Find the vector, not with determinants, but by using properties of cross products: $(\mathbf{i} + \mathbf{j}) \times (\mathbf{i} - \mathbf{j}).$

Neetesh Kumar | December 17, 2024

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This is the solution to Math 1C
Assignment: 12.4 Question Number 6
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Step-by-step solution:

We are tasked with computing the cross product $(\mathbf{i} + \mathbf{j}) \times (\mathbf{i} - \mathbf{j})$ using properties of cross products.

Step 1: Expand the expression:

Using the distributive property of cross products:

$(\mathbf{i} + \mathbf{j}) \times (\mathbf{i} - \mathbf{j}) = \mathbf{i} \times \mathbf{i} - \mathbf{i} \times \mathbf{j} + \mathbf{j} \times \mathbf{i} - \mathbf{j} \times \mathbf{j}.$

Step 2: Simplify each term using cross product rules:

From the standard cross product rules of unit vectors:

1. $\mathbf{i} \times \mathbf{i}$:
   The cross product of any vector with itself is zero:
   $\mathbf{i} \times \mathbf{i} = \mathbf{0}.$

2. $\mathbf{i} \times \mathbf{j}$:
   By definition:
   $\mathbf{i} \times \mathbf{j} = \mathbf{k}.$

3. $\mathbf{j} \times \mathbf{i}$:
   Using anti-commutativity:
   $\mathbf{j} \times \mathbf{i} = -(\mathbf{i} \times \mathbf{j}) = -\mathbf{k}.$

4. $\mathbf{j} \times \mathbf{j}$:
   The cross product of any vector with itself is zero:
   $\mathbf{j} \times \mathbf{j} = \mathbf{0}.$

Step 3: Combine the results:

Substitute the simplified terms into the expanded expression:

$(\mathbf{i} + \mathbf{j}) \times (\mathbf{i} - \mathbf{j}) = \mathbf{0} - \mathbf{k} + (-\mathbf{k}) - \mathbf{0}.$

Simplify further:

$(\mathbf{i} + \mathbf{j}) \times (\mathbf{i} - \mathbf{j}) = -\mathbf{k} - \mathbf{k}.$

Combine like terms:

$(\mathbf{i} + \mathbf{j}) \times (\mathbf{i} - \mathbf{j}) = -2\mathbf{k}.$

Final Answer:

The resulting vector is:

$(\mathbf{i} + \mathbf{j}) \times (\mathbf{i} - \mathbf{j}) = \boxed{-2\mathbf{k}}$

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