3.1 Three-Dimensional Geometry

1. 3D Coordinate Geometry

A point in a 2D plane is $P(x,y)$. In 3D space, a point is $P(x,y,z)$.

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

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

The Midpoint Formula: The midpoint $M$ of line segment $[AB]$ is the average of 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

Given $A(1,0,5)$ and $B(2,3,1)$, find:

(a) 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) Distance between origin $O(0,0,0)$ and B:

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

(c) Midpoint M of $[AB]$:

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

(d) Point C, if B is the midpoint of $[AC]$:
The coordinates of A, B, and C form arithmetic sequences.

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

Result: $\mathbf{C(3, 6, -3)}$.

2. Volumes and Surface Areas of Known Solids

The volume ($V$) and surface area ($S$) of 5 solids are:

Solid	Volume ($V$)	Surface Area ($S$)
Cuboid Notation: $x, y, z$	$V = xyz$	$S = 2xy + 2yz + 2zx$
Pyramid	$V = \dfrac{1}{3}(\text{Base Area}) \times h$	$S = \text{Sum of face areas}$
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 derivations for volume ($V$) and surface area ($S$) formulas are:

1. Cuboid

Volume: $V = xyz$
Explanation: The base area ($x \times y$) multiplies by the height ($z$). For a cube, $x=y=z$, yielding $V = x^3$.

Surface Area: $S = 2xy + 2yz + 2zx$
Explanation: A cuboid has 6 faces in 3 pairs: top/bottom ($2xy$), left/right ($2yz$), and front/back ($2zx$). For a cube, $S = 6x^2$.

2. Pyramid

Volume: $V = \dfrac{1}{3} \times (\text{Base Area}) \times h$
Explanation: A pyramid fills one-third the volume of a prism with the same base and height $h$.

Surface Area: $S = (\text{Base Area}) + (\text{Sum of face areas})$
Explanation: Sum the areas of the 2D shapes forming the 3D net.

3. Cylinder

Volume: $V = \pi r^2 h$
Explanation: The circular base area ($\pi r^2$) multiplies by the height ($h$).

Surface Area: $S = 2\pi rh + 2\pi r^2$
Explanation: Add the areas of two circular caps ($2\pi r^2$) and the lateral wall. Unrolled, the wall is a rectangle with height $h$ and width $2\pi r$, yielding $2\pi rh$.

4. Cone

Volume: $V = \dfrac{1}{3}\pi r^2 h$
Explanation: A cone occupies one-third the volume of a cylinder with the same base and height.

Surface Area: $S = \pi r^2 + \pi r L$
Explanation: Add the base area ($\pi r^2$) and the sector area ($\pi r L$). Slant height $L$ forms a right triangle with $h$ and $r$, so $L = \sqrt{r^2 + h^2}$.

5. Sphere

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

Surface Area: $S = 4\pi r^2$
Explanation: The surface area equals the area of four great circles.

EXAMPLE 3

If the volume of a cylinder is 25, then

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

$$ 25 = V = \pi r^2 h \implies h = \dfrac{25}{\pi r^2} $$

(b) Express surface area in terms of $r$:

$$ S = 2\pi rh + 2\pi r^2 = 2\pi r\left(\dfrac{25}{\pi r^2}\right) + 2\pi r^2 = \dfrac{50}{r} + 2\pi r^2 $$

EXAMPLE 4

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

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

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

(b) Express volume in terms of $r$:

$$ V = \pi r^2 h = \pi r^2\left(\dfrac{50 - r^2}{r}\right) = \pi r(50 - r^2) = 50\pi r - \pi r^3 $$

EXAMPLE 5 (Pyramid Architecture)

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

Solution:

Height $h = 4$.

Slant height (AM): Use the Pythagorean theorem on $\triangle ANM$.

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

Area of one triangular face (e.g., $\triangle AED$):

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

Total Volume ($V$):

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

Total Surface Area ($S$):

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

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 planes $ADE$ and $BCDE$ = angle $A\hat{M}N$
Angle between planes $ACB$ and $ADE$ = angle $M\hat{A}M' = 2 \times M\hat{A}N$