Vector Algebra
Vector triple product
Grade 12
Question:
<p>Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be three unit vectors such that \(\vec{a}\times(\vec{b}\times\vec{c})=\dfrac{\sqrt{3}}{2}(\vec{b}+\vec{c})\). If \(\vec{b}\) is not parallel to \(\vec{c}\), then the angle between \(\vec{a}\) and \(\vec{b}\) is</p>
<p>\(\dfrac{5\pi}{6}\)</p>
<p>\(\dfrac{3\pi}{4}\)</p>
<p>\(\dfrac{\pi}{2}\)</p>
<p>\(\dfrac{2\pi}{3}\)</p>
Step-by-Step Solution
Key Concept: Use the vector triple product formula to expand the left side, then compare coefficients with the right side. Since all vectors are unit vectors, the dot products are bounded by [-1, 1], which constrains the possible angles.
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: Set this equal to the given expression: $(\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c} = \frac{\sqrt{3}}{2}(\vec{b}+\vec{c})$ Step 3: Since $\vec{b}$ and $\vec{c}$ are not parallel (linearly independent), compare coefficients: Coefficient of $\vec{b}$: $\vec{a}\cdot\vec{c} = \frac{\sqrt{3}}{2}$ Coefficient of $\vec{c}$: $-\vec{a}\cdot\vec{b} = \frac{\sqrt{3}}{2}$ Step 4: From Step 3: $\vec{a}\cdot\vec{b} = -\frac{\sqrt{3}}{2}$ Step 5: Since $\vec{a}$ and $\vec{b}$ are unit vectors: $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta = \cos\theta$ Step 6: Therefore: $\cos\theta = -\frac{\sqrt{3}}{2}$, which gives $\theta = 150°$ or $\frac{5\pi}{6}$ ∴ Answer: A
Correct Answer: A