<p>If the volume of parallelopiped formed by the vectors \(\hat{i} + \lambda\hat{j} + \hat{k}\), \(\hat{j} + \lambda\hat{k}\) and \(\lambda\hat{i} + \hat{k}\) is minimum, then \(\lambda\) is equal to ______ (up to three decimal places).</p>
Step-by-Step Solution
Key Concept: The volume of a parallelepiped equals the absolute value of the scalar triple product of three vectors. To minimize volume, take the derivative of |V| with respect to λ, set it equal to zero, and solve—this requires recognizing that the minimum occurs at a critical point of the determinant function.
Step 1: Set up the scalar triple product. The three vectors are a = î + λĵ + k̂, b = ĵ + λk̂, c = λî + k̂. The volume is V = | a · ( b × c )|. Step 2: Compute the determinant: V = |det([1, λ, 1; 0, 1, λ; λ, 0, 1])| Expanding along the first row: det = 1(1·1 - λ·0) - λ(0·1 - λ·λ) + 1(0·0 - 1·λ) det = 1(1) - λ(-λ^2) + 1(-λ) = 1 + λ^3 - λ Step 3: To minimize V = |1 + λ^3 - λ|, find critical points by differentiating f(λ) = 1 + λ^3 - λ: f'(λ) = 3λ^2 - 1 = 0 λ^2 = 1/3 λ = ±1/√3 = ±√3/3 ≈ ±0.577 Step 4: Evaluate |f(λ)| at both critical points: At λ = 1/√3: f = 1 + (1/√3)^3 - (1/√3) = 1 + 1/(3√3) - 1/√3 = 1 - 2/(3√3) ≈ 0.615 At λ = -1/√3: f = 1 - 1/(3√3) + 1/√3 = 1 + 2/(3√3) ≈ 1.385 Step 5: The minimum volume occurs at λ = 1/√3 = √3/3. ∴ Answer: 0.577
Correct Answer: 0