3.1 Three-dimensional geometryIB Maths: Analysis and Approaches HL: Revision notes

Points in 3D have three coordinates $(x,y,z)$. The distance between $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ comes from applying Pythagoras twice: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}.$ The midpoint averages each coordinate: $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right).$ Example: $A(2,-1,5)$ and $B(8,3,-7)$ give $AB=\sqrt{36+16+144}=14$ and $M=(5,1,-1)$. If a distance is given but a coordinate is unknown, square both sides of the distance formula and solve — you often get two values, and the context decides which to keep.

All of these are in the formula booklet:

• Right pyramid: $V=\frac13Ah$, where $A$ is the base area and $h$ the perpendicular height.
• Right cone: $V=\frac13\pi r^2h$.
• Sphere: $V=\frac43\pi r^3$, so a hemisphere is $\frac23\pi r^3$.

For a composite solid, add (or subtract) the volumes of the parts. A cone of radius 3 and height 4 on a hemisphere of radius 3 has volume $12\pi+18\pi=30\pi$ cm$^3$. Give exact answers as multiples of $\pi$ unless a decimal is asked for.

• Cone: curved surface $\pi rl$, where $l$ is the slant height, $l=\sqrt{r^2+h^2}$. Add $\pi r^2$ for the base only if it is exposed.
• Sphere: $4\pi r^2$; hemisphere curved surface $2\pi r^2$ (plus $\pi r^2$ if the flat face is exposed).
• Pyramid: base plus the triangular faces; each face's height is the slant height of that face, not the pyramid's vertical height.

For a composite solid, include only the surfaces you could touch. Where two solids are joined along a flat face, that face is hidden and is not counted.

At SL, trigonometry in 3D uses right-angled triangles only. The skill is to spot a triangle that is right-angled and contains the length or angle you want. In a right pyramid with a square base of side 10 and height 12:

• The half-diagonal is $OA=5\sqrt2$, so the lateral edge is $VA=\sqrt{12^2+50}=\sqrt{194}$.
• The distance from the centre to the midpoint of a side is 5, so the height of a triangular face is $VN=\sqrt{12^2+5^2}=13$. In a cuboid $a\times b\times c$ the space diagonal is $\sqrt{a^2+b^2+c^2}$. Name each triangle by its vertices (for example triangle $VOA$, right-angled at $O$) so the examiner can follow your method.

The angle between a line and a plane is the angle between the line and its projection onto the plane. To find it:

1. Take a point on the line and drop a perpendicular from it to the plane.
2. Join the foot of the perpendicular to the point where the line meets the plane. This is the projection.
3. The line, the perpendicular and the projection form a right-angled triangle; the required angle is where the line meets the plane.

Example: camera at $C(0,0,9)$, corner $Q(20,12,0)$ on the floor. $O$ is directly below $C$, so the angle is $C\hat{Q}O$ with $\sin\theta=\frac{9}{25}$, $\theta=21.1^\circ$. The angle between two intersecting lines is found the same way, from a right-angled triangle containing both lines.