3.1 Geometry and trigonometry - SL content

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.