Vector Algebra
Angle between lines
Grade 12
Question:
<p>Both the lines pass through origin. Line <em>L</em><sub>1</sub> is parallel to the vector <br>
\(\vec{V}_1 = (\cos\theta + \sqrt{3})\,\hat{i} + (\sqrt{2}\sin\theta)\,\hat{j} + (\cos\theta - \sqrt{3})\,\hat{k}\)<br>
and <em>L</em><sub>2</sub> is parallel to the vector <br>
\(\vec{V}_2 = a\hat{i} + b\hat{j} + c\hat{k}\)<br>
If the angle \(\alpha\) between the lines is independent of \(\theta\), find \(\alpha\).</p>
<p>\(\dfrac{\pi}{3}\)</p>
<p>\(\dfrac{\pi}{6}\)</p>
<p>\(\dfrac{\pi}{4}\)</p>
<p>\(\dfrac{\pi}{2}\)</p>
Step-by-Step Solution
Key Concept: For the angle between two lines to be independent of θ, the direction vectors must satisfy a constraint that eliminates θ from the dot product formula. This occurs when V₁ has constant magnitude and V₂ is orthogonal to the derivative of V₁ with respect to θ.
Step 1: Find |V_1|^2 |V_1|^2 = (cos θ + √3)^2 + (√2 sin θ)^2 + (cos θ - √3)^2 = cos^2θ + 2√3 cos θ + 3 + 2sin^2θ + cos^2θ - 2√3 cos θ + 3 = 2cos^2θ + 2sin^2θ + 6 = 2 + 6 = 8 So |V_1| = 2√2 (constant, independent of θ) ✓ Step 2: For angle to be independent of θ, find dV_1/dθ dV_1/dθ = (-sin θ)î + (√2 cos θ)ĵ + (-sin θ)k̂ Step 3: Apply orthogonality condition For cos α = (V_1·V_2)/(|V_1||V_2|) to be independent of θ, V_2 must be perpendicular to dV_1/dθ: V_2 · (dV_1/dθ) = 0 -a sin θ + √2 b cos θ - c sin θ = 0 -(a+c) sin θ + √2 b cos θ = 0 This holds for all θ only if: a + c = 0 and b = 0 Step 4: Calculate the angle With V_2 = a î - a k̂ (let a = 1): V_2 = î - k̂, so |V_2| = √2 V_1 · V_2 = (cos θ + √3)(1) + (√2 sin θ)(0) + (cos θ - √3)(-1) = cos θ + √3 - cos θ + √3 = 2√3 cos α = (2√3)/(2√2 · √2) = 2√3/4 = √3/2 ∴ α = 30° or π/6 Answer: B
Correct Answer: B