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To find the distance between the points \(\mathbf{a} = \begin{pmatrix} -1 \\ -3 \\ 1 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} 0 \\ 0 \\ -2 \end{pmatrix}\), we use the distance formula in 3-dimensional space:

\[
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]

Substituting the coordinates of points \(\mathbf{a}\) and \(\mathbf{b}\):

\[
\text{Distance} = \sqrt{(0 - (-1))^2 + (0 - (-3))^2 + (-2 - 1)^2}
\]

Simplifying the terms inside the square root:

\[
\text{Distance} = \sqrt{(1)^2 + (3)^2 + (-3)^2}
\]

\[
\text{Distance} = \sqrt{1 + 9 + 9}
\]

\[
\text{Distance} = \sqrt{19}
\]

Therefore, the distance between the points \(\mathbf{a}\) and \(\mathbf{b}\) is \(\sqrt{19}\).

Would you like a detailed explanation or have any questions?

Here are some related questions:
1. How is the distance formula derived in 3D space?
2. What is the geometrical interpretation of the distance between two points in 3D?
3. How would the formula change if it were in 2D instead of 3D?
4. What other methods can be used to verify the distance between two points in space?
5. How does the concept of distance extend to higher-dimensional spaces?

**Tip:** It's important to visualize points in 3D space to better understand the concept of distance, which is the straight-line length between the points.

Find the distance between the points a = (-1, -3, 1) and b = (0, 0, -2).

This problem demonstrates how to find the distance between two points (-1, -3, 1) and (0, 0, -2) in 3D space using the distance formula. Step-by-step solution provided, covering key concepts from 3D coordinate geometry. Suitable for Grades 9-11.

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