Let's break this down into 2 steps, with explanations as requested. --- ### **Step 1: Prove that if \( u \) and \( v \) are orthogonal, then \( \|u+v\| = \|u-v\| \).** #### **Explanation:** Recall, vectors \( u \) and \( v \) are **orthogonal** if \( \langle u, v \rangle = 0 \). The **norm squared** of a sum or difference of two vectors is: \[ \|u+v\|^2 = \langle u+v, u+v \rangle = \langle u, u \rangle + 2\langle u, v \rangle + \langle v, v \rangle \] \[ \|u-v\|^2 = \langle u-v, u-v \rangle = \langle u, u \rangle - 2\langle u, v \rangle + \langle v, v \rangle \] If \( u \) and \( v \) are orthogonal, \( \langle u, v \rangle = 0 \), so: \[ \|u+v\|^2 = \|u\|^2 + \|v\|^2 \] \[ \|u-v\|^2 = \|u\|^2 + \|v\|^2 \] So, \( \|u+v\| = \|u-v\| \). --- ### **Step 2: Prove the converse. If \( \|u+v\| = \|u-v\| \), then \( u \) and \( v \) are orthogonal.** #### **Explanation:** If \( \|u+v\| = \|u-v\| \), then \[ \|u+v\|^2 = \|u-v\|^2 \] Expanding both sides: \[ (\|u\|^2 + 2\langle u, v \rangle + \|v\|^2) = (\|u\|^2 - 2\langle u, v \rangle + \|v\|^2) \] \[ 2\langle u, v \rangle = -2\langle u, v \rangle \] \[ 4\langle u, v \rangle = 0 \implies \langle u, v \rangle = 0 \] Thus, \( u \) and \( v \) are orthogonal. --- ## **Final Answer:** **We have shown, in both directions, that** \[ \boxed{\|u+v\| = \|u-v\| \text{ if and only if } u \text{ and } v \text{ are orthogonal.}} \] --- ### **Summary Table for Related Questions:** | **If u and v are orthogonal, the Pythagorean Theorem gives:** | \(\|u\|^2 + \|v\|^2 = \|u + v\|^2\) | |--------------------------------------------------------------|-------------------------------------| | **If \( \|u+v\| = \|u-v\| \), then** | \( \langle u, v \rangle = 0 \) |

Let's systematically address each part of your question step by step, with clear explanations. --- ### **Part 1: Prove that \( |\langle u, v \rangle| = \|u\|\|v\| \) if and only if \( u \) and \( v \) are orthogonal.** **Note:** Actually, the statement as given is not standard; typically, the Cauchy-Schwarz inequality states: \[ |\langle u, v \rangle| \leq \|u\| \|v\| \] and equality holds **if and only if** \( u \) and \( v \) are **linearly dependent**, **not orthogonal**. However, from your context, it seems you're referring to the property that: \[ \text{If } |\langle u, v \rangle| = 0, \text{ then } u \text{ and } v \text{ are orthogonal} \] and vice versa. --- ### **Part 2: Prove that if \( u \) and \( v \) are orthogonal, then \( \|u+v\| = \|u-v\| \).** **Step 1: Show that orthogonality implies \( \|u+v\| = \|u-v\| \).** **Explanation:** If \( u \perp v \), then \( \langle u, v \rangle = 0 \). Recall that: \[ \|u \pm v\|^2 = \langle u \pm v, u \pm v \rangle = \|u\|^2 \pm 2 \langle u, v \rangle + \|v\|^2 \] Since \( \langle u, v \rangle = 0 \): \[ \|u + v\|^2 = \|u\|^2 + \|v\|^2 \] \[ \|u - v\|^2 = \|u\|^2 + \|v\|^2 \] Thus: \[ \|u + v\| = \|u - v\| \] --- **Part 3: Conversely, prove that if \( \|u+v\| = \|u-v\| \), then \( u \) and \( v \) are orthogonal.** **Step 2: Show that equality of norms implies orthogonality.** Starting from: \[ \|u + v\|^2 = \|u - v\|^2 \] Expanding: \[ \|u\|^2 + 2 \langle u, v \rangle + \|v\|^2 = \|u\|^2 - 2 \langle u, v \rangle + \|v\|^2 \] Subtracting: \[ 2 \langle u, v \rangle = - 2 \langle u, v \rangle \] \[ 4 \langle u, v \rangle = 0 \Rightarrow \langle u, v \rangle = 0 \] which confirms that \( u \) and \( v \) are orthogonal. --- ### **Final conclusion:** **The two statements are equivalent:** \[ \boxed{ \text{Vectors } u, v \text{ are orthogonal} \quad \Longleftrightarrow \quad \|u + v\| = \|u - v\| } \] --- ### **Additional notes on the other parts:** - **Pythagorean Theorem in inner product spaces:** If \( u \perp v \), then: \[ \|u + v\|^2 = \|u\|^2 + \|v\|^2 \] - **Using the properties of inner products:** The derivation relies on expanding norms and applying the definition of orthogonality and properties of inner products. --- **In summary:** - **If \( u \) and \( v \) are orthogonal, then \( \|u+v\| = \|u-v\| \).** - **Conversely, if \( \|u+v\| = \|u-v\| \), then \( u \) and \( v \) are orthogonal.** This completes the proof and understanding of the relationships you asked about.