Question

If vectors \( \vec{a}, \vec{b} \) and \( 2\vec{a} + 3\vec{b} \) are unit vectors, then find the angle between \( \vec{a} \) and \( \vec{b} \).

Hint:

For problems involving unit vectors and dot products, remember that the dot product gives \( \cos \theta \). When working with combined vectors, always expand the magnitude using the dot product formula and equate terms systematically.

Answer 1

Given that \( \vec{a} \) and \( \vec{b} \) are unit vectors: \[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1. \] It is also given that \( 2\vec{a} + 3\vec{b} \) is a unit vector, which implies: \[ |2\vec{a} + 3\vec{b}| = 1. \] Using the formula for the magnitude of a vector: \[ |2\vec{a} + 3\vec{b}|^2 = (2\vec{a} + 3\vec{b}) \cdot (2\vec{a} + 3\vec{b}). \] Expanding the dot product: \[ |2\vec{a} + 3\vec{b}|^2 = 4|\vec{a}|^2 + 12 \vec{a} \cdot \vec{b} + 9|\vec{b}|^2. \] Substituting \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \): \[ |2\vec{a} + 3\vec{b}|^2 = 4 + 12 \vec{a} \cdot \vec{b} + 9 = 13 + 12 \vec{a} \cdot \vec{b}. \] Since \( |2\vec{a} + 3\vec{b}| = 1 \), we equate: \[ 13 + 12 \vec{a} \cdot \vec{b} = 1. \] Solving for \( \vec{a} \cdot \vec{b} \): \[ 12 \vec{a} \cdot \vec{b} = -12 \quad \Rightarrow \quad \vec{a} \cdot \vec{b} = -1. \] The dot product \( \vec{a} \cdot \vec{b} \) is related to the angle \( \theta \) between \( \vec{a} \) and \( \vec{b} \) as: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = \cos \theta. \] Substituting \( \vec{a} \cdot \vec{b} = -1 \): \[ \cos \theta = -1 \quad \Rightarrow \quad \theta = \pi. \] Final Answer: The angle between \( \vec{a} \) and \( \vec{b} \) is: \[ \boxed{\theta = \pi \text{ radians.}} \]