The scalar product, angles between vectors and components
Week 10 · The scalar product: definition, properties, coordinates, the angle between two vectors, the component and the projection of one vector on another · 3 lessons
1
The scalar product: definition, special cases and properties
Pure Mathematics Sec 3 Term 1, Unit 3, Lesson 3
GoalDefine the scalar product A · B = ‖A‖‖B‖cos θ, know its special values, and use its properties.
New words
scalar product · A · B = ‖A‖‖B‖cos θ, where θ is the angle between the vectors; the result is a numberperpendicular vectors · non-zero vectors with A · B = 0, that is θ = 90°angle between vectors · the angle θ (0° to 180°) between the two arrows when both start at the same point
Explanation
There are two ways to multiply vectors. One gives a vector (next week). The other gives a number and is called the scalar product, or dot product (in Arabic الضرب القياسي). Draw A and B from the same point. If one starts elsewhere, slide it without turning it. Let θ be the angle between them, with 0° ≤ θ ≤ 180°. Then A · B = ‖A‖‖B‖ cos θ. Its sign follows the cosine: positive for an acute angle, negative for an obtuse angle, zero for a right angle. Special cases: same direction (θ = 0°) gives ‖A‖‖B‖; opposite directions (θ = 180°) give −‖A‖‖B‖; and perpendicular non-zero vectors give 0, and conversely, if A · B = 0 and neither vector is zero, then they are perpendicular. In particular A · A = ‖A‖². Properties: A · B = B · A, A · (B + C) = A · B + A · C and (kA) · B = A · (kB) = k(A · B). For the unit vectors of the axes: i · i = j · j = k · k = 1 and i · j = j · k = k · i = 0.
Worked examples
Let ‖A‖ = 5 and ‖B‖ = 6. For θ = 60°: A · B = 30 cos 60° = 15. For θ = 120°: 30 cos 120° = −15. For θ = 90°: 0. For θ = 0°: 30.
Rectangle ABCD with AB = 8 and BC = 6, so the diagonal AC = 10 and cos ∠BAC = 8/10. AB · AC = 8 × 10 × 0.8 = 64. AB · DC = 64 (the same direction, 8 × 8). AB · BC = 0 (perpendicular). AB · CA = 8 × 10 × (−0.8) = −64, because CA points the other way along the diagonal, so the angle is 180° − ∠BAC.
If ‖A‖ = 3, ‖B‖ = 4 and A · B = 6, then cos θ = 6/12 = 1/2 and θ = 60°. By properties ‖A + B‖² = (A + B) · (A + B) = A · A + 2A · B + B · B = 9 + 12 + 16 = 37 and ‖A − B‖² = 9 − 12 + 16 = 13.
Class activity
Angles with a protractor. Draw two arrows from a point with an angle of 40° between them and lengths 5 cm and 4 cm. Calculate A · B = 20 cos 40° ≈ 15.3. Then change the angle to 140° and see how the sign changes.
Practice
1
Find A · B when ‖A‖ = 7, ‖B‖ = 2 and θ = 135°.
14 cos 135° = −7√2 ≈ −9.9.
2
Equilateral triangle ABC of side 6 cm. Find AB · AC, AB · BC and AB · CB.
AB · AC = 36 cos 60° = 18. For AB · BC the two arrows start at different points; the angle between them is 120°, so 36 cos 120° = −18. AB · CB: CB = −BC, so the answer is +18.
3
Show that (A + B) · (A − B) = ‖A‖² − ‖B‖² and use it when ‖A‖ = 5 and ‖B‖ = 3.
Expand: A · A − A · B + B · A − B · B = ‖A‖² − ‖B‖² (the two middle terms cancel since A · B = B · A). For 5 and 3 the value is 25 − 9 = 16.
4
A and B are perpendicular unit vectors. Find (A − 2B) · (3A + B).
3A · A + A · B − 6B · A − 2B · B = 3 + 0 − 0 − 2 = 1.
