3 Geometry and trigonometry

In three-dimensional coordinates, a point is (x,y,z). A displacement vector records the change in each coordinate, so distance and direction can be calculated component by component.

For points P and Q, PQ=Q−P and |PQ|=√((Δx)²+(Δy)²+(Δz)²). The midpoint averages corresponding coordinates. These formulas are geometric statements, not just calculator rules.

From P(1,−2,3) to Q(4,2,−1), the displacement is (3,4,−4) and the distance is √41. The sign tells direction; squaring removes it only when measuring length.

Do not drop the z-coordinate or confuse a displacement component with the total distance. Keep units and point order clear when interpreting direction.

The midpoint of $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ is $((x_1+x_2)/2,(y_1+y_2)/2,(z_1+z_2)/2)$. For solids, use the correct base area and perpendicular height: prism $V=Ah$, pyramid or cone $V=Ah/3$, sphere $V=4\pi r^3/3$; total surface area includes every exposed face. In a cuboid, identify a face diagonal first, then use a second right triangle to obtain a space length or an angle between a line and a plane.

Pythagoras links the sides of a right triangle: a²+b²=c², where c is opposite the right angle. Sine, cosine and tangent link an acute angle to the opposite, adjacent and hypotenuse sides.

Label the right angle and the chosen angle first. Use Pythagoras when all you need is a side length; use a trigonometric ratio when an angle is involved. The inverse ratio finds the angle after the side relationship is formed.

If the opposite side is 6 and the hypotenuse 10, sin θ=0.6, so θ≈36.9°. The same triangle gives the adjacent side √64=8, providing a consistency check.

SOHCAHTOA depends on the chosen angle. Do not use the hypotenuse as the adjacent side or round before the final step.

For any triangle, $a/\sin A=b/\sin B=c/\sin C$, $c^2=a^2+b^2-2ab\cos C$, and $Area=\tfrac12ab\sin C$. Example: with $a=7$, $b=9$, $C=60^\circ$, $c^2=67$ and the area is $63\sqrt3/4$ square units. Choose the sine rule for a known opposite side–angle pair, the cosine rule for SAS or SSS, and note that the ambiguous sine-rule case is not included at SL.

Applied trigonometry converts bearings, elevation, depression and written spatial relationships into right or non-right triangles that can be solved with Pythagoras, trigonometric ratios, the sine rule or the cosine rule.

Draw the north line for bearings, mark horizontal sight lines for elevation or depression, label known sides and angles, and decide whether triangles share a length. Keep bearings as three-figure clockwise angles from north.

Example

A point is 80 m horizontally from a tower and the angle of elevation to the top is $32^\circ$. Then $\tan32^\circ=h/80$, so $h=80\tan32^\circ\approx50.0$ m. State whether eye height must be added in the actual context.

Do not take a bearing from the wrong north line or use an angle of depression as an angle inside a triangle without transferring it through parallel horizontals. A labelled diagram is part of the reasoning.

At SL, an angle $\theta$ in degrees selects the fraction $\theta/360$ of a full circle. Therefore arc length is $s=(\theta/360)(2\pi r)$ and sector area is $A=(\theta/360)(\pi r^2)$.

Use the radius, not the diameter, and keep the angle in degrees. A perimeter of a sector includes the curved arc plus two radii, while the area formula includes only the sector region.

Example

For $r=6$ cm and $\theta=120^\circ$, $s=(120/360)(12\pi)=4\pi$ cm and $A=(120/360)(36\pi)=12\pi$ cm$^2$. The sector perimeter is $12+4\pi$ cm.

Radians are not required at SL. Do not use $s=r\theta$ unless $\theta$ is in radians, and do not report area in linear units.

The perpendicular bisector of a segment is the line through its midpoint at 90°, and every point on it is equally distant from the two endpoints.

Find the midpoint, calculate the segment's gradient, take the negative reciprocal for the perpendicular gradient, then use point–gradient form. The equal-distance property is often more useful than the equation itself.

For A(1,2) and B(5,4), the midpoint is (3,3) and AB has gradient 1/2, so the bisector has gradient −2: y−3=−2(x−3).

Perpendicular slopes multiply to −1 only when both are finite. Do not use the segment's midpoint with the original gradient.

A Voronoi diagram partitions a region so that every point in a cell is closer to its generating site than to any other site. Boundaries lie on perpendicular bisectors between competing sites.

Construct the relevant bisectors, keep only the boundaries that separate nearest-site regions, then use the cell to answer location questions. A new site can remove parts of neighbouring cells rather than simply adding an isolated shape.

If three clinics are sites, a point in Clinic A's cell is predicted to be served by A under a nearest-distance rule. A barrier or travel-time difference would invalidate that simple model.

A Voronoi boundary means equal distance under the chosen metric, not equal demand or guaranteed service quality. State the metric and context assumption.

Terminology: each generator is a site; a cell contains points closest to one site; an edge is an equal-distance boundary between two sites; a vertex is where three or more edges meet. Nearest-neighbour interpolation assigns an unknown point the value of its cell's site. When adding a site, use the supplied perpendicular bisectors to trim neighbouring cells. In the standard toxic-waste-dump task, the solution lies at an intersection of three edges; region areas may require coordinate geometry.

One radian is the angle subtended by an arc whose length equals the radius. A full turn is 2π radians, so radians connect angle directly to circle geometry.

Convert degrees using θ(rad)=θ°π/180. In radian measure, arc length is s=rθ and sector area is ½r²θ, with θ in radians.

For r=5 and θ=1.2, the arc length is 6 units and the sector area is 15 square units. Using 1.2° instead would give a completely different scale.

Do not substitute degrees into s=rθ or ½r²θ. Check the angle unit before using a formula.

Solving a trigonometric equation means finding every angle in the stated interval that produces the target value. Periodicity creates repeated solutions.

Use the unit circle or graph to identify the reference angle and the quadrants with the correct sign, then add the function's period. A calculator's principal value is only one solution unless the interval makes it complete.

For sin x=0.5 on 0≤x≤2π, x=π/6 and 5π/6. Stopping at π/6 misses the second intersection of the sine curve with 0.5.

The number of solutions depends on the interval and period. Check endpoints and do not use the cosine rule for a sine equation.

Unit-circle definitions are $\cos heta=x$ and $\sin heta=y$ for the point $(x,y)$ on the unit circle, so $\cos^2 heta+\sin^2 heta=1$ and $\tan heta=\sin heta/\cos heta$ when $\cos heta\ne0$. The sine rule's ambiguous case can give two triangles because $\sin heta=\sin(\pi- heta)$; accept only solutions that satisfy the side lengths and stated interval. Exact trig values are helpful but are not directly assessed.

A 2×2 matrix maps a point vector to its image: x′=Ax. The columns show where the basis vectors go, so the matrix encodes the geometric action.

Apply the matrix to the column vector in the stated order. Determinant magnitude gives area scale; a negative determinant reverses orientation. To recover an original point, solve with A⁻¹ when it exists.

The matrix [[0,−1],[1,0]] sends (1,0) to (0,1) and (0,1) to (−1,0): a 90° anticlockwise rotation. Multiplying by the row vector would describe a different operation.

Do not read rows as images of basis vectors without checking the convention. A zero determinant means the transformation has no inverse.

The syllabus also includes affine maps $\begin{pmatrix}x'\y'\end{pmatrix}=A\begin{pmatrix}x\y\end{pmatrix}+\begin{pmatrix}e\f\end{pmatrix}$, so translation requires the added vector and cannot be represented by a $2\times2$ matrix alone. In a composition, the rightmost transformation acts first; $BAx$ means apply $A$, then $B$. Iterating one or more transformation rules from an initial shape or point can generate a fractal, while $|\det A|$ remains the area scale for the linear part.

A vector has magnitude and direction. Two non-zero vectors are parallel when one is a scalar multiple of the other; the scalar's sign tells whether their directions agree or oppose.

Compare components rather than relying on a sketch. The length is √(a·a), and addition represents successive displacements. In 3D, every component must satisfy the same scalar relationship.

(2,−4,6) is 2(1,−2,3), so the vectors are parallel and point the same way. (−2,4,−6) is a negative multiple and points oppositely.

Equal length does not mean parallel, and a single matching component is not enough. Check all components and exclude the zero vector when using direction tests.

A unit vector in the direction of non-zero $v$ is $\hat v=v/|v|$; rescaling to magnitude $s$ gives $s\hat v$. For example, a particle moving at $7\text{ m s}^{-1}$ in direction $3i+4j$ has unit direction $(3i+4j)/5$ and velocity $(21i+28j)/5\text{ m s}^{-1}$. A resultant is the vector sum, and the zero vector has magnitude zero but no defined direction.

A line through point a with direction vector d is r=a+td, where t is a real parameter. Changing t moves along the line without changing its direction.

Use two points P and Q to form d=Q−P. To test whether X lies on the line, solve the component equations for one common t; inconsistent values mean X is not on it.

Through (1,0,2) and (3,4,2), r=(1,0,2)+t(2,4,0). The point (2,2,2) has t=1/2 in every component, so it lies on the line.

Matching one coordinate is not enough. Check the same parameter across all components and distinguish a line from a finite segment.

A vector kinematics model writes position as r(t)=r₀+vt for constant velocity, or uses the corresponding component equations when acceleration is present.

To test whether two moving objects meet, solve their position vectors for a common time in the allowed interval. A closest approach is different: minimise the squared separation rather than forcing an exact intersection.

If r₁=(0,0)+t(3,1) and r₂=(6,4)+t(−1,−1), solving gives t=2 and position (6,2); the same time in both equations is the evidence of an intersection.

Equal coordinates at different times do not mean collision. State the time domain and distinguish position from velocity.

For variable velocity in two dimensions, $v(t)=r'(t)$ and $r(t)=\int v(t)\,dt+C$, with the initial position fixing $C$; speed is $|v(t)|$. Relative position of $B$ from $A$ is $r_B-r_A$. To find closest approach, minimize $|r_B-r_A|^2$ over the allowed time interval and check endpoints as well as stationary values; projectile and circular motion are special cases of this vector model.

The scalar product a·b=|a||b|cosθ measures alignment and is zero for perpendicular vectors. The vector product a×b is perpendicular to both and has magnitude |a||b|sinθ, the area of the parallelogram they span.

Use a·b for angles, projections and perpendicular tests; use a×b for normals, orientation and areas. Order matters for the cross product: b×a=−(a×b).

For a=(1,0,0) and b=(0,2,0), a·b=0 and a×b=(0,0,2), so the vectors are perpendicular and span area 2.

The cross product is not a scalar and the dot product is not a vector. Check the dimension and interpretation of the requested result.

The scalar component of $a$ in the direction of non-zero $b$ is $a\cdot b/|b|=|a|\cos heta$; the magnitude of the component perpendicular to $b$ in their plane is $|a\times b|/|b|=|a|\sin heta$. Thus $|a\times b|$ is parallelogram area and half of it is triangle area. For line directions, use the acute angle by replacing $\cos heta$ with $|a\cdot b|/(|a||b|)$.

A graph consists of vertices and edges representing objects and connections. Edges may be directed or undirected and weighted or unweighted, depending on the relationship being modelled.

The degree counts incident edges; a path records a route and a cycle returns to its start. Choose the graph type to match whether direction, capacity or cost matters.

In a delivery network, a one-way road needs a directed edge and travel time needs a weight. Treating it as an unweighted undirected graph can permit an impossible or misleading route.

A graph edge is not automatically a physical straight line. State what vertices and edges mean before interpreting an algorithm's result.

Graph checklist: a simple graph has no loops or parallel edges; a complete graph joins every pair of distinct vertices; a subgraph uses selected vertices and edges; a tree is connected and has no cycles. In a directed graph distinguish in-degree from out-degree. Connected means every vertex pair is linked by a path in an undirected graph; strongly connected means directed paths exist in both directions between every pair.

An adjacency matrix records which vertices are directly connected. For an unweighted graph, Aᵢⱼ=1 means an edge from i to j under the chosen row/column convention.

The entry (A^k)ᵢⱼ counts walks of length k from i to j. Matrix multiplication works because each intermediate vertex is summed over, linking consecutive steps.

If two different two-step routes connect A to C, the corresponding entry of A² is 2. In a weighted matrix, the same multiplication may represent a different quantity, so do not mix conventions.

A walk may revisit vertices; it is not automatically a simple path. Check whether the graph is directed and what the matrix entries mean.

A weighted adjacency table stores a cost, distance or time rather than just 0 or 1. A transition matrix converts each vertex's outgoing weights or links into probabilities whose relevant row or column sums to 1 under the declared convention; repeated multiplication models movement through a strongly connected graph, as in a simplified PageRank model. Do not interpret powers of a weighted cost matrix as counts of walks unless that convention is explicitly justified.

A graph algorithm is a rule for extracting a route, assignment or bound from network data. A heuristic such as nearest neighbour chooses a locally short next edge; it is fast but not generally optimal.

Record the starting vertex and tie rule, then compare the resulting tour with a lower or upper bound when evaluating quality. Deleted-vertex or spanning-tree reasoning can provide a bound without proving the exact optimum.

A nearest-neighbour tour can choose the closest next city and later be forced into one very long final edge. A different early choice may produce a shorter total route, so the first route is a candidate, not a proof.

Greedy does not mean optimal. State the algorithm, its assumptions and whether the conclusion is a route, a bound or a proven minimum.

Algorithm map: an Eulerian trail uses every edge once (exactly two odd vertices) and an Eulerian circuit has all vertices even; Hamiltonian paths or cycles visit every vertex once. Kruskal selects globally smallest non-cycling edges, while Prim grows a minimum spanning tree from a chosen vertex. For a Chinese postman route, pair up to four odd vertices using least-distance paths, duplicate the minimum-total pairing, then take an Eulerian circuit. For a complete weighted travelling-salesman graph, nearest neighbour gives an upper bound and deleted-vertex plus MST reasoning gives a lower bound; complete a least-distance table first when the practical graph is not complete. State and justify the selected algorithm.