https://file.aiquickdraw.com/mathbot/file/172678652388022c0mjf8.jpg

Solve the problem provided in the screenshot and provide a step-by-step solution

The given expression in the image is:

\[
\mathbf{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}
\]

We are asked to find \( a_0 \).

In vector notation, a vector \( \mathbf{a} \) can be written as:

\[
\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}
\]

Comparing this with the given vector \( \mathbf{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \), we can observe that:

- The component along \( \mathbf{i} \) (denoted \( a_1 \)) is 1.
- The component along \( \mathbf{j} \) (denoted \( a_2 \)) is 2.
- The component along \( \mathbf{k} \) (denoted \( a_3 \)) is 3.

Now, the term \( a_0 \) often refers to the magnitude of the vector \( \mathbf{a} \), which can be calculated using the formula for the magnitude of a 3-dimensional vector:

\[
|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}
\]

Substituting the values \( a_1 = 1 \), \( a_2 = 2 \), and \( a_3 = 3 \), we get:

\[
|\mathbf{a}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}
\]

Thus, \( a_0 = |\mathbf{a}| = \sqrt{14} \).

### Final Answer:
\[
a_0 = \sqrt{14}
\]

Let me know if you'd like any further details or explanations!

### Related Questions:
1. What is the geometric interpretation of the magnitude of a vector?
2. How do we generalize the formula for the magnitude of a vector in n-dimensions?
3. Can you find the unit vector corresponding to \( \mathbf{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \)?
4. How does the direction of a vector relate to its components?
5. What happens if one or more of the components of the vector \( \mathbf{a} \) were negative?

### Tip:
Remember that when dealing with vectors, the magnitude (or norm) represents the "length" of the vector in Euclidean space!

This page explains how to calculate the magnitude of a 3D vector with components i, 2j, and 3k. The problem covers essential concepts from vector algebra, such as finding the magnitude of a vector using the Pythagorean theorem in 3D space. Suitable for students in Grades 10-12.

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