Question

If vectors $ \mathbf{u}, \mathbf{v}, $ and $ \mathbf{w} $ satisfy $ \mathbf{u} + \mathbf{v} + \mathbf{w} = 0 $, and $ \mathbf{u} $ and $ \mathbf{v} $ are unit vectors, while $ |\mathbf{w}| = \sqrt{3} $, then the angle between $ \mathbf{v} $ and $ \mathbf{w} $ is:

• A. \( 90^\circ \)
• B. \( 60^\circ \)
• C. \( 120^\circ \)
• D. \( 45^\circ \)

Hint:

When vectors satisfy conditions like \( \mathbf{u} + \mathbf{v} + \mathbf{w} = 0 \), use the relationship between the vectors to express one vector in terms of the others and then calculate the required angle using the dot product formula.

Answer 1

The Correct Option is C

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