<p>Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) lie in the \(x - y\) plane. Let \(\vec{a} = \hat{i}\), \(\vec{b} = -\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}\) and \(\vec{c} = -\frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}\). If \(\vec{p} = \lambda\vec{a}\), \(\vec{q} = \mu\vec{b}\), \(\vec{r} = \nu\vec{c}\) for scalars \(\lambda, \mu, \nu\), then \(|\vec{p}+\vec{q}+\vec{r}|\) can take the value(s):</p>
Step-by-Step Solution
Key Concept: The vectors a, b, c are unit vectors equally spaced at 120° angles (forming angles 0°, 120°, 240°); the sum |p+q+r| = |λa + μb + νc| is minimized when coefficients balance these angular separations, achieving minimum value 0.
Step 1: Identify the vectors. We have: <ul><li>a = î (angle 0°)</li><li>b = -½î + (√3/2)ĵ (angle 120°)</li><li>c = -½î - (√3/2)ĵ (angle 240°)</li></ul> These are unit vectors separated by 120° angles. Step 2: Verify that a + b + c = 0: a + b + c = î + (-½î + (√3/2)ĵ) + (-½î - (√3/2)ĵ) = 0 ✓ Step 3: Express p + q + r: p + q + r = λa + μb + νc = λa + μb + (−λa − μb) when ν adjusts appropriately, or more generally for any λ, μ, ν: Step 4: Rewrite using the constraint a + b + c = 0: |p + q + r|^2 = |λa + μb + νc|^2 Expanding: = λ^2|a|^2 + μ^2|b|^2 + ν^2|c|^2 + 2λμ(a·b) + 2μν(b·c) + 2νλ(c·a) Since a, b, c are unit vectors: a·b = cos(120°) = -½, b·c = -½, c·a = -½ = λ^2 + μ^2 + ν^2 + 2λμ(-½) + 2μν(-½) + 2νλ(-½) = λ^2 + μ^2 + ν^2 - λμ - μν - νλ Step 5: Find the range. Setting λ = μ = ν = t: = t^2 + t^2 + t^2 - t^2 - t^2 - t^2 = 0 → |p + q + r| = 0 Setting λ = 1, μ = ν = 0: |p + q + r|^2 = 1 → |p + q + r| = 1 By varying λ, μ, ν, the expression λ^2 + μ^2 + ν^2 - λμ - μν - νλ achieves all values ≥ 0. ∴ Answer: D (the set of all non-negative values, or specific value determined by context)
Correct Answer: D