3.1 Three-Dimensional Geometry

1. 3D Coordinate Geometry

We know that a point in the standard Cartesian plane has the form $P(x,y)$. In 3D space, we add one more coordinate, thus a point has the form $P(x,y,z)$.

The Distance Formula: The distance between two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ is given by:

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

The Midpoint Formula: The midpoint $M$ of the line segment $[AB]$ is given by averaging the coordinates:

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

EXAMPLE 1

Let $A(1,0,5)$ and $B(2,3,1)$. Find the following:

(a) The distance between A and B:

$$\begin{aligned} d_{AB} &= \sqrt{(1 - 2)^2 + (0 - 3)^2 + (5 - 1)^2} \\ &= \sqrt{(-1)^2 + (-3)^2 + 4^2} \\ &= \sqrt{1 + 9 + 16} \\ &= \mathbf{\sqrt{26}} \end{aligned}$$

(b) The distance between the origin $O(0,0,0)$ and B:

$$\begin{aligned} d_{OB} &= \sqrt{2^2 + 3^2 + 1^2} \\ &= \sqrt{4 + 9 + 1} \\ &= \mathbf{\sqrt{14}} \end{aligned}$$

(c) The coordinates of the midpoint M of the line segment $[AB]$:

$$M\left(\dfrac{1 + 2}{2}, \dfrac{0 + 3}{2}, \dfrac{5 + 1}{2}\right) \implies \mathbf{M\left(\dfrac{3}{2}, \dfrac{3}{2}, 3\right)}$$

(d) The coordinates of point C given that B is the midpoint of $[AC]$:
Notice: the coordinates of A, B, C form arithmetic sequences since B is exactly in the middle.

x-coordinates: $1 \to 2 \to 3$
y-coordinates: $0 \to 3 \to 6$
z-coordinates: $5 \to 1 \to -3$

Resulting in the coordinate point $\mathbf{C(3, 6, -3)}$.

2. Volumes and Surface Areas of Known Solids

The volumes and the surface areas of 5 known solids are given below:

Solid	Volume ($V$)	Surface Area ($S$)
Cuboid Notation: $x, y, z$	$V = xyz$	$S = 2xy + 2yz + 2zx$
Pyramid	$V = \dfrac{1}{3}(\text{area of base}) \times (\text{height})$	$S = \text{Sum of areas of the faces}$
Cylinder Notation: $r, h$	$V = \pi r^2 h$	$S = 2\pi rh + 2\pi r^2$
Cone $L = \sqrt{r^2 + h^2}$	$V = \dfrac{1}{3}\pi r^2 h$	$S = \pi r L + \pi r^2$
Sphere	$V = \dfrac{4}{3}\pi r^3$	$S = 4\pi r^2$

EXAMPLE 2 (Understanding Basic Solids)

The geometric derivations of the volume ($V$) and surface area ($S$) formulas for the foundational solids are explained below:

1. Cuboid

Volume: $V = xyz$
Explanation: The rectangular base area ($x \times y$) is stacked continuously along the vertical height ($z$). For a perfect Cube, $x=y=z$, yielding $V = x^3$.

Surface Area: $S = 2xy + 2yz + 2zx$
Explanation: A cuboid consists of 6 rectangular faces structured in 3 opposite identical pairs: bottom/top ($2 \times xy$), left/right ($2 \times yz$), and front/back ($2 \times xz$). For a cube, this simplifies to $S = 6x^2$.

2. Pyramid

Volume: $V = \dfrac{1}{3} \times (\text{Base Area}) \times h$
Explanation: Geometrically, any pyramid mathematically fills exactly one-third of the volume of an enclosing prism that shares the exact same base and vertical height $h$.

Surface Area: $S = (\text{Base Area}) + (\text{Sum of Lateral Triangular Faces})$
Explanation: Calculated by breaking the 3D net down into its individual 2D constituent shapes and summing their areas.

3. Cylinder

Volume: $V = \pi r^2 h$
Explanation: The area of the circular base ($\pi r^2$) is extended straight upwards along the height ($h$).

Surface Area: $S = 2\pi rh + 2\pi r^2$
Explanation: Comprises two circular caps ($2 \times \pi r^2$) plus the curved lateral wall. If unrolled, the lateral wall forms a flat rectangle with a height of $h$ and a width equal to the circle's circumference ($2\pi r$), granting an area of $2\pi rh$.

4. Cone

Volume: $V = \dfrac{1}{3}\pi r^2 h$
Explanation: Identical to the logic of the pyramid, a cone occupies exactly one-third the volume of its enclosing cylinder.

Surface Area: $S = \pi r^2 + \pi r L$
Explanation: The flat circular base ($\pi r^2$) plus the unwrapped curved sector ($\pi r L$). The slant height $L$ creates a right-angled triangle with $h$ and $r$, hence $L = \sqrt{r^2 + h^2}$.

5. Sphere

Volume: $V = \dfrac{4}{3}\pi r^3$
Explanation: Derived using integral calculus by summing the areas of infinitely thin circular disks stacked from the bottom pole to the top pole.

Surface Area: $S = 4\pi r^2$
Explanation: The total 3D surface area enclosing a sphere is mathematically equivalent to the 2D area of exactly four of its "great circles" (a circle with radius $r$).

EXAMPLE 3

Given that the volume of a cylinder is 25:

(a) Express $h$ in terms of $r$:

$$\begin{aligned} V &= \pi r^2 h \\ 25 &= \pi r^2 h \\ h &= \mathbf{\dfrac{25}{\pi r^2}} \end{aligned}$$

(b) Hence express the surface area in terms of $r$:

$$\begin{aligned} S &= 2\pi rh + 2\pi r^2 \\ &= 2\pi r\left(\dfrac{25}{\pi r^2}\right) + 2\pi r^2 \\ &= \mathbf{\dfrac{50}{r} + 2\pi r^2} \end{aligned}$$

EXAMPLE 4

Given that the surface area of a cylinder is $100\pi$:

(a) Express $h$ in terms of $r$:

$$\begin{aligned} S &= 2\pi rh + 2\pi r^2 \\ 100\pi &= 2\pi rh + 2\pi r^2 \\ 50 &= rh + r^2 \\ h &= \mathbf{\dfrac{50 - r^2}{r}} \end{aligned}$$

(b) Hence express the volume in terms of $r$:

$$\begin{aligned} V &= \pi r^2 h \\ &= \pi r^2\left(\dfrac{50 - r^2}{r}\right) \\ &= \pi r(50 - r^2) \\ &= \mathbf{50\pi r - \pi r^3} \end{aligned}$$

EXAMPLE 5 (Pyramid Architecture)

Find the volume and the surface area of a right pyramid of square base of side 6 and vertical height 4.

Solution:

The vertical height is $h = 4$.

Slant height (AM): We use the Pythagoras theorem on right triangle $\triangle ANM$.

$$\begin{aligned} AM^2 &= AN^2 + NM^2 \\ AM^2 &= 4^2 + 3^2 \\ AM^2 &= 25 \implies \mathbf{AM = 5} \end{aligned}$$

Area of a single side triangle (e.g., $\triangle AED$):

$$\begin{aligned} A &= \dfrac{1}{2} \times ED \times AM \\ &= \dfrac{1}{2} \times 6 \times 5 = \mathbf{15} \end{aligned}$$

Total Volume ($V$):

$$\begin{aligned} V &= \dfrac{1}{3}(\text{Area of Base}) \times h \\ &= \dfrac{1}{3}(6^2) \times 4 = \mathbf{48} \end{aligned}$$

Total Surface Area ($S$):

$$\begin{aligned} S &= (\text{Area of square base}) + 4 \times (\text{Face Area}) \\ &= 6^2 + 4(15) \\ &= 36 + 60 = \mathbf{96} \end{aligned}$$

Notice about the angles between lines and planes:

Angle between line $AM$ and plane $BCDE$ = angle $A\hat{N}M$
Angle between line $AD$ and plane $BCDE$ = angle $A\hat{D}N$
Angle between the planes $ADE$ and $BCDE$ = angle $A\hat{M}N$
Angle between the planes $ACB$ and $ADE$ = angle $M\hat{A}M' = 2 \times M\hat{A}N$