Three-dimensional geometry: distance, volume, surface area and angles

Welcome to 3D Geometry

Welcome! In everyday life, we live in a three-dimensional world—from the screen in front of you to ice cream cones, basketballs, and the pyramids of Egypt. In this chapter, we extend our two-dimensional geometry skills into 3D space. You will master four essential skills outlined in Topic SL 3.1: finding distance and midpoints in three dimensions, calculating the volume of standard and composite 3D solids, finding their surface area, and determining the angle between two lines or between a line and a plane.

Cross-reference note: Solving for angles inside triangles relies on right-angled trigonometry, the sine rule, and the cosine rule, which are covered thoroughly in our chapter on right-angled and non-right-angled trigonometry.

1. 3D Coordinates: Distance and Midpoint

In two dimensions, a point is described by two coordinates, \((x, y)\). In three dimensions, we add a third axis, \(z\), perpendicular to the other two. Any point in space is represented as an ordered triplet: \((x, y, z)\).

An easy way to visualize this: imagine the corner of your room on the floor as the origin \((0, 0, 0)\). Moving along the wall to your right is the \(x\)-axis, moving along the wall forward is the \(y\)-axis, and moving straight up toward the ceiling is the \(z\)-axis!

Distance Between Two Points in 3D

To find the straight-line distance \(d\) between two points \(A(x_1, y_1, z_1)\) and \(B(x_2, y_2, z_2)\), we apply the 3D version of Pythagoras's theorem:

\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\)

Step-by-step example: Find the distance between \(A(1, -2, 4)\) and \(B(4, 2, -8)\).

1. Find the horizontal and vertical differences:
\(x_2 - x_1 = 4 - 1 = 3\)
\(y_2 - y_1 = 2 - (-2) = 4\)
\(z_2 - z_1 = -8 - 4 = -12\)

2. Square each difference:
\(3^2 = 9\), \(4^2 = 16\), \((-12)^2 = 144\)

3. Sum the squares and take the square root:
\(d = \sqrt{9 + 16 + 144} = \sqrt{169} = 13\)

Midpoint Between Two Points in 3D

The midpoint \(M\) is simply the balance point or the mathematical average of the coordinates of each axis:

\(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)\)

Example: Using the same points \(A(1, -2, 4)\) and \(B(4, 2, -8)\):

\(M = \left(\frac{1 + 4}{2}, \frac{-2 + 2}{2}, \frac{4 + (-8)}{2}\right) = \left(\frac{5}{2}, 0, -2\right) = (2.5, 0, -2)\)

Key Takeaway: 3D distance and midpoint formulas are direct extensions of their 2D counterparts—just calculate the third component (\(z\)) the exact same way!

2. Volumes and Surface Areas of 3D Solids

The IB syllabus specifically focuses on right pyramids, right cones, spheres, hemispheres, and combinations of these shapes.

A. Right Pyramids

A right pyramid has its apex (top vertex) aligned directly above the centre of its base. The base is typically a square or rectangle.

Volume:
\(V = \frac{1}{3} \times \text{base area} \times h\)
where \(h\) is the vertical (perpendicular) height from the base to the apex.

Surface Area:
\(\text{Total Surface Area} = \text{Area of base} + \text{Area of triangular faces}\)

Watch out: Do not confuse vertical height \(h\) with slant height \(l\). The vertical height goes straight down the inside of the pyramid to the centre of the base, while the slant height runs along the surface of a triangular face. They form a right-angled triangle with the half-width of the base:
\(l^2 = h^2 + \left(\frac{w}{2}\right)^2\)

B. Right Cones

A right circular cone has a circular base of radius \(r\), a vertical height \(h\), and a slant height \(l\). By Pythagoras's theorem, \(l = \sqrt{r^2 + h^2}\).

Volume:
\(V = \frac{1}{3}\pi r^2 h\)

Surface Area:
\(\text{Curved Surface Area} = \pi r l\)
\(\text{Total Surface Area} = \pi r l + \pi r^2\)

Memory aid: Think of the volume of a cone as exactly one-third of the volume of a cylinder with the same base and height!

C. Spheres and Hemispheres

For a sphere of radius \(r\):

Volume of a Sphere:
\(V = \frac{4}{3}\pi r^3\)

Surface Area of a Sphere:
\(A = 4\pi r^2\)

For a hemisphere (half of a sphere):

Volume:
\(V = \frac{1}{2} \left(\frac{4}{3}\pi r^3\right) = \frac{2}{3}\pi r^3\)

Curved Surface Area:
\(A_{\text{curved}} = 2\pi r^2\)

Total Surface Area of a Solid Hemisphere:
A solid hemisphere includes its flat circular base!
\(A_{\text{total}} = 2\pi r^2 + \pi r^2 = 3\pi r^2\)

Common Mistake Alert: When asked for the surface area of a solid hemisphere, students frequently write \(2\pi r^2\) and forget the circular base of area \(\pi r^2\). Always check whether a shape is described as "open" or "solid"!

D. Composite Solids

A composite solid combines two or more basic shapes, such as a silo (cylinder + hemisphere) or a toy spinning top (cone + hemisphere).

Finding Volume: Add or subtract the separate volumes directly.
\(V_{\text{total}} = V_1 + V_2\)

Finding Surface Area: Be careful! You only count exposed outer surfaces. Any faces that are glued together inside the solid do not form part of the external surface area.

Step-by-step example: An ice-cream cone toy is made of a cone of radius \(r = 3\text{ cm}\) and height \(h = 4\text{ cm}\), topped with a solid hemisphere of radius \(r = 3\text{ cm}\).
1. Slant height of cone: \(l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 4^2} = 5\text{ cm}\).
2. Volume:
\(V_{\text{cone}} = \frac{1}{3}\pi (3)^2(4) = 12\pi\)
\(V_{\text{hemisphere}} = \frac{2}{3}\pi (3)^3 = 18\pi\)
\(V_{\text{total}} = 12\pi + 18\pi = 30\pi \approx 94.2\text{ cm}^3\)
3. Exposed Surface Area:
The circular flat faces join internally, so they are not included.
\(\text{Area} = \text{Curved area of cone} + \text{Curved area of hemisphere}\)
\(\text{Area} = \pi r l + 2\pi r^2 = \pi(3)(5) + 2\pi(3)^2 = 15\pi + 18\pi = 33\pi \approx 104\text{ cm}^2\)

Key Takeaway: For volume, add the components up. For surface area, sketch the object and tally only the outer, touchable faces.

3. Angles in Three Dimensions

Don't worry if calculating angles in 3D feels intimidating at first! The secret to solving any 3D angle problem is dimension reduction: locate the right 2D triangle embedded inside the 3D shape, draw it flat on your page, and solve it using basic trigonometry.

A. Angle Between Two Intersecting Lines

When two straight lines intersect at a point in 3D space, they lie together in a single flat plane.

To find the angle between them:

1. Identify a triangle that contains the two lines meeting at the target angle \(\theta\).
2. Calculate the lengths of all three sides of this triangle using Pythagoras's theorem or the 3D distance formula.
3. If the triangle has a right angle, use basic trigonometric ratios:
\(\sin\theta = \frac{\text{opp}}{\text{hyp}}\), \(\cos\theta = \frac{\text{adj}}{\text{hyp}}\), \(\tan\theta = \frac{\text{opp}}{\text{adj}}\)
4. If the triangle is non-right-angled, use the cosine rule:
\(\cos\theta = \frac{a^2 + b^2 - c^2}{2ab}\) where \(c\) is the side opposite angle \(\theta\).

B. Angle Between a Line and a Plane

The angle between a line and a plane is defined as the angle between the line and its orthogonal projection onto that plane.

Think of it using a real-world analogy: imagine a sunny day with the sun directly overhead. The orthogonal projection is the flat shadow that a slanted stick casts on the ground!

Step-by-Step Procedure to Find the Angle:

1. Identify the line: Let the line be \(AB\), where point \(A\) lies on the plane and point \(B\) is elevated above the plane.
2. Drop a perpendicular: From point \(B\), drop a perpendicular line straight down to the plane, meeting the plane at point \(C\). The line segment \(BC\) is perpendicular to every line on the plane passing through \(C\).
3. Connect the projection: Connect point \(A\) to point \(C\). The line segment \(AC\) is the projection of \(AB\) on the plane.
4. Identify the angle: The angle between the line \(AB\) and the plane is \(\angle BAC\).
5. Solve the right-angled triangle: Triangle \(ABC\) has a right angle at \(C\). Calculate the required lengths (\(AC\), \(BC\), or \(AB\)) and use right-angled trigonometry:
\(\tan(\angle BAC) = \frac{\text{Height } BC}{\text{Base projection } AC}\) or \(\sin(\angle BAC) = \frac{BC}{AB}\)

Worked Example: Consider a rectangular cuboid with base dimensions \(AB = 8\text{ cm}\), \(BC = 6\text{ cm}\), and vertical height \(AE = 5\text{ cm}\). Find the angle that the body diagonal \(EC\) makes with the base plane \(ABCD\).

1. Find the projection on the base: Point \(C\) is already on the base. Point \(E\) projects vertically down to point \(A\). Thus, the projection of diagonal \(EC\) onto the base is the face diagonal \(AC\).
2. Calculate length of base diagonal \(AC\):
Triangle \(ABC\) is a right-angled triangle on the bottom face:
\(AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ cm}\)
3. Form the vertical right-angled triangle \(EAC\):
This triangle has vertices at \(E\), \(A\), and \(C\), with the right angle at \(A\).
Height \(EA = 5\text{ cm}\), base \(AC = 10\text{ cm}\).
4. Calculate the angle \(\theta = \angle ECA\):
\(\tan\theta = \frac{EA}{AC} = \frac{5}{10} = 0.5\)
\(\theta = \arctan(0.5) \approx 26.6^\circ\) (or \(0.464\text{ rad}\)).

Key Takeaway: Never try to calculate angles directly in 3D perspective. Always drop a perpendicular, extract a 2D right-angled triangle, draw it flat, and apply standard trigonometry.

4. Examination Tips and Common Traps

1. Clean Working & Method Marks: In both Paper 1 and Paper 2, marks are awarded for method. Always state the formula you are using, show your substitutions clearly, and draw a labelled 2D triangle representing your slice of the 3D solid.

2. Calculator Notation: On Paper 2, never write calculator shorthand like \(5.2\text{E}3\) in your final answers. Always write proper mathematical scientific notation: \(5.2 \times 10^3\).

3. Accuracy: Unless specified otherwise in the question, intermediate working should carry extra precision, and final non-exact answers should be rounded to 3 significant figures.

4. Slant Height vs. Vertical Height: Always double-check whether the problem has given you the perpendicular height \(h\) or the slant height \(l\) before substituting into cone or pyramid formulas.