Question

Find the distance between the following pairs of points. a. \(\left[\begin{array}{r}3 \\ -1 \\ 0\end{array}\right]\) and \(\left[\begin{array}{r}2 \\ -1 \\ 1\end{array}\right]\) b. \(\left[\begin{array}{r}2 \\ -1 \\ 2\end{array}\right]\) and \(\left[\begin{array}{l}2 \\ 0 \\ 1\end{array}\right]\) c. \(\left[\begin{array}{r}-3 \\ 5 \\ 2\end{array}\right]\) and \(\left[\begin{array}{l}1 \\ 3 \\ 3\end{array}\right]\) d. \(\left[\begin{array}{r}4 \\ 0 \\ -2\end{array}\right]\) and \(\left[\begin{array}{l}3 \\ 2 \\ 0\end{array}\right]\)

Short answer

a) \(\sqrt{2}\), b) \(\sqrt{2}\), c) \(\sqrt{21}\), d) 3.

Step-by-step solution

Understanding the Distance Formula for Points in 3D

The distance between two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) in 3D space is found using the formula: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\].

Calculating Distance for Part (a)

Substitute the coordinates from the points \[\begin{bmatrix} 3 \ -1 \ 0 \end{bmatrix}\] and \[\begin{bmatrix} 2 \ -1 \ 1 \end{bmatrix}\] into the distance formula: \[d = \sqrt{(2 - 3)^2 + (-1 + 1)^2 + (1 - 0)^2} = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{1 + 0 + 1} = \sqrt{2}.\] Thus, the distance is \sqrt{2}.\.

Calculating Distance for Part (b)

Use the points \[\begin{bmatrix} 2 \ -1 \ 2 \end{bmatrix}\] and \[\begin{bmatrix} 2 \ 0 \ 1 \end{bmatrix}\] in the formula: \[d = \sqrt{(2 - 2)^2 + (0 + 1)^2 + (1 - 2)^2} = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2}.\] The distance is \sqrt{2}.\.

Calculating Distance for Part (c)

Apply the distance formula to the points \[\begin{bmatrix} -3 \ 5 \ 2 \end{bmatrix}\] and \[\begin{bmatrix} 1 \ 3 \ 3 \end{bmatrix}\]: \[d = \sqrt{(1 + 3)^2 + (3 - 5)^2 + (3 - 2)^2} = \sqrt{(4)^2 + (-2)^2 + 1^2} = \sqrt{16 + 4 + 1} = \sqrt{21}.\] The distance is \sqrt{21}.\.

Calculating Distance for Part (d)

Use the distance formula for the points \[\begin{bmatrix} 4 \ 0 \ -2 \end{bmatrix}\] and \[\begin{bmatrix} 3 \ 2 \ 0 \end{bmatrix}\]: \[d = \sqrt{(3 - 4)^2 + (2 - 0)^2 + (0 + 2)^2} = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3.\] The distance is 3.

Key concepts

Distance Formula

When calculating the distance between two points in three-dimensional space, we rely on the distance formula. This formula is an extension of the Pythagorean Theorem into three dimensions. Given two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \), the distance \( d \) between them is calculated as: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \].
This equation helps us find the shortest path between two points in a 3D space by adding up the squares of the differences between their corresponding coordinates.

• The part \( (x_2 - x_1)^2 \) accounts for the horizontal difference.
• Likewise, \( (y_2 - y_1)^2 \) takes care of the vertical difference.
• Finally, \( (z_2 - z_1)^2 \) deals with the depth or z-axis difference.

These squared values are then summed and square rooted to result in the 3D Euclidean distance.

3D Vectors

In mathematics and physics, 3D vectors play an instrumental role. These vectors are quantities that have both magnitude and direction in three-dimensional space. Each vector is represented by three components, usually written as \( \langle x, y, z \rangle \), where each component corresponds to an axis in 3D space.When we consider points in 3D space, such as \[ \begin{bmatrix} x \ y \ z \end{bmatrix} \], we can think of these points as vectors originating from the origin \( (0, 0, 0) \) to the point. This visualization is helpful when using the distance formula, as the differences between coordinates resemble the calculation of a vector's magnitude.
Understanding vectors can:

• Help you in physics to calculate forces, velocities, and other vector quantities.
• Offer insight when examining geometric problems within multiple dimensions.

Therefore, navigating calculations involving 3D vectors requires a solid grip on operations like addition, subtraction, and finding magnitudes.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, provides a bridge between algebra and geometry through graphs and equations. This branch of mathematics involves plotting points, lines, and curves within a coordinate plane, which is extended into three dimensions in 3D space.The significance of coordinate geometry lies in its ability to model and solve geometric problems using an algebraic approach. The principles of coordinate geometry allow us to:

• Visualize complex problems in a spatial context.
• Compute distances, midpoints, and other properties of geometric figures directly from algebraic equations.

In three-dimensional coordinate geometry, each point is defined by an \( x \), \( y \), and \( z \) value, which makes analyzing shapes and spatial relationships possible. By applying the distance formula in 3D, we can solve problems involving the shortest path between points, assess shapes, and even predict spatial interactions in fields like physics and engineering.