3.1 Three-Dimensional Geometry
1. 3D Coordinate Geometry
A point in the standard 2D Cartesian plane is defined by the coordinate pair $P(x,y)$. When extended into 3D space, a third coordinate is added, assigning the point the functional form $P(x,y,z)$.
The Distance Formula: The spatial distance between two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ is mathematically defined by:
$d_{AB} = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}$
The Midpoint Formula: The midpoint $M$ defining the exact center of the line segment AB is calculated by averaging the respective coordinates:
$M\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)$
EXAMPLE 1
Given the spatial points $A(1,0,5)$ and $B(2,3,1)$, evaluate the following spatial properties:
(a) The distance between A and B:
$d_{AB} = \sqrt{(1 - 2)^2 + (0 - 3)^2 + (5 - 1)^2} = \sqrt{1 + 9 + 16} = \mathbf{\sqrt{26}}$.
$d_{AB} = \sqrt{(1 - 2)^2 + (0 - 3)^2 + (5 - 1)^2} = \sqrt{1 + 9 + 16} = \mathbf{\sqrt{26}}$.
(b) The distance between the origin O(0,0,0) and B:
$d_{OB} = \sqrt{2^2 + 3^2 + 1^2} = \mathbf{\sqrt{14}}$.
$d_{OB} = \sqrt{2^2 + 3^2 + 1^2} = \mathbf{\sqrt{14}}$.
(c) The coordinates of the midpoint M of [AB]:
$M\left(\frac{1 + 2}{2}, \frac{0 + 3}{2}, \frac{5 + 1}{2}\right) \Rightarrow \mathbf{M(1.5, 1.5, 3)}$.
$M\left(\frac{1 + 2}{2}, \frac{0 + 3}{2}, \frac{5 + 1}{2}\right) \Rightarrow \mathbf{M(1.5, 1.5, 3)}$.
(d) The coordinates of point C given that B is the midpoint of [AC]:
The coordinates of A, B, and C logically form arithmetic sequences due to equal spacing.
x-coordinates: $1 \rightarrow 2 \rightarrow 3$
y-coordinates: $0 \rightarrow 3 \rightarrow 6$
z-coordinates: $5 \rightarrow 1 \rightarrow -3$
Resulting in coordinate point $\mathbf{C(3, 6, -3)}$.
The coordinates of A, B, and C logically form arithmetic sequences due to equal spacing.
x-coordinates: $1 \rightarrow 2 \rightarrow 3$
y-coordinates: $0 \rightarrow 3 \rightarrow 6$
z-coordinates: $5 \rightarrow 1 \rightarrow -3$
Resulting in coordinate point $\mathbf{C(3, 6, -3)}$.
2. Volumes and Surface Areas of Known Solids
| Solid | Volume ($V$) | Surface Area ($S$) |
|---|---|---|
| Cuboid (Sides x, y, z) |
$V = xyz$ | $S = 2xy + 2yz + 2zx$ |
| Pyramid | $V = \frac{1}{3}(\text{Area of Base}) \times h$ | $S = \text{Sum of areas of all faces}$ |
| Cylinder (Radius r, Height h) |
$V = \pi r^2 h$ | $S = 2\pi rh + 2\pi r^2$ |
| Cone (Slant height L) |
$V = \frac{1}{3}\pi r^2 h$ | $S = \pi r L + \pi r^2$ where $L = \sqrt{r^2 + h^2}$ |
| Sphere | $V = \frac{4}{3}\pi r^3$ | $S = 4\pi r^2$ |
EXAMPLE 2 (Basic Solids)
Cube of side $x$: $V = x \cdot x \cdot x = \mathbf{x^3}$ and $S = \mathbf{6x^2}$.
Cuboid with square base $x$ and height $y$: $V = \mathbf{x^2y}$ and $S = \mathbf{2x^2 + 4xy}$.
EXAMPLE 3 & 4 (Algebraic Manipulation)
Example 3: Given that the volume of a cylinder is fixed at $25$.
(a) Express $h$ strictly in terms of $r$: $V = \pi r^2 h \Rightarrow 25 = \pi r^2 h \Rightarrow \mathbf{h = \frac{25}{\pi r^2}}$.
(b) Express Surface Area in terms of $r$: $S = 2\pi rh + 2\pi r^2 = 2\pi r\left(\frac{25}{\pi r^2}\right) + 2\pi r^2 = \mathbf{\frac{50}{r} + 2\pi r^2}$.
(a) Express $h$ strictly in terms of $r$: $V = \pi r^2 h \Rightarrow 25 = \pi r^2 h \Rightarrow \mathbf{h = \frac{25}{\pi r^2}}$.
(b) Express Surface Area in terms of $r$: $S = 2\pi rh + 2\pi r^2 = 2\pi r\left(\frac{25}{\pi r^2}\right) + 2\pi r^2 = \mathbf{\frac{50}{r} + 2\pi r^2}$.
Example 4: Given the surface area of a cylinder is fixed at $100\pi$.
(a) Express $h$ strictly in terms of $r$: $S = 2\pi rh + 2\pi r^2 \Rightarrow 100\pi = 2\pi rh + 2\pi r^2 \Rightarrow 50 = rh + r^2 \Rightarrow \mathbf{h = \frac{50 - r^2}{r}}$.
(b) Express Volume in terms of $r$: $V = \pi r^2 h = \pi r^2\left(\frac{50 - r^2}{r}\right) = \pi r(50 - r^2) = \mathbf{50\pi r - \pi r^3}$.
(a) Express $h$ strictly in terms of $r$: $S = 2\pi rh + 2\pi r^2 \Rightarrow 100\pi = 2\pi rh + 2\pi r^2 \Rightarrow 50 = rh + r^2 \Rightarrow \mathbf{h = \frac{50 - r^2}{r}}$.
(b) Express Volume in terms of $r$: $V = \pi r^2 h = \pi r^2\left(\frac{50 - r^2}{r}\right) = \pi r(50 - r^2) = \mathbf{50\pi r - \pi r^3}$.
EXAMPLE 5 (Pyramid Architecture)
Determine the volume and surface area of a right pyramid possessing a square base of side 6 and a vertical height of 4.
Vertical height: $h = 4$.
Slant height (AM): Applying Pythagoras' theorem to the interior triangle formed by the height and half the base length ($3$):
$AM^2 = 4^2 + 3^2 \Rightarrow AM^2 = 25 \Rightarrow \mathbf{AM = 5}$.
$AM^2 = 4^2 + 3^2 \Rightarrow AM^2 = 25 \Rightarrow \mathbf{AM = 5}$.
Area of single triangular face: $A = \frac{1}{2} \times \text{base} \times \text{slant height} = \frac{1}{2} \times 6 \times 5 = \mathbf{15}$.
Total Volume: $V = \frac{1}{3}(\text{Area of Base}) \times h = \frac{1}{3}(6^2) \times 4 = \mathbf{48}$.
Total Surface Area: $S = (\text{Area of square base}) + 4 \times (\text{Face Area}) = 36 + 4(15) = \mathbf{96}$.