Distance between two points in 3D Space

By the end of this lesson, students will be able to:
1. Understand the concept of three-dimensional space.
2. Apply the formula to find the distance between two points in 3D space.
3. Solve problems involving distances between points in 3D coordinates.

1. Introduction to 3D Coordinate System**

In two-dimensional space, points are represented by coordinates \( (x, y) \), with each point lying on a plane. In three-dimensional (3D) space, a point is represented by three coordinates \( (x, y, z) \), where:

•  \(x\) is the horizontal distance (along the x-axis).
•  \(y\) is the vertical distance (along the y-axis).
• \(z\) is the depth (along the z-axis).

The three axes (x, y, and z) are mutually perpendicular, and every point in 3D space can be identified by its unique set of \( (x, y, z) \) coordinates.

2. Definition: Distance Between Two Points in 3D Space

The distance between two points in 3D space, \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \), is the length of the straight line segment that connects these points

3. Formula for Distance Between Two Points

The formula to find the distance between two points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \) in 3D space is:

\[
d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}
\]This formula is an extension of the Pythagorean theorem, which calculates the distance between two points in a plane, applied to three dimensions.

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Interactive graph

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4. Worked Examples

Example 1:

Find the distance between the points \( A(2, 3, 5) \) and \( B(6, 7, 9) \).

Solution:

Using the formula:

\[
d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}
\]

Substitute the coordinates of \( A \) and \( B \):

\[
d = \sqrt{(6 – 2)^2 + (7 – 3)^2 + (9 – 5)^2}
\]
\[
d = \sqrt{4^2 + 4^2 + 4^2}
\]
\[
d = \sqrt{16 + 16 + 16}
\]
\[
d = \sqrt{48}
\]
\[
d = 4\sqrt{3} \approx 6.93
\]

So, the distance between \( A(2, 3, 5) \) and \( B(6, 7, 9) \) is approximately **6.93 units**.

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Examples : 2
Find the distance between the points \( C(-1, -3, 4) \) and \( D(3, -1, -2) \).

Solution:

Using the formula:

\[
d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}
\]

Substitute the coordinates of \( C \) and \( D \):

\[
d = \sqrt{(3 – (-1))^2 + (-1 – (-3))^2 + (-2 – 4)^2}
\]
\[
d = \sqrt{(3 + 1)^2 + (-1 + 3)^2 + (-2 – 4)^2}
\]
\[
d = \sqrt{4^2 + 2^2 + (-6)^2}
\]
\[
d = \sqrt{16 + 4 + 36}
\]
\[
d = \sqrt{56}
\]
\[
d \approx 7.48
\]

So, the distance between \( C(-1, -3, 4) \) and \( D(3, -1, -2) \) is approximately 7.48 units.

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Practice Problems :

1. Find the distance between the points \( P(1, 4, 7) \) and \( Q(5, 8, 12) \).

2. Calculate the distance between the points \( R(0, 0, 0) \) and \( S(3, 6, 9) \).

3. Find the distance between \( T(-2, -3, 1) \) and \( U(2, 0, -1) \).

4. If the points \( A(4, 2, 6) \) and \( B(x, 5, 9) \) are 5 units apart, find the value of \(x\). 

Answers to practice worksheet