To show that if $\| \mathbf{u} \| = \| \mathbf{v} \|$ for vectors $\mathbf{u}$ and $\mathbf{v}$, then $\mathbf{u} + \mathbf{v}$ and $\mathbf{u} - \mathbf{v}$ are orthogonal, we need to prove that the dot product $(\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}) = 0$.

Step 1: Expand the dot product

The expression we need to evaluate is:

$(\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}).$

Using the distributive property of the dot product, this becomes:

$\mathbf{u} \cdot \mathbf{u} - \mathbf{u} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{u} - \mathbf{v} \cdot \mathbf{v}.$

Step 2: Simplify the terms

We know the following properties of the dot product:

• $\mathbf{u} \cdot \mathbf{u} = \| \mathbf{u} \|^2$,
• $\mathbf{v} \cdot \mathbf{v} = \| \mathbf{v} \|^2$,
• The dot product is commutative, so $\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}$.

Thus, the expression becomes:

$\| \mathbf{u} \|^2 - \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v} - \| \mathbf{v} \|^2.$

Step 3: Cancel out terms

The terms $- \mathbf{u} \cdot \mathbf{v}$ and $+ \mathbf{u} \cdot \mathbf{v}$ cancel out, leaving us with:

$\| \mathbf{u} \|^2 - \| \mathbf{v} \|^2.$

Step 4: Use the given condition

We are given that $\| \mathbf{u} \| = \| \mathbf{v} \|$, which implies that $\| \mathbf{u} \|^2 = \| \mathbf{v} \|^2$. Therefore, the expression simplifies to:

$\| \mathbf{u} \|^2 - \| \mathbf{v} \|^2 = 0.$

Conclusion

Since the dot product $(\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}) = 0$, it follows that the vectors $\mathbf{u} + \mathbf{v}$ and $\mathbf{u} - \mathbf{v}$ are orthogonal.

Would you like more details or have any questions?

Here are 5 related questions:

1. What is the geometric interpretation of orthogonal vectors?
2. How does the equality $\| \mathbf{u} \| = \| \mathbf{v} \|$ influence the result?
3. Can this result be extended to higher dimensions or complex vectors?
4. How does the distributive property of the dot product help in simplifying expressions?
5. What other conditions can guarantee orthogonality of two vectors?

Tip: When proving vector properties, always check if the dot product can be expanded, as it often simplifies the process.