Given vectors \( \mathbf{u}, \mathbf{v}, \) and \( \mathbf{w} \) satisfy \( \mathbf{u} + \mathbf{v} + \mathbf{w} = 0 \). This implies \( \mathbf{w} = -(\mathbf{u} + \mathbf{v}) \). The vectors \( \mathbf{u} \) and \( \mathbf{v} \) are unit vectors, so \( |\mathbf{u}| = 1 \) and \( |\mathbf{v}| = 1 \). Additionally, \( |\mathbf{w}| = \sqrt{3} \).Step 1: Calculate the magnitude of \( \mathbf{w} \). The magnitude of \( \mathbf{w} \) is \( |\mathbf{w}| = |-(\mathbf{u} + \mathbf{v})| = |\mathbf{u} + \mathbf{v}| \). Squaring both sides gives \( |\mathbf{w}|^2 = |\mathbf{u} + \mathbf{v}|^2 \). Substituting known values: \( 3 = |\mathbf{u}|^2 + |\mathbf{v}|^2 + 2 \mathbf{u} \cdot \mathbf{v} \). Since \( |\mathbf{u}|^2 = 1 \) and \( |\mathbf{v}|^2 = 1 \), we have \( 3 = 1 + 1 + 2 \mathbf{u} \cdot \mathbf{v} \), which simplifies to \( 2 \mathbf{u} \cdot \mathbf{v} = 1 \), so \( \mathbf{u} \cdot \mathbf{v} = \frac{1}{2} \).Step 2: Determine the angle between \( \mathbf{v} \) and \( \mathbf{w} \). Let \( \theta \) be the angle between \( \mathbf{v} \) and \( \mathbf{w} \). Using the dot product formula, \( \mathbf{v} \cdot \mathbf{w} = |\mathbf{v}| |\mathbf{w}| \cos \theta \). Substituting known values: \( \mathbf{v} \cdot \mathbf{w} = 1 \times \sqrt{3} \times \cos \theta = \sqrt{3} \cos \theta \). Substituting \( \mathbf{w} = -(\mathbf{u} + \mathbf{v}) \): \( \mathbf{v} \cdot \mathbf{w} = \mathbf{v} \cdot (-(\mathbf{u} + \mathbf{v})) = - (\mathbf{v} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v}) = - \left( \frac{1}{2} + 1 \right) = - \frac{3}{2} \). Equating the two expressions for \( \mathbf{v} \cdot \mathbf{w} \): \( \sqrt{3} \cos \theta = - \frac{3}{2} \), which yields \( \cos \theta = - \frac{1}{2} \). Therefore, \( \theta = 120^\circ \). The angle between \( \mathbf{v} \) and \( \mathbf{w} \) is \( 120^\circ \).