Question

(c) If the unit vectors \( \vec{a}, \vec{b}, \vec{c} \) are such that \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \), find the value of \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \):

Hint:

Use unit vector properties and dot product identities for simplifications.

Answer 1

Given \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \), we have: \[ \vec{c} = -(\vec{a} + \vec{b}). \] Now calculate the dot products: \[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = \vec{a} \cdot \vec{b} + \vec{b} \cdot (-\vec{a} - \vec{b}) + (-\vec{a} - \vec{b}) \cdot \vec{a}. \] Simplify: \[ \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} - \vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{a} = -(\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b}). \] Since \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors: \[ \vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = 1, \quad \text{so } \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -2. \]