<p>It is given that \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are three unit vectors such that<br/>
\[\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2}(\vec{b}+\vec{c})\]<br/>
Using this condition, find the angle between \(\vec{a}\) and \(\vec{b}\) (or relevant quantity as per full question).</p>
Step-by-Step Solution
Key Concept: Use the vector triple product formula: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$. Since $\vec{a}$, $\vec{b}$, $\vec{c}$ are unit vectors with specific constraints, compare coefficients to establish relationships between 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: Given that this equals $\frac{\sqrt{3}}{2}(\vec{b}+\vec{c})$: $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{\sqrt{3}}{2}\vec{b} + \frac{\sqrt{3}}{2}\vec{c}$ Step 3: Since $\vec{b}$ and $\vec{c}$ are linearly independent unit vectors, compare coefficients: Coefficient of $\vec{b}$: $\vec{a} \cdot \vec{c} = \frac{\sqrt{3}}{2}$ Coefficient of $\vec{c}$: $-(\vec{a} \cdot \vec{b}) = \frac{\sqrt{3}}{2}$, so $\vec{a} \cdot \vec{b} = -\frac{\sqrt{3}}{2}$ Step 4: Since $\vec{a}$ and $\vec{b}$ are unit vectors: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = \cos\theta = -\frac{\sqrt{3}}{2}$ Step 5: Therefore $\theta = 150°$ or $\frac{5\pi}{6}$ radians. ∴ Answer: A
Correct Answer: A