3.1 Three-dimensional geometryIB Maths: Applications and Interpretation HL: Revision notes

For points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$: $AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2},\qquad M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right).$ The distance is Pythagoras' theorem applied twice. The midpoint is the average of each coordinate. Example: $A(1,-2,5)$ and $B(4,2,17)$ give $AB=\sqrt{9+16+144}=\sqrt{169}=13$ and midpoint $(2.5,0,11)$.

The formulae are in the data booklet. With base area $A$, height $h$, radius $r$ and slant height $l$:

• right pyramid: $V=\frac13Ah$
• right cone: $V=\frac13\pi r^2h$, curved surface area $\pi rl$
• sphere: $V=\frac43\pi r^3$, surface area $4\pi r^2$
• hemisphere: half of the sphere, so $V=\frac23\pi r^3$ and curved surface area $2\pi r^2$ (adding the flat face gives a total of $3\pi r^2$). The slant height of a cone is found from $l^2=r^2+h^2$. Example: a cone with $r=5$ and $h=12$ has $l=13$, $V=100\pi=314$ cm$^3$ and curved surface area $65\pi=204$ cm$^2$. A sphere with $r=3$ has $V=36\pi=113$ cm$^3$.

For a solid made from several parts, find the volume of each part and add. For the surface area, add only the surfaces that are exposed: a face where two parts join is hidden. Example: an ice-cream cone of radius 3 cm and height 10 cm, with a hemisphere of radius 3 cm on top. Volume: cone $\frac13\pi(3)^2(10)=30\pi$, hemisphere $\frac23\pi(3)^3=18\pi$, total $48\pi=151$ cm$^3$. Curved surface: slant height $\sqrt{3^2+10^2}=\sqrt{109}$, so cone $\pi(3)\sqrt{109}=98.4$ and hemisphere $2\pi(3)^2=56.5$, total $155$ cm$^2$.

To find lengths and angles in a solid, identify a right-angled triangle inside it, draw it separately and label the known lengths. In SL examinations only right-angled trigonometry is set for 3D shapes. Example: a cuboid with base 6 cm by 8 cm and height 5 cm. The base diagonal is $\sqrt{6^2+8^2}=10$, and the space diagonal is $\sqrt{10^2+5^2}=\sqrt{125}=11.2$ cm. Use $\tan$, $\sin$ or $\cos$ in the same triangles to find angles. In a right pyramid, the triangle formed by the height, half the base diagonal and the sloping edge is right-angled.

The angle between a line and a plane is the angle between the line and its projection (shadow) on the plane. The angle between two intersecting lines is the angle at the point where they meet, found in a triangle containing both lines. Example (cuboid above): the line from corner $A$ to the opposite top corner $G$ projects onto the base as the base diagonal $AC=10$. With $CG=5$ the angle is $\tan^{-1}\left(\frac{5}{10}\right)=26.6^\circ$. If the triangle is isosceles but not right-angled, split it into two right-angled triangles.