Vector Algebra
Vector triple product
Grade 12
Question:
<p>Given \(\vec{a} \times (\vec{b} \times \vec{c}) = \dfrac{1}{2}\vec{b}\), where \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are unit vectors. If \(\alpha\) is the angle between \(\vec{a}\) and \(\vec{c}\) and \(\beta\) is the angle between \(\vec{a}\) and \(\vec{b}\), find \(|\alpha - \beta|\).</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 equate coefficients since $\vec{b}$ and \vec{c}$ are linearly independent unit vectors.
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: Since this equals $\frac{1}{2}\vec{b}$, we have: $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{2}\vec{b}$ Step 3: Since $\vec{b}$ and $\vec{c}$ are unit vectors and linearly independent, equate 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: Convert dot products to angles: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\beta = 1 \cdot 1 \cdot \cos\beta = 0$ ∴ $\cos\beta = 0$ → $\beta = 90°$ $\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}|\cos\alpha = 1 \cdot 1 \cdot \cos\alpha = \frac{1}{2}$ ∴ $\cos\alpha = \frac{1}{2}$ → $\alpha = 60°$ Step 5: Calculate the angle difference: $|\alpha - \beta| = |60° - 90°| = 30°$ ∴ Answer: 30
Correct Answer: 30