This note covers distance and midpoint in three dimensions, the volume and surface area of pyramids, cones, spheres and hemispheres, and how to find angles between lines and between a line and a plane. These ideas turn a 3D picture into flat right-angled triangles that you can solve with trigonometry.
Key ideas
A point in 3D space has three coordinates \((x, y, z)\). The distance between two points is found by applying Pythagoras twice, which gives one neat formula. The midpoint is found by averaging each coordinate.
- Solids are named by their base and apex. A pyramid has a polygon base and triangular faces meeting at an apex. A cone has a circular base. A sphere is the set of points at a fixed distance from a centre. A hemisphere is half a sphere.
- For a cone, the slant height \(l\), the vertical height \(h\) and the radius \(r\) form a right-angled triangle with \(l^2 = r^2 + h^2\).
- To find an angle in a 3D solid, first draw the right-angled triangle that contains it. Redraw this triangle on its own, label the known lengths, then use \(\sin\), \(\cos\) or \(\tan\).
- The angle between a line and a plane is the angle between the line and its shadow (its projection) on the plane. It is the smallest angle between the line and the plane.
- The angle between two intersecting lines is found by making a triangle from the two lines. Work out its three side lengths, then use the cosine rule to get the angle.
The right-angled triangle for a line meeting a plane is usually made from three lengths: the line itself (hypotenuse), the vertical drop to the plane, and the distance along the plane between the two feet.
Useful patterns to spot in 3D diagrams:
- The diagonal across a rectangular face comes from Pythagoras on that face. The space diagonal of a cuboid with edges \(a\), \(b\), \(c\) is \(\sqrt{a^2 + b^2 + c^2}\), which is the 3D distance formula in disguise.
- In a pyramid with a rectangular base and its apex directly above the centre, the foot of the vertical height is the point where the base diagonals cross. Half a base diagonal is often one side of your right-angled triangle.
- Volumes are in cubic units (cm³, m³) and surface areas are in square units (cm², m²). Check the units match the question before you finish.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\) | Distance between \((x_1,y_1,z_1)\) and \((x_2,y_2,z_2)\). |
| \(M = \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}, \dfrac{z_1+z_2}{2}\right)\) | Midpoint of the line segment joining two points. |
| \(V = \dfrac{1}{3} \times \text{base area} \times h\) | Volume of any pyramid, where \(h\) is the perpendicular height. |
| \(V = \dfrac{1}{3}\pi r^2 h\) | Volume of a cone. |
| \(A_{\text{curved}} = \pi r l\) | Curved surface area of a cone with slant height \(l\). |
| \(V = \dfrac{4}{3}\pi r^3\), \(A = 4\pi r^2\) | Volume and surface area of a sphere. |
| Hemisphere: \(V = \dfrac{2}{3}\pi r^3\), curved area \(2\pi r^2\) | Half of a sphere. Its total surface area includes the flat circle: \(3\pi r^2\). |
| Total surface area | Add the areas of every face, including the base. Check whether the question includes the base. |
Worked examples
Example 1 A survey drone hovers at \(A(2, -1, 3)\) and a second drone at \(B(8, 3, -9)\), with all lengths in metres. Find the distance between them and the midpoint of \(AB\).
Find the differences: \(8-2 = 6\), \(3-(-1) = 4\), \(-9-3 = -12\).
\(d = \sqrt{6^2 + 4^2 + (-12)^2} = \sqrt{36 + 16 + 144} = \sqrt{196} = 14\).
Midpoint: \(\left(\dfrac{2+8}{2}, \dfrac{-1+3}{2}, \dfrac{3+(-9)}{2}\right) = (5, 1, -3)\).
Answer: distance 14 m; midpoint \((5, 1, -3)\).
Example 2 A traffic cone is modelled as a cone with base radius 5 cm and slant height 13 cm. Find its volume and its total surface area.
Find the vertical height: \(h = \sqrt{13^2 - 5^2} = \sqrt{144} = 12\) cm.
Volume: \(V = \dfrac{1}{3}\pi (5)^2 (12) = 100\pi \approx 314\) cm³.
Curved area: \(\pi r l = \pi (5)(13) = 65\pi\).
Base area: \(\pi r^2 = 25\pi\).
Total surface area: \(65\pi + 25\pi = 90\pi \approx 283\) cm².
Answer: \(V \approx 314\) cm³; total surface area \(\approx 283\) cm².
Example 3 A storage box is a cuboid \(ABCDEFGH\). The base \(ABCD\) measures \(AB = 8\) m and \(BC = 6\) m, and the height \(CG = 5\) m. Find the length of the diagonal \(AG\) and the angle it makes with the base.
First find the diagonal of the base: \(AC = \sqrt{8^2 + 6^2} = 10\) m.
Triangle \(ACG\) has a right angle at \(C\), because \(CG\) is vertical and \(AC\) lies in the base.
\(AG = \sqrt{10^2 + 5^2} = \sqrt{125} \approx 11.2\) m.
The shadow of \(AG\) on the base is \(AC\), so the angle required is \(\angle GAC\).
\(\tan \angle GAC = \dfrac{5}{10} = 0.5\), so \(\angle GAC = \tan^{-1}(0.5) \approx 26.6^\circ\).
Answer: \(AG \approx 11.2\) m; angle with the base \(\approx 26.6^\circ\).
Common errors
- Using the slant height in place of the vertical height in the cone or pyramid volume formula. The volume needs the perpendicular height.
- Forgetting the one-third in the volume of a pyramid or cone.
- Using the angle at the wrong vertex. Name the angle with three letters, such as \(\angle GAC\), and check that the line and its shadow meet at the middle letter.
- Calling a total surface area "curved area". A cone has a base, and a hemisphere has a flat face, which both need adding when "total" is asked for.
- Using degree mode or radian mode wrongly on the calculator. Here angles are in degrees, so check the mode.
Exam tip: Sketch the solid, then draw the single right-angled triangle you need as a separate flat diagram with its lengths marked. This stops you from mixing up edges that lie in different planes.
Check yourself
- Find the distance between \(P(1, 2, 3)\) and \(Q(-3, 5, 15)\).
Answer
Differences: \(-4, 3, 12\). \(d = \sqrt{16 + 9 + 144} = \sqrt{169} = 13\). - A sphere has radius 6 cm. Find its volume and surface area.
Answer
\(V = \frac{4}{3}\pi(6)^3 = 288\pi \approx 905\) cm³. \(A = 4\pi(6)^2 = 144\pi \approx 452\) cm². - A square-based pyramid has base side 10 cm and vertical height 12 cm, with its apex directly above the centre of the base. Find its volume and the angle between a sloping edge and the base.
Answer
\(V = \frac{1}{3}(100)(12) = 400\) cm³. Half the base diagonal is \(\frac{10\sqrt{2}}{2} = 5\sqrt{2} \approx 7.07\) cm. The angle is \(\tan^{-1}\left(\frac{12}{5\sqrt{2}}\right) \approx 59.5^\circ\). - Find the total surface area of a solid hemisphere of radius 4 cm.
Answer
Curved part \(2\pi(4)^2 = 32\pi\), flat circle \(\pi(4)^2 = 16\pi\). Total \(= 48\pi \approx 151\) cm².