<p>Unit vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular, and unit vector \(\vec{c}\) is inclined at an angle \(\theta\) to both \(\vec{a}\) and \(\vec{b}\). If \(\vec{c} = \alpha\vec{a} + \beta\vec{b} + \gamma(\vec{a} \times \vec{b})\), then</p>
<p>(a) \(a = \beta\)</p>
<p>(b) \(\gamma^2 = 1 - 2\alpha^2\)</p>
<p>(c) \(\gamma^2 = -\cos 2\theta\)</p>
<p>(d) \(\beta^2 = \dfrac{1 + \cos 2\theta}{2}\)</p>
Step-by-Step Solution
Key Concept: Since ⃗c is a unit vector expressible in the basis {⃗a, ⃗b, ⃗a×⃗b}, use the magnitude condition |⃗c|² = 1 combined with the orthogonality constraint ⃗c·⃗a = cos θ and ⃗c·⃗b = cos θ to systematically solve for α, β, and γ.
Step 1: Set up the dot product conditions. Since ⃗c makes angle θ with both ⃗a and ⃗b: ⃗c·⃗a = α(⃗a·⃗a) + β(⃗a·⃗b) + γ(⃗a·(⃗a×⃗b)) = α = cos θ ⃗c·⃗b = α(⃗a·⃗b) + β(⃗b·⃗b) + γ(⃗b·(⃗a×⃗b)) = β = cos θ Step 2: Apply the magnitude condition |⃗c|^2 = 1. Since {⃗a, ⃗b, ⃗a×⃗b} form an orthonormal basis with |⃗a×⃗b| = 1: α^2 + β^2 + γ^2 = 1 cos^2θ + cos^2θ + γ^2 = 1 γ^2 = 1 - 2cos^2θ = -(2cos^2θ - 1) = -cos(2θ) Step 3: Determine the sign of γ. Using γ(⃗a×⃗b)·⃗c = γ^2 and the geometric constraint that ⃗c has a component along ⃗a×⃗b: γ = ±√(1 - 2cos^2θ) = ±sin(√(2(1-cos^2θ))) when 1 - 2cos^2θ ≥ 0 For valid geometry: γ = ±√(sin^2θ - cos^2θ) or γ^2 = sin^2(2θ)/2 depending on interpretation Standard result: α = β = cos θ and γ = ±√(1 - 2cos^2θ) (or γ = 0 when 2cos^2θ = 1, i.e., θ = 45°) ∴ Answer: A, B, C (all options typically verify the above relationships with specific angle values)
Correct Answer: A,B,C