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The problem asks to find the vector \((\mathbf{i} \times \mathbf{j}) \times \mathbf{k}\) using the properties of cross products rather than using determinants.

Let's break it down:

1. **Step 1: Compute \(\mathbf{i} \times \mathbf{j}\)**  
   From the standard basis vectors in 3D, we know that:
   \[
   \mathbf{i} \times \mathbf{j} = \mathbf{k}
   \]
   So, \((\mathbf{i} \times \mathbf{j}) = \mathbf{k}\).

2. **Step 2: Compute \(\mathbf{k} \times \mathbf{k}\)**  
   Now, we need to compute \(\mathbf{k} \times \mathbf{k}\). The cross product of any vector with itself is always the zero vector:
   \[
   \mathbf{k} \times \mathbf{k} = \mathbf{0}
   \]

Therefore, the result of \((\mathbf{i} \times \mathbf{j}) \times \mathbf{k}\) is the **zero vector**:
\[
(\mathbf{i} \times \mathbf{j}) \times \mathbf{k} = \mathbf{0}
\]

Would you like more details on cross product properties or further explanation?

Here are five related questions to expand on this topic:
1. What is the geometric interpretation of the cross product?
2. How do the properties of the cross product change in higher dimensions?
3. How can you compute the cross product using determinants?
4. What is the scalar triple product involving cross products?
5. How do we calculate the magnitude of a cross product?

**Tip**: Remember, the cross product of two parallel vectors is always the zero vector!

Solve the vector (i × j) × k using the properties of cross products. Learn how the cross product between standard basis vectors leads to the zero vector. Suitable for Grades 10-12.

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