Vector Algebra
Vector Triple Product
Grade 12
Question:
<p><strong>69.</strong> Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be three unit vectors, out of which vectors \(\vec{b}\) and \(\vec{c}\) are non-parallel. If \(\alpha\) and \(\beta\) are the angles which vector \(\vec{a}\) makes with vectors \(\vec{b}\) and \(\vec{c}\) respectively and \(\vec{a} \times (\vec{b} \times \vec{c}) = \dfrac{1}{2}\vec{b}\), then \(|\alpha - \beta|\) is equal to ________ °.</p>
Step-by-Step Solution
Key Concept: Use the vector triple product formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$ and compare coefficients since $\vec{b}$ and $\vec{c}$ are non-parallel to establish relationships between the dot products.
Step 1: Apply the vector triple product formula. $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$ Step 2: Use the given condition $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{2}\vec{b}$. $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{2}\vec{b}$ Step 3: Since $\vec{b}$ and $\vec{c}$ are non-parallel (linearly independent), compare coefficients: Coefficient of $\vec{b}$: $\vec{a} \cdot \vec{c} = \frac{1}{2}$ Coefficient of $\vec{c}$: $\vec{a} \cdot \vec{b} = 0$ Step 4: Since all vectors are unit vectors: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\alpha = 1 \cdot 1 \cdot \cos\alpha = 0$ $\therefore \cos\alpha = 0 \implies \alpha = 90°$ Step 5: Similarly: $\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}|\cos\beta = 1 \cdot 1 \cdot \cos\beta = \frac{1}{2}$ $\therefore \cos\beta = \frac{1}{2} \implies \beta = 60°$ Step 6: Calculate the difference: $|\alpha - \beta| = |90° - 60°| = 30°$ ∴ Answer: 30
Correct Answer: 30