, 0, 0) metres. For P = (3, 1, 0) m, find the full-line foot, decide whether it lies on the beam, and find the nearest beam point and distance.
Give me a hint
The line parametrisation A + t(B − A) uses segment bounds 0 ≤ t ≤ 1.
Compare my reasoning
- The full-line foot is H = (3, 0, 0), at t = 3/2, with line distance 1 m.
- It lies outside the segment. On 0 ≤ t ≤ 1 the squared separation (3 − 2t)² + 1 decreases to the endpoint t = 1.
- The nearest beam point is B = (2, 0, 0), at distance √[(3 − 2)² + 1²] = √2 m.
Full-line foot H = (3, 0, 0), outside the beam; nearest beam point B = (2, 0, 0), distance √2 m.
Look for these in your work
- Check the foot against the physical parameter range.
- Distinguish full-line and finite-segment distances.
Common mistakes
- Use the difference p − a in the numerator, not p by itself unless a is the origin.
- Check line membership as well as a zero scalar product.
- A full-line foot outside a segment is not the nearest point of that segment.
What can you explain now?
Which scalar product must be zero at a perpendicular foot H on a line with direction d?
Compare with the explanation
(p − h) ⋅ d.
The point-to-foot separation must be perpendicular to the non-zero line direction.
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Make it stick
Derive the foot parameter from a zero scalar product, verify membership and perpendicularity, and check whether a finite segment changes the nearest-point answer.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.