3D Geometry
Direction Cosines
Grade 12
Question:
<p>If \(\cos^2\theta + \cos^2\theta + \cos^2\alpha = 1\) (sum of squares of direction cosines = 1), find the range of \(\theta\) given that \(0 \leq 2\cos^2\theta \leq \frac{1}{2}\).</p>
<p>\(\theta \in \left[0, \frac{\pi}{4}\right]\)</p>
<p>\(\theta \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right]\)</p>
<p>\(\theta \in \left[0, \frac{\pi}{2}\right]\)</p>
<p>\(\theta \in \left[\frac{\pi}{6}, \frac{\pi}{3}\right]\)</p>
Step-by-Step Solution
Key Concept: Direction cosines must satisfy l² + m² + n² = 1. Here, with two equal direction cosines (cos²θ), the constraint 2cos²θ + cos²α = 1 determines α in terms of θ, and the given inequality directly restricts θ's range.
Step 1: Apply the direction cosine property: cos^2θ + cos^2θ + cos^2α = 1, which gives 2cos^2θ + cos^2α = 1. Step 2: From the given constraint 0 ≤ 2cos^2θ ≤ 1/2, we have 0 ≤ cos^2θ ≤ 1/4. Step 3: Taking square roots: 0 ≤ |cosθ| ≤ 1/2, which means -1/2 ≤ cosθ ≤ 1/2. Step 4: For cosθ ∈ [-1/2, 1/2], the angle θ must satisfy: θ ∈ [π/3, 2π/3] (considering θ ∈ [0, π]). Step 5: Verify: At θ = π/3: cos^2(π/3) = 1/4, so 2cos^2θ = 1/2 ✓. At θ = π/2: cos^2(π/2) = 0, so 2cos^2θ = 0 ✓. ∴ Answer: B (θ ∈ [π/3, 2π/3])
Correct Answer: B