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

We are given that the vectors \( \mathbf{u}, \mathbf{v}, \) and \( \mathbf{w} \) satisfy the equation: \[ \mathbf{u} + \mathbf{v} + \mathbf{w} = 0 \] From this, we can express \( \mathbf{w} \) as: \[ \mathbf{w} = -(\mathbf{u} + \mathbf{v}) \] Next, we are given that \( \mathbf{u} \) and \( \mathbf{v} \) are unit vectors, so: \[ |\mathbf{u}| = 1 \quad \text{and} \quad |\mathbf{v}| = 1 \] Also, we are given that \( |\mathbf{w}| = \sqrt{3} \).
Step 1: Find the magnitude of \( \mathbf{w} \) The magnitude of \( \mathbf{w} \) is: \[ |\mathbf{w}| = |-(\mathbf{u} + \mathbf{v})| = |\mathbf{u} + \mathbf{v}| \] Now, square both sides: \[ |\mathbf{w}|^2 = |\mathbf{u} + \mathbf{v}|^2 \] \[ 3 = |\mathbf{u}|^2 + |\mathbf{v}|^2 + 2 \mathbf{u} \cdot \mathbf{v} \] Since \( |\mathbf{u}|^2 = 1 \) and \( |\mathbf{v}|^2 = 1 \), we get: \[ 3 = 1 + 1 + 2 \mathbf{u} \cdot \mathbf{v} \] \[ 3 = 2 + 2 \mathbf{u} \cdot \mathbf{v} \] \[ 2 \mathbf{u} \cdot \mathbf{v} = 1 \] \[ \mathbf{u} \cdot \mathbf{v} = \frac{1}{2} \] Step 2: Find the angle between \( \mathbf{v} \) and \( \mathbf{w} \) We need to find the angle between \( \mathbf{v} \) and \( \mathbf{w} \). Using the formula for the dot product: \[ \mathbf{v} \cdot \mathbf{w} = |\mathbf{v}| |\mathbf{w}| \cos \theta \] Substitute the known values: \[ \mathbf{v} \cdot \mathbf{w} = 1 \times \sqrt{3} \times \cos \theta = \sqrt{3} \cos \theta \] Now, substitute \( \mathbf{w} = -(\mathbf{u} + \mathbf{v}) \) into the dot product: \[ \mathbf{v} \cdot \mathbf{w} = \mathbf{v} \cdot (-(\mathbf{u} + \mathbf{v})) = - (\mathbf{v} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v}) \] \[ \mathbf{v} \cdot \mathbf{w} = - \left( \frac{1}{2} + 1 \right) = - \frac{3}{2} \] Equating the two expressions for \( \mathbf{v} \cdot \mathbf{w} \): \[ \sqrt{3} \cos \theta = - \frac{3}{2} \] \[ \cos \theta = - \frac{1}{2} \] Thus, the angle \( \theta \) is: \[ \theta = 120^\circ \] Thus, the angle between \( \mathbf{v} \) and \( \mathbf{w} \) is \( 120^\circ \).