Find the vector, not with determinants, but by using properties of cross products.
(j − k) × (k − i)

We are tasked with calculating the cross product \((\mathbf{j} - \mathbf{k}) \times (\mathbf{k} - \mathbf{i})\). To solve this, we use the distributive property of the cross product, which states that:

\[
\mathbf{A} \times (\mathbf{B} + \mathbf{C}) = (\mathbf{A} \times \mathbf{B}) + (\mathbf{A} \times \mathbf{C})
\]
and similarly
\[
(\mathbf{A} + \mathbf{B}) \times \mathbf{C} = (\mathbf{A} \times \mathbf{C}) + (\mathbf{B} \times \mathbf{C})
\]

Let’s apply this property to expand the expression:

\[
(\mathbf{j} - \mathbf{k}) \times (\mathbf{k} - \mathbf{i}) = \mathbf{j} \times \mathbf{k} - \mathbf{j} \times \mathbf{i} - \mathbf{k} \times \mathbf{k} + \mathbf{k} \times \mathbf{i}
\]

Now, calculate each term:

1. \(\mathbf{j} \times \mathbf{k} = \mathbf{i}\)
2. \(\mathbf{j} \times \mathbf{i} = -\mathbf{k}\)
3. \(\mathbf{k} \times \mathbf{k} = \mathbf{0}\) (any vector crossed with itself is 0)
4. \(\mathbf{k} \times \mathbf{i} = -\mathbf{j}\)

Substituting these into the expression:

\[
(\mathbf{j} - \mathbf{k}) \times (\mathbf{k} - \mathbf{i}) = \mathbf{i} - (-\mathbf{k}) - \mathbf{0} + (-\mathbf{j})
\]

Simplifying:

\[
= \mathbf{i} + \mathbf{k} - \mathbf{j}
\]

So, the result of \((\mathbf{j} - \mathbf{k}) \times (\mathbf{k} - \mathbf{i})\) is:

\[
\mathbf{i} - \mathbf{j} + \mathbf{k}
\]

Let me know if you'd like more details, or if you have any questions! Here are some related questions:

1. What are the properties of cross products that can simplify calculations?
2. How do the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) interact in cross products?
3. Why does the cross product of a vector with itself result in zero?
4. How does the distributive property apply to vector cross products?
5. What is the geometric meaning of the result \(\mathbf{i} - \mathbf{j} + \mathbf{k}\)?

**Tip:** Remember that cross products are anti-commutative, meaning \(\mathbf{A} \times \mathbf{B} = -(\mathbf{B} \times \mathbf{A})\).

Find the vector, not with determinants, but by using properties of cross products. (j − k) × (k − i)

Solve the vector cross product (j − k) × (k − i) using properties of cross products. This problem illustrates how to apply the distributive property and the interactions between unit vectors. Ideal for students in Grades 11-12 studying vector algebra.

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