Introduction

Three-dimensional geometry is the branch of mathematics that deals with the study of shapes and figures in three-dimensional space. It involves understanding and analyzing the properties of 3D shapes such as cubes, spheres, cylinders, cones, and prisms. In this topic, we explore concepts such as volume, surface area, and spatial relationships of these geometric objects.

Coordinates in Three-Dimensional Space

Key Terms:

• Coordinates: Three numbers $(x, y, z)$ that locate a point in 3D space.
• Origin: The point $(0, 0, 0)$ where the x, y, and z axes intersect.

Example:

Find the coordinates of point P in 3D space if it is located 3 units to the right, 2 units up, and 5 units in front of the origin.

Solution: Given that P is located 3 units to the right (x-axis), 2 units up (y-axis), and 5 units in front (z-axis), the coordinates of P will be $(3, 2, 5)$.

Distance and Midpoint in 3D Space

Key Terms:

• Distance Formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$
• Midpoint Formula: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)$

Example:

Find the distance between points A(1, 2, 3) and B(4, 5, 6).

Solution: Using the distance formula, we have: $d = \sqrt{(4 - 1)^2 + (5 - 2)^2 + (6 - 3)^2} = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}$ units.

Equations of Lines and Planes in 3D Space

Key Terms:

• Vector Form of a Line: $\vec{r} = \vec{a} + \lambda\vec{b}$
• Scalar Equation of a Plane: $ax + by + cz = d$

Example:

Find the vector equation of a line passing through point A(1, 2, 3) and parallel to vector $\vec{b} = \langle 2, -1, 4 \rangle$.

Solution: Since the line is parallel to $\vec{b}$, the vector equation of the line is: $\vec{r} = \langle 1, 2, 3 \rangle + \lambda\langle 2, -1, 4 \rangle$

Intersection of Lines and Planes

Key Terms:

• Intersection: The point where two lines or planes meet.
• Coincident Lines/Planes: Lines or planes that lie on top of each other.

Example:

Find the point of intersection of the lines: $\vec{r} = \langle 1, 2, 3 \rangle + \lambda\langle 2, -1, 4 \rangle$ and $\vec{s} = \langle -1, 3, 2 \rangle + \mu\langle 3, 1, -2 \rangle$.

Solution: Equating the x, y, and z components of the two lines, we can solve for $\lambda$ and $\mu$ to find the point of intersection.

Common Mistakes

• Forgetting to account for all dimensions in 3D space when solving problems.
• Misinterpreting the vector form of lines or planes.
• Incorrectly applying the distance formula due to errors in subtraction or squaring.

Key Points

• 3D geometry involves visualizing objects in three-dimensional space.
• Understanding coordinates, distance, and midpoint calculations is crucial.
• Equations of lines and planes help in determining their positions and relationships in 3D space.

Practice Questions

1. Find the midpoint of the line segment with endpoints P(2, 4, 6) and Q(8, 10, 12).

Solution: The midpoint M is given by $\left(\frac{2 + 8}{2}, \frac{4 + 10}{2}, \frac{6 + 12}{2}\right) = (5, 7, 9)$.

2. Determine the distance between the points A(-1, 3, 2) and B(4, -2, 7).

Solution: The distance formula gives $d = \sqrt{(-1 - 4)^2 + (3 - (-2))^2 + (2 - 7)^2} = \sqrt{25 + 25 + 25} = 5\sqrt{3}$ units.

3. Write the vector equation of a line passing through point P(3, 1, -2) and parallel to vector $\langle 2, 3, -1 \rangle$.

Solution: The vector equation is $\vec{r} = \langle 3, 1, -2 \rangle + \lambda\langle 2, 3, -1 \rangle$.

4. Find the point of intersection of the lines: $\vec{r} = \langle 2, -1, 3 \rangle + \lambda\langle 1, 2, -1 \rangle$ and $\vec{s} = \langle -1, 3, 2 \rangle + \mu\langle 2, 1, -3 \rangle$.

Solution: Solving for $\lambda$ and $\mu$, we get the point of intersection as (1, 1, 2).