In this section, we will explore how to apply Pythagoras' theorem and trigonometry in three-dimensional (3D) spaces. Understanding these concepts is crucial for solving problems related to finding angles between lines and planes, as well as calculating distances in 3D. These skills are essential for tackling real-world problems and are a key part of the CIE IGCSE syllabus.

Understanding 3D Geometry

Before diving into complex calculations, it's important to grasp the basics of 3D geometry. In a 3D space, points are represented by coordinates $(x, y, z)$, where $z$ represents the third dimension.

Key Concepts

• Points in 3D: Coordinates $(x, y, z)$ used to locate points in space.
• Distance Formula: The formula to calculate the distance between two points in 3D.
• Vectors: Representing direction and magnitude in 3D.

Pythagoras' Theorem in 3D

Pythagoras' theorem is not limited to 2D spaces. In 3D, it helps us calculate the distance between two points, which is crucial for many geometric and trigonometric applications.

Image courtesy of MME Revise

Formula and Application

• Distance Between Two Points: The formula for calculating the distance between points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Example 1: Calculating Distance in 3D

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

Solution:

Using the distance formula, the distance is:

$d = \sqrt{(4-1)^2 + (5-2)^2 + (6-3)^2}$$d = \sqrt{27} $<p></p>$d = 3\sqrt{3} \text{ units}.$<h2 id="trigonometry-in-3-d"><strong>Trigonometry in 3D</strong></h2><p>Trigonometry extends into 3D by using angles between vectors and planes. This is fundamental for finding angles in 3D structures and for understanding the spatial relationships between different geometric entities.</p><img src="https://tutorchase-production.s3.eu-west-2.amazonaws.com/e2b421c2-09d7-4c6b-951a-f874694efe24-file.png" alt="Trigonometry in 3D" style="width: 500px; height: 411px" width="500" height="411"><p></p><h3><strong>Key Trigonometric Ratios</strong></h3><ul><li><strong>Sine, Cosine, and Tangent</strong> in 3D involve angles between lines and planes or between two intersecting lines.</li></ul><h3><strong>Example 2: Finding the Angle between a Line and a Plane</strong></h3><p>Imagine a line intersecting a plane at a point, and we wish to find the angle of intersection.</p><p></p><h4><strong>Solution</strong>: </h4><p>First, determine the direction vector of the line and the normal vector of the plane. The angle$\theta$between the line and the plane can be found using the dot product formula and trigonometric identities.</p><p></p><h2 id="solving-3-d-problems"><strong>Solving 3D Problems</strong></h2><p>When solving 3D problems, it's essential to visualise the geometric configuration and apply the correct formulae to find distances or angles.</p><p></p><h3><strong>Example 3: Calculating Distances in 3D Space</strong></h3><p>Given points$C(7, -2, 1)$and$D(-1, 4, 5)$, calculate the distance between them.</p><p></p><h4><strong>Solution</strong>: </h4><p>Using the distance formula, the distance is</p><p></p>$d = \sqrt{(-1-7)^2 + (4+2)^2 + (5-1)^2}$<p></p>$d = \sqrt{64 + 36 + 16}$<p></p><p></p>$d = \sqrt{116}$<p></p>$d = 2\sqrt{29} \text { units}$<p></p><h2 id="practice-questions"><strong>Practice Questions</strong></h2><h3><strong>Problem 1:</strong> </h3><p>Find the distance between the points$E(2, -3, 5)$and$F(-3, 2, -4)$.</p><p></p><h4><strong>Solution:</strong></h4><p>To find the distance between two points in 3D, you use the formula:</p><p></p>$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$<p></p><p>Plugging in the coordinates of$E$and$F$yields:</p><p></p>$d = \sqrt{(-3 - 2)^2 + (2 + 3)^2 + (-4 - 5)^2}$<p></p>$d = \sqrt{25 + 25 + 81}$<p></p>$d = \sqrt{131}$<p></p><h3><strong>Problem 2:</strong> </h3><p>Determine the angle between the line passing through points$G(1, 1, 1)$and$H(4, 4, 4)$and the plane defined by the equation$x + y + z = 12$.</p><p></p><h4><strong>Solution:</strong></h4><p>The angle$\theta$between a line and a plane can be found using the dot product of the line's direction vector and the plane's normal vector. For the line through$G$and$H$, the direction vector$d$is$(3, 3, 3)$(the difference between$H$and$G$).</p><p>The normal vector$N$of the plane$x + y + z = 12$is$(1, 1, 1)$. The angle is given by: </p><p></p>$\cos(\theta) = \frac{d \cdot N}{|d||N|}$<p></p>$d \cdot N = (3)(1) + (3)(1) + (3)(1) = 9$<p></p>$|d| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27}$<p></p>$|N| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$<p></p>$\cos(\theta) = \frac{9}{\sqrt{27}\sqrt{3}}$<p></p>$\cos (\theta) = \frac{9}{\sqrt{81}} = \frac{9}{9} = 1$<p></p><p>The result implies that$\theta = 0^\circ$(or equivalently, the line and the plane's normal vector are parallel), which aligns with the conclusion that there was a misunderstanding in my previous explanation. The accurate interpretation is that when the cosine of the angle equals 1, it means the angle between the line direction vector and the plane's normal vector is$0^\circ$, indicating they are parallel. This is a unique situation given the specific vectors involved and does not generally represent the angle between a line and a plane in 3D space.