Scalar multiples, unit vectors, i j k, the vector AB and direction cosines
Week 9 · Multiplying a vector by a number, equal vectors, unit vectors, the vectors i, j, k, the vector AB, direction angles and direction cosines · 3 lessons
3
Direction angles and direction cosines, and resolving a vector into components
Pure Mathematics Sec 3 Term 1, Unit 3, Lesson 2
GoalFind direction angles and direction cosines, and resolve a vector or a force into components.
New words
direction angles · the angles θₓ, θᵧ, θz (0° to 180°) that a vector makes with the positive x, y, z axesdirection cosines · cos θₓ, cos θᵧ, cos θz: the components of the unit vectorresolving · finding the components of a vector from its size and direction
Explanation
A vector A makes angles θₓ, θᵧ and θz with the positive axes. A component is the norm times a cosine: Aₓ = ‖A‖cos θₓ, Aᵧ = ‖A‖cos θᵧ and Az = ‖A‖cos θz. So A = ‖A‖(cos θₓ i + cos θᵧ j + cos θz k), and dividing by ‖A‖ shows that the three direction cosines (in Arabic جيوب تمام الاتجاه) are exactly the components of the unit vector A/‖A‖. A unit vector has norm 1, therefore cos²θₓ + cos²θᵧ + cos²θz = 1. Use it to find a missing angle: the missing cosine squared is 1 minus the other two squares, which gives two possible angles, one acute and one obtuse. To resolve a force F of known norm along a known direction, multiply F by the unit vector of that direction. The picture shows a rope pulling along the diagonal of a box.
Worked examples
Direction cosines and angles of A = (3, 4, 12). The norm is 13, so the cosines are 3/13, 4/13, 12/13. Check: (9 + 16 + 144)/169 = 1. The angles are cos⁻¹(3/13) ≈ 76.7°, cos⁻¹(4/13) ≈ 72.1° and cos⁻¹(12/13) ≈ 22.6°.
A vector makes 45° with the x-axis and 60° with the y-axis. Find the angle with the z-axis. cos²θz = 1 − cos²45° − cos²60° = 1 − 1/2 − 1/4 = 1/4, so cos θz = ±1/2 and θz = 60° or 120°. Both are possible.
A rope pulls with a force of 35 N along the diagonal of a box 6 m × 2 m × 3 m, away from the corner O. The diagonal is (6, 2, 3) with norm √(36 + 4 + 9) = 7, so the unit vector is (6/7, 2/7, 3/7). The force is F = 35(6/7, 2/7, 3/7) = (30, 10, 15) N. Its direction angles are cos⁻¹(6/7) ≈ 31.0°, cos⁻¹(2/7) ≈ 73.4° and cos⁻¹(3/7) ≈ 64.6°.
Class activity
Compass angles. Hold a pencil so that it points up and a little to the front-right. Estimate the three angles it makes with the edges of a corner of your room. Do the cosines of your three estimates have squares that add to about 1?
Practice
1
A vector of norm 20 makes 30° with the z-axis, and its shadow on the xy-plane makes 45° with the x-axis. Find its components.
Vertical part 20 cos 30° ≈ 17.32. The shadow has length 20 sin 30° = 10, so Aₓ = 10 cos 45° ≈ 7.07 and Aᵧ = 10 sin 45° ≈ 7.07. Components (7.07, 7.07, 17.32).
2
Find the angle between (1, −4, 8) and the y-axis.
The norm is 9, so cos θᵧ = −4/9 and θᵧ = cos⁻¹(−4/9) ≈ 116.4° (obtuse because the y-component is negative).
3
A vector makes 60° with the x-axis and 60° with the y-axis. Find θz.
cos²θz = 1 − 1/4 − 1/4 = 1/2, so cos θz = ±√2/2 and θz = 45° or 135°.
4
Find the vector of norm 12 that makes equal angles with the positive axes.
The direction cosines are equal, so 3c² = 1 and c = 1/√3. The vector is 12(1/√3)(1, 1, 1) = 4√3(1, 1, 1), with norm 4√3 × √3 = 12.
5
A force of 90 N acts along the direction of (1, 2, 2). Find its components and its angle with the z-axis.
The unit vector is (1/3, 2/3, 2/3), so F = (30, 60, 60) N. The angle with the z-axis is cos⁻¹(2/3) ≈ 48.2°.
