Space coordinates, distance, midpoint and the first vectors
Week 8 · The three-dimensional coordinate system, the distance and the midpoint in space, and vectors in space: position vector, norm and addition · 3 lessons
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Vectors in space: position vector, components, norm and addition
Pure Mathematics Sec 3 Term 1, Unit 3, Lessons 1–2
GoalDescribe a vector by its components, find its norm, and add vectors.
New words
vector · a quantity with a size and a direction, drawn as an arrowposition vector · the vector from the origin O to a point A, written (Aₓ, Aᵧ, Az)norm · the length of a vector, ‖A‖ = √(Aₓ² + Aᵧ² + Az²)
Explanation
Some quantities have a size only, like a mass of 5 kg. Others need a direction too: a force, a velocity, a displacement. We draw them as arrows and call them vectors (in Arabic متجهات). In the plane, “3 m east and 4 m north” is the vector (3, 4); its length is √(9 + 16) = 5 m. In space we use three components. The position vector of a point A(Aₓ, Aᵧ, Az) is the arrow from O to A, written A = (Aₓ, Aᵧ, Az). The numbers are its components along the three axes. Its length is the norm ‖A‖ = √(Aₓ² + Aᵧ² + Az²), which is just the distance OA. Two vectors are added component by component: (a, b, c) + (d, e, f) = (a + d, b + e, c + f). In a picture, place the second arrow at the tip of the first (or complete the parallelogram, as in the plane picture) and the sum goes from the start to the end. Addition behaves like addition of numbers: A + B is again a vector of space, A + B = B + A, (A + B) + C = A + (B + C), the zero vector (0, 0, 0) changes nothing, and every A has an opposite −A = (−Aₓ, −Aᵧ, −Az) with A + (−A) = (0, 0, 0). In general ‖A + B‖ ≤ ‖A‖ + ‖B‖.
Worked examples
Norms. A = (3, −4, 12) has ‖A‖ = √(9 + 16 + 144) = 13. B = (2, 3, −6) has ‖B‖ = √(4 + 9 + 36) = 7. The point P(−2, 5, 1) has position vector (−2, 5, 1).
Add the vectors of Example 1. A + B = (3 + 2, −4 + 3, 12 − 6) = (5, −1, 6). Its norm is √(25 + 1 + 36) = √62 ≈ 7.9, which is less than ‖A‖ + ‖B‖ = 20.
A drone makes two moves, u = (6, 2, 3) m and v = (−2, 4, 6) m. Where does it end, and how far is it from the start? The total move is u + v = (4, 6, 9), so it ends at the point (4, 6, 9) and its distance from the start is √(16 + 36 + 81) = √133 ≈ 11.5 m.
Class activity
Walk it. On squared paper draw the plane vectors a = (3, 1) and b = (1, 3) from the origin, then draw the parallelogram they make. Check that the diagonal from the origin ends at (4, 4).
Practice
1
Find the norm of (2, −6, 9).
√(4 + 36 + 81) = √121 = 11.
2
Add A = (4, 1, −2) and B = (−3, 0, 5).
(4 − 3, 1 + 0, −2 + 5) = (1, 1, 3).
3
Does the vector (0, −4, 3) lie in a coordinate plane? Find its norm.
Its x-component is 0, so it lies in the yz-plane. Norm √(16 + 9) = 5.
4
Find the vector C with A + C = (0, 0, 0) for A = (5, −2, 1).
C = −A = (−5, 2, −1), the additive inverse.
5
Two forces F₁ = (2, 6, 3) N and F₂ = (2, −2, 4) N pull on a ring. Find the resultant force and its norm.
F₁ + F₂ = (4, 4, 7) N and its norm is √(16 + 16 + 49) = 9 N.