Angles between lines and perpendiculars

Angle between two lines

Use the direction vectors only, not the position vectors. If cos θ comes out negative, the angle is obtuse; the acute angle between the lines is 180° - θ. (Equivalently, use |a · b| in the formula.)

Foot of a perpendicular

To find the point P on the line r = a + λb closest to a point A:

1. Write P in terms of λ.

2. Find AP = p - a in terms of λ.

3. Solve AP · b = 0.

4. Substitute λ to get P.

Shortest distance

The shortest distance from A to the line is |AP|, where P is the foot of the perpendicular. Exam questions usually guide you through these steps.

Worked example

Find the foot of the perpendicular from A(4, 3, 2) to r = (i + 2k) + λ(i + 2j + 2k), and the shortest distance.

1. P = (1 + λ, 2λ, 2 + 2λ), so AP = (λ - 3, 2λ - 3, 2λ)
2. AP · (1, 2, 2) = λ - 3 + 4λ - 6 + 4λ = 9λ - 9 = 0, so λ = 1
3. P = (2, 2, 4) and AP = (-2, -1, 2)

Answer: P(2, 2, 4); shortest distance |AP| = 3

Check yourself

1. Find the acute angle between lines with directions i + 2j + 2k and 2i - j + 2k, to 1 decimal place.
   Show answer

   63.6 °. Scalar product = 2 - 2 + 4 = 4; both lengths are 3. cos θ = 4/9, so θ = 63.6°.

2. Lines have directions i + j and -i + k. Find the acute angle between the lines.
   Show answer

   60 °. Scalar product = -1; lengths sqrt(2) each. cos θ = -1/2 gives 120°, so the acute angle is 180° - 120° = 60°.

3. To find the angle between r = a + λb and r = c + μd, which vectors do you use?
   • a and c
   • a and d
   • a - c and b - d
   • b and d
   Show answer

   b and d. The angle depends only on the directions of the lines.

4. A(5, 3, -1) and the line r = (j + 2k) + λ(2i + j - 2k). Find the value of λ at the foot of the perpendicular from A.
   Show answer

   2. P = (2λ, 1 + λ, 2 - 2λ), so AP = (2λ - 5, λ - 2, 3 - 2λ). AP · (2, 1, -2) = 4λ - 10 + λ - 2 - 6 + 4λ = 9λ - 18 = 0, so λ = 2.

5. The foot of the perpendicular from A(5, 3, -1) to a line is P(4, 3, -2). Find the shortest distance from A to the line, to 3 decimal places.
   Show answer

   1.414. AP = (-1, 0, -1), so |AP| = sqrt(2) = 1.414.

6. What condition locates the foot of the perpendicular P from a point A to the line r = a + λb?
   • AP · b = 0
   • AP · a = 0
   • |AP| = |b|
   • AP = λb
   Show answer

   AP · b = 0. AP must be perpendicular to the direction of the line.

7. Find the angle between the line r = λ(i + 2j + 2k) and the x-axis, to 1 decimal place.
   Show answer

   70.5 °. The x-axis has direction i. cos θ = 1/(3 × 1) = 1/3, so θ = 70.5°.

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3)
• Pearson Edexcel IAL Pure Mathematics 4 scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.