3D Geometry
Angle between lines
Grade 12
Question:
<p>\(L_1\) and \(L_2\) are two lines whose vector equations are<br>\(L_1 : \vec{r} = \lambda\left[\left(\cos\theta + \sqrt{3}\right)\hat{i} + \left(\sqrt{2}\sin\theta\right)\hat{j} + \left(\cos\theta - \sqrt{3}\right)\hat{k}\right]\)<br>\(L_2 : \vec{r} = \mu\left(a\hat{i} + b\hat{j} + c\hat{k}\right)\), where \(\lambda\) and \(\mu\) are scalars and \(\alpha\) is the acute angle between \(L_1\) and \(L_2\). If the angle \(\alpha\) is independent of \(\theta\), then the value of \(\alpha\) is</p>
<p>\(\dfrac{\pi}{6}\)</p>
<p>\(\dfrac{\pi}{4}\)</p>
<p>\(\dfrac{\pi}{3}\)</p>
<p>\(\dfrac{\pi}{2}\)</p>
Step-by-Step Solution
Key Concept: For the angle between two lines to be independent of a parameter θ, the direction vector of L₁ must have constant magnitude and a specific relationship with L₂'s direction vector. We use the angle formula cos α = |d₁·d₂|/(|d₁||d₂|) and require this to be independent of θ.
Step 1: Identify the direction vector of L_1 The direction vector of L_1 is: d_1 = (cos θ + √3)î + (√2 sin θ)ĵ + (cos θ - √3)k̂ Step 2: Calculate |d_1|^2 |d_1|^2 = (cos θ + √3)^2 + 2sin^2θ + (cos θ - √3)^2 Expanding: = cos^2θ + 2√3 cos θ + 3 + 2sin^2θ + cos^2θ - 2√3 cos θ + 3 = 2cos^2θ + 2sin^2θ + 6 = 2(cos^2θ + sin^2θ) + 6 = 2(1) + 6 = 8 Therefore: |d_1| = 2√2 (constant, independent of θ) Step 3: Express the angle formula For the angle α between L_1 and L_2: cos α = |d_1 · d_2|/(|d_1||d_2|) = |d_1 · d_2|/(2√2|d_2|) Step 4: Condition for independence from θ For α to be independent of θ, the expression |d_1 · d_2| must be independent of θ. Since |d_1| is already constant, we need d_1 · d_2 to be constant (independent of θ). Given that d_2 = (a, b, c) is fixed, and d_1 contains θ terms, the only way their dot product is constant is if the coefficients of the varying terms (cos θ and sin θ) in d_1 · d_2 sum to zero. Step 5: Analyze the constraint d_1 · d_2 = a(cos θ + √3) + b(√2 sin θ) + c(cos θ - √3) = (a + c)cos θ + √2 b sin θ + √3(a - c) For this to be independent of θ: a + c = 0 and √2 b = 0 Therefore: c = -a and b = 0 Step 6: Calculate d_1 · d_2 under the constraint d_1 · d_2 = √3(a - c) = √3(a - (-a)) = 2√3 a |d_1 · d_2| = 2√3 |a| Step 7: Determine |d_2| |d_2| = √(a^2 + 0^2 + c^2) = √(a^2 + a^2) = √(2a^2) = √2 |a| Step 8: Calculate cos α cos α = |2√3 a|/(2√2 · √2|a|) = 2√3|a|/(2 · 2|a|) = 2√3/4 = √3/2 Therefore: α = arccos(√3/2) = π/3 ∴ Answer: C
Correct Answer: C