Flashcards · IB Diploma Programme Mathematics: Applications and Interpretation · Topic 3

Geometry and trigonometry flashcards

81 cards on Topic 3, subtopics 3.1 to 3.16. Pick a subtopic or study them all.

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3.1 3D distance and midpoint; volume and surface area of pyramids, cones, spheres and combinations; angles between lines/planes

Distance and midpoint in 3D: d = √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²) Midpoint ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2) Example: (1, 2, 3) to (4, 6, 15) is √(9 + 16 + 144) = 13.
Volumes of a right pyramid, right cone and sphere: Pyramid V = (1/3)Ah, A = base area. Cone V = (1/3)πr²h. Sphere V = (4/3)πr³; hemisphere (2/3)πr³.
Surface areas of a cone, sphere and hemisphere: Cone curved surface πrl, slant height l = √(r² + h²). Sphere 4πr². Hemisphere curved 2πr², plus πr² for the flat face (3πr² in total).
How do you find the angle between a line and a plane?: 1. Drop a perpendicular from a point on the line to the plane. 2. Join its foot to where the line meets the plane (the projection). 3. Solve the right-angled triangle: line = hypotenuse, perpendicular = opposite.
Surface area of a combined solid: the key rule: List every face, cross out faces that are joined together, and add the rest. Use slant height for cones and triangular faces.

3.2 Right-angled trigonometry; the sine rule, cosine rule and area of a triangle

Sine rule: a/sin A = b/sin B = c/sin C
Cosine rule, and its form for finding an angle: c² = a² + b² − 2ab cos C cos C = (a² + b² − c²)/(2ab)
Area of a triangle from two sides and the included angle: Area = (1/2)ab sin C
Which rule? Two angles and a side / two sides and the included angle / three sides: Two angles and a side: sine rule. Two sides and the included angle: cosine rule. Three sides: cosine rule rearranged for cos. Right angle present: SOH CAH TOA or Pythagoras.
A triangle has sides 7, 9 and 12. Find the largest angle: cos C = (49 + 81 − 144)/(2 × 7 × 9) = −14/126, so C = 96.4°. A negative cosine means an obtuse angle.

3.3 Applications of right- and non-right-angled trigonometry, including Pythagoras’ theorem and bearings

Angles of elevation and depression: Both measured from the horizontal: elevation looking up, depression looking down. The angle of depression from A to B equals the angle of elevation from B to A (alternate angles).
How is a bearing measured and written?: Clockwise from north, written as three figures, such as 077°.
The bearing of B from A is 062°. Find the bearing of A from B: Back bearing = 062° + 180° = 242°.
A ship sails on 070°, then turns onto 150°. Find the interior angle at the turn: Back bearing at the turn = 070° + 180° = 250°. Interior angle = 250° − 150° = 100°. Draw a north line at every point first.
From the top of a 60 m cliff the angle of depression of a buoy is 25°. How far is the buoy from the foot of the cliff?: tan 25° = 60/d, so d = 60/tan 25° = 129 m (3 s.f.).

3.4 The circle: length of an arc and area of a sector

Arc length and sector area (θ in degrees): Arc length = (θ/360) × 2πr Sector area = (θ/360) × πr²
Perimeter of a sector: Arc length + 2r. Do not forget the two radii.
Area of a segment: Sector area − triangle area, where the triangle area is (1/2)r² sin θ.
A sector has radius 6 cm and angle 120°. Find its arc length and area: Arc = (120/360) × 12π = 4π = 12.6 cm. Area = (120/360) × 36π = 12π = 37.7 cm².

3.5 Equations of perpendicular bisectors

Perpendicular bisector of AB: The line through the midpoint of AB at right angles to AB. Every point on it is equidistant from A and B.
How do you find the perpendicular bisector of two points?: 1. Find the midpoint M. 2. Find the gradient m of AB. 3. Perpendicular gradient = −1/m. 4. y − y_M = (−1/m)(x − x_M), then rearrange.
Find the perpendicular bisector of (1, −2) and (5, 6) in the form ax + by + d = 0: M = (3, 2), gradient 2, so m⊥ = −1/2. y − 2 = −(1/2)(x − 3) gives x + 2y − 7 = 0.
Perpendicular bisector of a horizontal or a vertical segment: Horizontal segment: a vertical bisector x = x_M. Vertical segment: a horizontal bisector y = y_M. No −1/m needed.
A segment lies on 3x + y = 11 with midpoint (2, 5). Find its perpendicular bisector: The segment's gradient is −3, so m⊥ = 1/3. y − 5 = (1/3)(x − 2) gives x − 3y + 13 = 0.

3.6 Voronoi diagrams: sites, vertices, edges, cells; nearest-neighbour interpolation; the "toxic waste dump" problem

Voronoi diagram: site and cell: Site: a given point the diagram is built around. Cell: all points closer to that site than to any other. One cell per site.
Voronoi diagram: edge and vertex: Edge: the boundary between two cells, part of the perpendicular bisector of their sites. Vertex: where edges meet; where three edges meet, it is equidistant from three sites.
Nearest neighbour interpolation: Estimating a value at a point by the value at the nearest site. The estimate is constant in each cell and jumps at the edges.
Toxic waste dump problem: where does the dump go?: At the vertex that is furthest from its nearest sites. List the vertices, find each vertex's distance to one of its three sites, and choose the largest. In exams it is at an intersection of three edges.
How do you add a new site to a Voronoi diagram?: 1. Find the cell containing the new site. 2. Draw the bisector of the new site and that cell's site to the cell's edges. 3. Cross into the next cell and repeat. 4. Stop when the new cell closes or reaches the boundary. 5. Delete old edges inside it.
How do you show which site is closest to a point P?: Either calculate the distance from P to each nearby site and take the smallest, or substitute P into the edge equation and compare its side with each site's.

3.7 Radian measure; area of sector and arc length using radians (AHL only)

RadianHL: The angle at the centre of a circle subtended by an arc equal in length to the radius. 2π rad = 360°.
Converting between degrees and radiansHL: Degrees to radians: × π/180 (150° = 5π/6). Radians to degrees: × 180/π (2.4 rad = 137.5°).
Arc length and sector area with θ in radiansHL: l = rθ and A = ½r²θ. Example: r = 8, θ = 1.2 gives l = 9.6, A = 38.4, perimeter 9.6 + 16 = 25.6.
Which angle unit do HL papers assume?HL: Radians, unless the question says otherwise. Check the GDC mode before every HL question with angles.

3.8 Definitions of sinθ/cosθ via the unit circle; Pythagorean identity; ambiguous case of the sine rule (AHL only)

Definition of cos θ, sin θ and tan θ using the unit circleHL: The point at angle θ, measured anticlockwise from the positive x-axis, is (cos θ, sin θ). tan θ = sin θ/cos θ.
Pythagorean identity, and a use of itHL: cos²θ + sin²θ = 1. Example: sin θ = 0.6 with θ obtuse gives cos θ = −0.8 (negative in the second quadrant) and tan θ = −0.75.
How do you handle the ambiguous case of the sine rule?HL: 1. Find sin B with the sine rule. 2. Take the GDC value B₁ and also B₂ = 180° − B₁. 3. Keep B₂ only if B₂ + the given angle < 180°. 4. Finish each triangle separately.
a = 7, b = 9, A = 40°. Find the possible values of BHL: sin B = 9 sin 40°/7, so B = 55.7° or 180° − 55.7° = 124.3°. Both are possible, as 124.3° + 40° < 180°.
How do you solve a trigonometric equation graphically?HL: Set the GDC to the unit of the interval (usually radians), graph both sides over exactly that interval, and read off every intersection inside it.

3.9 Matrix transformations: reflections, stretches, enlargements, translations, rotations; determinant as area scale factor (AHL only)

Quick check: what are the columns of a 2 × 2 transformation matrix?HL: The images of (1, 0) and (0, 1).
Matrix for a rotation by θ anticlockwise about the originHL: Rows (cos θ, −sin θ) and (sin θ, cos θ). Determinant 1. For 90°: rows (0, −1) and (1, 0).
Matrices for reflections in the x-axis, the y-axis and y = xHL: x-axis: rows (1, 0), (0, −1). y-axis: rows (−1, 0), (0, 1). y = x: rows (0, 1), (1, 0). Each has determinant −1.
Matrices for stretches and an enlargement (centre O)HL: Horizontal stretch factor k: rows (k, 0), (0, 1). Vertical stretch factor k: rows (1, 0), (0, k). Enlargement factor k: rows (k, 0), (0, k), determinant k².
Transformation A is followed by transformation B. Which single matrix?HL: BA: the first transformation goes on the right.
Area of the image under a matrix transformationHL: Area of image = |det A| × area of object. Example: rows (2, 1), (1, 3) has det 5, so an area of 5 becomes 25. A translation never changes area.

3.10 Vectors: concept, components, base vectors, magnitude, position vectors, normalization (AHL only)

Scalar vs vector; base vectorsHL: A scalar has size only; a vector has size and direction. i, j, k are vectors of length 1 along the x-, y- and z-axes, so v = v₁i + v₂j + v₃k.
Magnitude of a vector; vector from A to BHL: |v| = √(v₁² + v₂² + v₃²). AB = b − a, where a and b are the position vectors of A and B.
Normalising a vectorHL: Dividing by its magnitude: v/|v| is the unit vector in the direction of v. Example: (6, −2, 3) has magnitude 7, so the unit vector is (6/7, −2/7, 3/7).
A drone moves at 15 m s⁻¹ in the direction 2i − 2j + k. Find its velocity vectorHL: |2i − 2j + k| = 3, so v = 15 × (2i − 2j + k)/3 = 10i − 10j + 5k m s⁻¹. A vector of length L along v is L × v/|v|.

3.11 Vector equation of a line in two and three dimensions (AHL only)

Vector equation of a lineHL: r = a + λb, where a is the position vector of a point on the line, b is a direction vector and λ is a parameter.
Parametric form of the line through (x₀, y₀, z₀) with direction (l, m, n)HL: x = x₀ + λl, y = y₀ + λm, z = z₀ + λn.
Does (7, −1, 4) lie on r = (1, 2, −2) + λ(2, −1, 2)?HL: x: 1 + 2λ = 7 gives λ = 3. y: 2 − 3 = −1 ✓. z: −2 + 6 = 4 ✓. The same λ fits all three, so yes.
How do you test whether two lines intersect?HL: 1. Write both in parametric form (λ for one, μ for the other). 2. Solve two of the equations for λ and μ. 3. Check the third. If it holds, substitute to find the point; if not, they do not meet.

3.12 Vector applications to kinematics: linear motion with constant and variable velocity (AHL only)

Motion with constant velocity: vector equationHL: r = r₀ + vt, where r₀ is the position at t = 0 and v is the velocity (the change in position per unit time).
Speed vs velocity in vector kinematicsHL: Velocity is the vector v; speed is its magnitude |v|, a scalar. Example: speed 10 m s⁻¹ in direction 3i − 4j (length 5) gives v = 6i − 8j m s⁻¹.
How do you find the closest approach of two objects?HL: 1. Write r_A and r_B in terms of t. 2. Form AB = r_B − r_A. 3. Find |AB|² as a quadratic in t. 4. Minimise it (differentiate or graph |AB|). 5. State the time and the distance, with units.
How do you find position from a variable velocity vector?HL: Integrate each component of v with respect to t, and use the position at t = 0 to find the constants.

3.13 Scalar and vector product of two vectors; angle between vectors; components of vectors (AHL only)

Scalar product: two formsHL: v · w = v₁w₁ + v₂w₂ + v₃w₃ = |v||w| cos θ. The answer is a number.
Angle between two vectors, and the test for perpendicularHL: cos θ = (v · w)/(|v||w|). Non-zero vectors are perpendicular if and only if v · w = 0.
Vector product v × wHL: (v₂w₃ − v₃w₂, v₃w₁ − v₁w₃, v₁w₂ − v₂w₁). A vector perpendicular to both v and w.
|v × w| and its geometric meaningHL: |v × w| = |v||w| sin θ = the area of the parallelogram with sides v and w. Example: (2, 0, 1) × (1, 3, 0) = (−3, 1, 6), area √46 = 6.78.
Component of a in the direction of b, and perpendicular to bHL: Along b: (a · b)/|b| = |a| cos θ. Perpendicular to b: |a × b|/|b| = |a| sin θ.
Angle between two lines vs angle between two vectorsHL: Vectors can give an obtuse angle. For the acute angle between two lines, use |v · w| with their direction vectors.

3.14 Graph theory: graphs, vertices, edges, degree, simple/complete/weighted/directed graphs, subgraphs and trees (AHL only)

Degree of a vertex, and the handshake resultHL: Degree: the number of edges at a vertex (a loop counts 2). Sum of degrees = 2 × number of edges, so there is always an even number of odd vertices.
Simple graph and complete graph KₙHL: Simple: no loops and no multiple edges. Kₙ: simple, with every pair of its n vertices joined; it has n(n − 1)/2 edges (K₇ has 21).
Weighted graph and directed graphHL: Weighted: edges carry values (distance, cost, time). Directed: edges have directions; in degree = arrows in, out degree = arrows out.
Subgraph and treeHL: Subgraph: some of the vertices and edges of a graph. Tree: connected with no cycles. A tree with n vertices has n − 1 edges.
Connected vs strongly connectedHL: Connected: a route exists between every pair of vertices. Strongly connected (directed graphs): every vertex can reach every other along the arrows.

3.15 Adjacency matrices, walks, weighted adjacency tables, transition matrices for strongly-connected graphs (AHL only)

Adjacency matrixHL: Entry aᵢⱼ is the number of edges from vertex i to vertex j. Symmetric for an undirected graph. For a directed graph, row sum = out degree and column sum = in degree.
Number of walks of length k, and of length up to k, from i to jHL: Length k: the (i, j)th entry of Aᵏ. Length up to k: the (i, j)th entry of A + A² + … + Aᵏ.
How do you build the transition matrix for a random walk on a graph?HL: 1. Find the degree (or out degree) of each vertex j. 2. In column j, put 1/degree in the row of each vertex j leads to, 0 elsewhere. 3. Check every column sums to 1.
Steady state of a graph's transition matrix, and PageRankHL: s with Ts = s and entries summing to 1: raise T to a high power on the GDC, or solve the equations. PageRank ranks pages (vertices) by their steady-state probabilities.
Adjacency matrix vs transition matrixHL: A counts edges, with rows = from. T holds probabilities, with columns = from, each column summing to 1.

3.16 Tree and cycle algorithms: Kruskal’s and Prim’s MST, Chinese postman and travelling salesman problems (AHL only)

Walk, trail, path, circuit, cycleHL: Walk: any sequence of adjacent edges. Trail: no edge repeated. Path: no vertex repeated. Circuit: a trail that returns to its start. Cycle: a circuit with no vertex repeated except start/end.
Eulerian test for a connected graphHL: 0 odd vertices: an Eulerian circuit. 2 odd vertices: an Eulerian trail from one odd vertex to the other. Otherwise: neither.
Eulerian vs HamiltonianHL: Eulerian: every edge exactly once. Hamiltonian: every vertex exactly once. There is no simple degree test for Hamiltonian; show one exists by writing it down.
Kruskal's algorithmHL: 1. Order the edges by weight. 2. Add the next cheapest edge unless it forms a cycle. 3. Stop at n − 1 edges. Write down the order, including rejected edges.
Prim's algorithm (matrix method)HL: 1. Choose a start vertex: delete its row, label its column. 2. Find the smallest undeleted entry in the labelled columns; its row is the new vertex. 3. Delete that row, label that column. 4. Repeat until every column is labelled.
Chinese postman problem: methodHL: 1. Total weight of all edges. 2. Find the odd vertices. 3. Two odd: add the shortest path between them. Four odd: check all three pairings and add the least total. 4. State which edges are repeated.
Travelling salesman: upper and lower boundsHL: Upper bound: nearest neighbour tour from a start vertex, returning to it. Lower bound: delete a vertex, find the MST of the rest, add the two shortest edges from the deleted vertex. Best interval: largest lower bound ≤ optimal ≤ smallest upper bound.
Table of least distances: when and why?HL: When the graph is not complete, or a direct edge is not the shortest route. Each entry is the shortest route between the two vertices (for example A–C–B = 9 replaces a direct AB of 12).