2D & 3D Vectors, Position Vectors, Geometric Proofs & Vector Kinematics
Vectors provide a unified coordinate-free framework for geometry and mechanics in two and three dimensions. In Year 2 A-Level Maths, you extend i-j vectors by adding the orthogonal unit vector k along the z-axis, calculating 3D magnitudes, unit vectors, and angles with the coordinate axes. Examiners test two demanding applications: pure geometric vector proofs (finding unknown ratios where two lines intersect by expressing a point's position vector in two independent ways and equating coefficients) and 2D/3D vector kinematics in Mechanics.
- 3D Coordinates, Magnitude, and Distance Between Two Points
- In three dimensions, the orthogonal unit vectors i, j, and k point along the positive x, y, and z axes respectively: i = (1, 0, 0)^T, j = (0, 1, 0)^T, k = (0, 0, 1)^T. By repeated application of Pythagoras' theorem, the magnitude (length) of a 3D vector a = xi + yj + zk is |a| = √(x² + y² + z²). If point A has position vector OA = (x₁, y₁, z₁)^T and point B has position vector OB = (x₂, y₂, z₂)^T, the displacement vector from A to B is AB = OB - OA = (x₂ - x₁)i + (y₂ - y₁)j + (z₂ - z₁)k, and the distance AB is |AB| = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²].
- Unit Vectors and Angles with the Coordinate Axes
- A unit vector has a magnitude of exactly 1. The unit vector in the direction of any non-zero vector a is â = a / |a|. If a 3D vector a = xi + yj + zk makes angles θ_x, θ_y, and θ_z with the positive x, y, and z axes respectively, its direction cosines are given by cos(θ_x) = x / |a|, cos(θ_y) = y / |a|, and cos(θ_z) = z / |a|. Notice that cos²(θ_x) + cos²(θ_y) + cos²(θ_z) = (x² + y² + z²) / |a|² = 1.
- Parallel, Collinear, and Coplanar Vectors
- Two non-zero vectors a and b are parallel if and only if one is a scalar multiple of the other: a = λb for some scalar λ ≠ 0 (if λ > 0 they point in the same direction; if λ < 0 they point in opposite directions). To prove that three points A, B, and C are collinear (lie on the same straight line), you must show two things: 1. AB = k BC (or AC = k AB) so the vectors are parallel; and 2. They share a common point (B or A).
2D & 3D Vectors, Position Vectors, Geometric Proofs & Vector Kinematics: 3D Molecular, Lattice, DNA & Vector Explorer
Drag directly on the 3D model to rotate it through any angle, or switch tabs to inspect VSEPR bond angles, crystal lattices, DNA base pairing, and 3D space diagonals.
Position Vector r = (4i + 3j + 5k)
- 1. Base Diagonal √(x² + y²):
- √(4² + 3²) = 5.00
- 2. 3D Magnitude |r| = √(x² + y² + z²):
- √50 = 7.07
- 3. Angle with Base tan⁻¹(z / base):
- 45.0°
To find the angle between the space diagonal OP and the horizontal base plane, form the shaded vertical right-angled triangle with adjacent = √(x² + y²) and opposite = z.
Key Concepts & Exam Definitions (6)
- 3D Coordinates, Magnitude, and Distance Between Two Points
- In three dimensions, the orthogonal unit vectors i, j, and k point along the positive x, y, and z axes respectively: i = (1, 0, 0)^T, j = (0, 1, 0)^T, k = (0, 0, 1)^T. By repeated application of Pythagoras' theorem, the magnitude (length) of a 3D vector a = xi + yj + zk is |a| = √(x² + y² + z²). If point A has position vector OA = (x₁, y₁, z₁)^T and point B has position vector OB = (x₂, y₂, z₂)^T, the displacement vector from A to B is AB = OB - OA = (x₂ - x₁)i + (y₂ - y₁)j + (z₂ - z₁)k, and the distance AB is |AB| = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²].
- Unit Vectors and Angles with the Coordinate Axes
- A unit vector has a magnitude of exactly 1. The unit vector in the direction of any non-zero vector a is â = a / |a|. If a 3D vector a = xi + yj + zk makes angles θ_x, θ_y, and θ_z with the positive x, y, and z axes respectively, its direction cosines are given by cos(θ_x) = x / |a|, cos(θ_y) = y / |a|, and cos(θ_z) = z / |a|. Notice that cos²(θ_x) + cos²(θ_y) + cos²(θ_z) = (x² + y² + z²) / |a|² = 1.
- Parallel, Collinear, and Coplanar Vectors
- Two non-zero vectors a and b are parallel if and only if one is a scalar multiple of the other: a = λb for some scalar λ ≠ 0 (if λ > 0 they point in the same direction; if λ < 0 they point in opposite directions). To prove that three points A, B, and C are collinear (lie on the same straight line), you must show two things: 1. AB = k BC (or AC = k AB) so the vectors are parallel; and 2. They share a common point (B or A).
- Dividing a Line Segment in a Given Ratio & Geometric Proofs
- If a point P divides the line segment AB in the ratio λ : μ (so AP : PB = λ : μ), then AP = [λ / (λ + μ)] AB, and the position vector of P relative to origin O is OP = OA + [λ / (λ + μ)](OB - OA) = [μ / (λ + μ)]OA + [λ / (λ + μ)]OB. To find the intersection point X of two lines in a triangle defined by non-parallel vectors a and b, write OX in two different ways using unknown scalar parameters λ and μ (e.g., OX = OA + λ AB and OX = OC + μ CD), express both in terms of a and b, and equate coefficients of a and b to solve simultaneous equations for λ and μ.
- 3D Geometric Shapes (Parallelograms, Rhombuses, Triangles)
- Given 3D points A, B, C, D: 1. ABCD is a parallelogram if opposite sides have equal displacement vectors: AB = DC (not CD). 2. ABCD is a rhombus if AB = DC and adjacent sides have equal lengths: |AB| = |BC|. 3. To find the angle θ inside triangle ABC at vertex A using 3D vectors (without using the Further Maths dot product), calculate the three side lengths |AB|, |AC|, and |BC|, and apply the Cosine Rule: cos A = (|AB|² + |AC|² - |BC|²) / (2 |AB| |AC|), then use Area = (1/2) |AB| |AC| sin A.
- Vector Kinematics: Constant Acceleration and Collisions
- In 2D or 3D Mechanics with constant acceleration a, the SUVAT equations v = u + at and r = r₀ + ut + (1/2)at² apply directly in vector form (where r₀ is the initial position vector at t = 0). A particle is moving parallel to a vector d (e.g., due East i or North-East i + j) when the ratio of its velocity components matches the ratio of d's components. Two particles P and Q collide at time t if and only if their position vectors are equal at the same instant: r_P(t) = r_Q(t) for both the i and j (and k) components simultaneously.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Given position vectors OA = a and OB = b, what is the displacement vector AB and what is the position vector of the midpoint M of AB?
How do you find the unit vector in the direction of a = 3i - 4j + 12k?
How do you find the angle θ_y that a 3D vector a = xi + yj + zk makes with the positive y-axis?
In a parallelogram ABCD (labelled in order around the perimeter), which vector equals AB?
How do you determine the time t when a moving particle with velocity v = (12 - 3t)i + (4 + 2t)j is moving parallel to the vector 2i + 3j?
Why can you equate the coefficients of a and b in λ₁a + μ₁b = λ₂a + μ₂b?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Relative to a fixed origin O, the points A, B, and C have position vectors OA = (2, -1, 4)^T, OB = (5, 3, -8)^T, and OC = (6, 2, 16)^T. (a) Find the unit vector in the direction of AB. (b) Find the acute angle that AB makes with the positive z-axis. (c) Show that triangle ABC is isosceles and find its exact area. (7 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Find displacement vector AB = OB - OA = (5 - 2, 3 - (-1), -8 - 4)^T = (3, 4, -12)^T [1 mark]. 2. Find magnitude |AB| = √(3² + 4² + (-12)²) = √(9 + 16 + 144) = √169 = 13; unit vector = (1/13)(3i + 4j - 12k) [1 mark]. 3. (b) Use cos(θ_z) = z / |AB| = -12 / 13 → obtuse angle with +z axis is arccos(-12/13) = 157.4°, so the acute angle with the z-axis is arccos(12/13) = 22.6° [1 mark]. 4. (c) Find AC and BC and their lengths: for OC = (6, 2, 16)^T, AC = OC - OA = (4, 3, 12)^T so |AC| = √(4² + 3² + 12²) = √(16 + 9 + 144) = √169 = 13. Since |AB| = |AC| = 13, triangle ABC is isosceles [2 marks]. 5. Use the scalar product to find cos A: AB · AC = (3)(4) + (4)(3) + (-12)(12) = 12 + 12 - 144 = -120, so cos A = -120 / (13 × 13) = -120/169 and sin A = √(1 - (-120/169)²) = 119/169. Area = (1/2)|AB||AC| sin A = (1/2)(169)(119/169) = 119/2 = 59.5 square units [2 marks].
- Q2
In triangle OAB, OA = a and OB = b. The point M is the midpoint of OA, and the point N lies on AB such that AN : NB = 1 : 2. The lines ON and BM intersect at the point X. By expressing OX in two different ways using scalar parameters λ and μ, find the position vector OX in terms of a and b, and determine the ratio OX : XN. (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. Find the position vectors of M and N in terms of a and b: OM = (1/2)a, and ON = OA + (1/3)AB = a + (1/3)(b - a) = (2/3)a + (1/3)b [1 mark]. 2. Since X lies on the line ON, write OX = λ ON = (2/3)λ a + (1/3)λ b for some scalar λ [1 mark]. 3. Find the vector BM = OM - OB = (1/2)a - b; since X also lies on BM, write OX = OB + μ BM = b + μ((1/2)a - b) = (1/2)μ a + (1 - μ)b [1 mark]. 4. Since a and b are non-parallel, equate the coefficients of a and b in the two expressions for OX: (2/3)λ = (1/2)μ (so μ = (4/3)λ) and (1/3)λ = 1 - μ [1 mark]. 5. Substitute μ = (4/3)λ into the second equation: (1/3)λ = 1 - (4/3)λ → (5/3)λ = 1 → λ = 3/5 (and μ = 4/5) [1 mark]. 6. Substitute λ = 3/5 into OX = λ ON: OX = (2/5)a + (1/5)b, and since OX = (3/5)ON, X divides ON in the ratio OX : XN = 3 : 2 [1 mark].
- Q3
At time t = 0, a ship P is at the point with position vector (2i + 8j) km and is moving with constant velocity (4i - j) km h⁻¹. At the same instant t = 0, a second ship Q is at the point with position vector (14i - 7j) km and is moving with constant velocity (-2i + 4j) km h⁻¹. (a) Write down expressions for the position vectors r_P and r_Q at time t hours. (b) Show that P and Q do not collide, and find the minimum distance between the two ships during their motion. (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Use r = r₀ + vt for each ship: r_P = (2 + 4t)i + (8 - t)j and r_Q = (14 - 2t)i + (-7 + 4t)j [1 mark]. 2. (b) Equate the i-components to find when their x-coordinates match: 2 + 4t = 14 - 2t → 6t = 12 → t = 2 hours [1 mark]. 3. Check the j-components at t = 2: for P, y_P = 8 - 2 = 6; for Q, y_Q = -7 + 4(2) = 1. Since 6 ≠ 1, the ships are not at the same position at t = 2, so they do not collide [1 mark]. 4. Find the relative displacement vector PQ = r_Q - r_P at general time t: PQ = (12 - 6t)i + (-15 + 5t)j [1 mark]. 5. Express the squared distance D² = |PQ|² as a quadratic in t: D² = (12 - 6t)² + (-15 + 5t)² = (144 - 144t + 36t²) + (225 - 150t + 25t²) = 61t² - 294t + 369 [1 mark]. 6. Differentiate D² with respect to t (or complete the square) to find the minimum: d(D²)/dt = 122t - 294 = 0 → t = 147 / 61 ≈ 2.41 hours; substituting t = 147/61 into D = √(61t² - 294t + 369) gives D_min = √(900 / 61) = 30 / √61 ≈ 3.84 km [1 mark].
- Q4
A particle of mass 0.5 kg is acted upon by three forces: F₁ = (3i - 2j + 5k) N, F₂ = (-i + 6j - 2k) N, and F₃ = (pi + qj + rk) N. Given that the particle accelerates with constant acceleration a = (8i + 4j - 6k) m s⁻², find the values of p, q, and r, and find the magnitude of F₃. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Apply Newton's Second Law in 3D vector form: ΣF = F₁ + F₂ + F₃ = ma [1 mark]. 2. Substitute m = 0.5 and a = (8i + 4j - 6k) to find the resultant force: ma = 0.5(8i + 4j - 6k) = (4i + 2j - 3k) N [1 mark]. 3. Equate components in (3 - 1 + p)i + (-2 + 6 + q)j + (5 - 2 + r)k = 4i + 2j - 3k: 2 + p = 4 → p = 2; 4 + q = 2 → q = -2; 3 + r = -3 → r = -6 [1 mark]. 4. Calculate the magnitude of F₃ = (2i - 2j - 6k) N: |F₃| = √(2² + (-2)² + (-6)²) = √(4 + 4 + 36) = √44 = 2√11 ≈ 6.63 N [1 mark].
Common Exam Questions (4)
Should I write vectors in i, j, k form or as column vectors in my working?
Always convert i, j, k vectors into column vectors for your working. Column vectors make it virtually impossible to accidentally add an i-component to a j-component or miss a zero component (e.g., writing 3i - 5k as (3, 0, -5)^T). Just write your final answer in whatever format the question requested.
What is the difference between a position vector and a displacement vector?
A position vector (like OA, OB, or r) always starts at the fixed origin O and specifies the location of a point in space. A displacement vector (like AB = OB - OA) specifies the relative movement from point A to point B.
How do I find the time when Particle P is due North of Particle Q?
If i is East and j is North, P is due North of Q when: 1. Their East (i) components are equal (x_P = x_Q), and 2. P's North (j) component is strictly greater than Q's North component (y_P > y_Q). Solve x_P(t) = x_Q(t) for t, then check that y_P(t) > y_Q(t) at that time.
Can I use the scalar (dot) product a · b = |a||b|cos θ in regular A-Level Maths to find the angle between two 3D vectors?
Yes. Even though the dot product is formally taught in Further Maths, AQA, Edexcel, and OCR A-Level Maths mark schemes explicitly allow full marks for either the Cosine Rule or the scalar product method when finding the angle between 3D vectors.
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