MAT2611 Assignment 07 2025 (Accurate Solutions) Due Friday, 11 July 2025

MAT2611 Assignment 07 Solutions
Due date: Friday, 11 July 2025

Problem 1
Problem Statement: Suppose that u, v, w are vectors such that ⟨u, v⟩ = 2,
⟨v, w⟩ = 2, ⟨u, w⟩ = 4, ∥u∥ = 1, ∥v∥ = 3, ∥w∥ = 5. Evaluate the following
expressions:

(a) ⟨u + v, v + w⟩

(b) ⟨2v − w, 3u + 2w⟩

(c) ∥2u + v∥

(d) d(2v, w)

(e) ∥u − 2v + w∥

Part (a): ⟨u + v, v + w⟩
Step 1: Use the bilinearity of the inner product:

⟨u + v, v + w⟩ = ⟨u, v⟩ + ⟨u, w⟩ + ⟨v, v⟩ + ⟨v, w⟩

Step 2: Substitute given values:

⟨u, v⟩ = 2, ⟨u, w⟩ = 4, ⟨v, w⟩ = 2, ∥v∥2 = ⟨v, v⟩ = 32 = 9

⟨u + v, v + w⟩ = 2 + 4 + 9 + 2
Step 3: Compute:
2 + 4 + 9 + 2 = 17
Final Answer: ⟨u + v, v + w⟩ = 17




1

, Part (b): ⟨2v − w, 3u + 2w⟩
Step 1: Expand using bilinearity:

⟨2v − w, 3u + 2w⟩ = 6⟨v, u⟩ + 4⟨v, w⟩ − 3⟨w, u⟩ − 2⟨w, w⟩

Step 2: Substitute values:

⟨v, u⟩ = 2, ⟨v, w⟩ = 2, ⟨w, u⟩ = 4, ⟨w, w⟩ = 52 = 25

= 6 · 2 + 4 · 2 − 3 · 4 − 2 · 25 = 12 + 8 − 12 − 50
Step 3: Compute:

12 + 8 − 12 − 50 = −42

Final Answer: ⟨2v − w, 3u + 2w⟩ = −42

Part (c): ∥2u + v∥
Step 1: Use the norm definition:
p
∥2u + v∥ = ⟨2u + v, 2u + v⟩

Step 2: Expand the inner product:

⟨2u + v, 2u + v⟩ = 4∥u∥2 + 4⟨u, v⟩ + ∥v∥2

Step 3: Substitute values:

∥u∥2 = 1, ⟨u, v⟩ = 2, ∥v∥2 = 9

= 4 · 1 + 4 · 2 + 9 = 4 + 8 + 9 = 21
Step 4: Compute the norm:
√
∥2u + v∥ = 21
√
Final Answer: ∥2u + v∥ = 21




2