Use scalar products to find angles between vectors and lines
Combine the component and geometric dot-product formulae, check perpendicularity and distinguish directed-vector angles from acute line angles.
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
- Calculate vector magnitudes and identify line directions.
- Use inverse cosine with a requested angle unit and retain exact values until rounding.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the magnitude of (1, 2, 2)ᵀ?
What you will learn
- Calculate scalar products in 2D and 3D and find angles between non-zero vectors.
- Use direction vectors for the acute angle between lines and recognise perpendicular directions.
The component formula produces a scalar
For a = (a₁, a₂, a₃)ᵀ and b = (b₁, b₂, b₃)ᵀ, a ⋅ b = a₁b₁ + a₂b₂ + a₃b₃. In two dimensions use the first two products. This scalar product is a number, not a new vector. Its geometric form is a ⋅ b = |a||b|cos θ for the angle 0 ≤ θ ≤ π between two non-zero directed vectors. Hence cos θ = (a ⋅ b)/(|a||b|). Calculate the dot product and both magnitudes separately before applying inverse cosine.
For a = (1, 2, 2)ᵀ and b = (2, 1, −2)ᵀ, the dot product is 2 + 2 − 4 = 0. Both magnitudes are 3, so cos θ = 0 and θ = 90°. A zero dot product establishes perpendicular directions only when both vectors are non-zero. The zero vector has zero dot product with every vector but supplies no defined angle or direction.
A line has no preferred direction
A line direction can be replaced by its negative without changing the full line. For the acute or right angle between two lines with directions d and e, use cos θ = |d ⋅ e|/(|d||e|), giving 0 ≤ θ ≤ 90°. For a specifically directed-vector angle, keep the sign of d ⋅ e and allow an obtuse angle. If the question explicitly asks for the obtuse angle between the lines, it is the supplement of the acute angle; state which angle you report.
Directions (1, 2, 2)ᵀ and (−2, 1, −2)ᵀ have dot product −4 and magnitudes 3. Their directed-vector angle is arccos(−4/9) ≈ 116.4°, while the acute angle between their lines is arccos(4/9) ≈ 63.6°. A line angle uses direction vectors, not arbitrary position vectors of points on the lines. Retain exact cosine values until the final rounding and use the requested degree or radian output.
Pause and explain
Two non-zero line directions have dot product −4 and both magnitudes 3. Which cosine gives their acute line angle?
Worked example
Find the angle between a = (1, 2, 2)ᵀ and b = (2, 1, −2)ᵀ.
Show the worked solution
- Calculate a ⋅ b = 1·2 + 2·1 + 2·(−2) = 0.
- The magnitudes are |a| = |b| = 3, both non-zero.
- Thus cos θ = 0/(3·3) = 0, giving θ = 90°.
Answer 90°; the two non-zero directions are perpendicular.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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 angle between u = (1, 1, 0)ᵀ and v = (1, 0, 1)ᵀ exactly.
Give me a hint
Calculate the scalar product and the two square-root magnitudes.
Compare my reasoning
- u ⋅ v = 1, while |u| = |v| = √2.
- Therefore cos θ = 1/(√2·√2) = 1/2.
- The directed-vector angle is θ = 60°, or π/3 radians.
60° = π/3 radians.
Look for these in your work
- Sum corresponding-component products.
- Use both non-zero magnitudes and an exact inverse-cosine value.
Lines have direction vectors (1, 2, 2)ᵀ and (−2, 1, −2)ᵀ. Find their acute angle and the angle between the stated directed vectors, to one decimal place.
Give me a hint
The negative dot product affects the directed-vector angle, while its absolute value gives the acute line angle.
Compare my reasoning
- The dot product is −4 and both magnitudes are 3.
- The acute line angle is arccos(4/9) ≈ 63.6°.
- The directed-vector angle is arccos(−4/9) ≈ 116.4°, the supplement of the acute angle.
Acute line angle 63.6°; stated directed-vector angle 116.4°.
Look for these in your work
- Use line directions rather than anchor positions.
- Keep the sign or absolute value appropriate to the requested angle.
A rectangular cuboid has perpendicular edges of lengths 4, 3 and 2 from O along the coordinate axes. Find the acute angle between its body diagonal with direction (4, 3, 2)ᵀ and its length-4 edge, to one decimal place.
Give me a hint
Use edge direction (4, 0, 0)ᵀ and calculate both magnitudes.
Compare my reasoning
- The dot product is 16; the diagonal magnitude is √29 and the edge magnitude is 4.
- The cosine is 16/(4√29) = 4/√29.
- The angle is arccos(4/√29) ≈ 42.0°, found from the 3D components rather than a perspective drawing.
arccos(4/√29) ≈ 42.0°.
Look for these in your work
- Construct the correct 3D diagonal and edge directions.
- Calculate the angle from components, not the apparent drawing angle.
Common mistakes
- A scalar product is a number; do not return a component vector.
- Use actual direction vectors for line angles and distinguish acute from directed-vector angles.
- A zero vector makes the angle formula undefined even though its dot product is zero.
What can you explain now?
Find (1, 1, 0)ᵀ ⋅ (1, 0, 1)ᵀ.
Compare with the explanation
1.
The matching-component products sum to 1 + 0 + 0.
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Make it stick
Use the component and geometric scalar-product formulae together. Explain the non-zero condition and the difference between a directed-vector angle and an acute line angle.
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.