Question

Let \(\hat{u}\) and \(\hat{v}\) be unit vectors inclined at an acute angle such that \(|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2}\). If \(\vec{A} = \lambda \hat{u} + \hat{v} + (\hat{u} \times \hat{v})\), then \(\lambda\) is equal to:

• A. \(\frac{4}{3}(\vec{A} \cdot \hat{u}) - \frac{2}{3}(\vec{A} \cdot \hat{v})\)
• B. \(\frac{2}{3}(\vec{A} \cdot \hat{u}) - \frac{1}{3}(\vec{A} \cdot \hat{v})\)
• C. \(\frac{4}{3}(\vec{A} \cdot \hat{u}) + \frac{2}{3}(\vec{A} \cdot \hat{v})\)
• D. \((\vec{A} \cdot \hat{u}) - \frac{1}{2}(\vec{A} \cdot \hat{v})\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
By taking the dot products of the given equation \(\vec{A}\) with the basis vectors \(\hat{u}\) and \(\hat{v}\) independently, we create two algebraic equations. Solving these equations reveals \(\lambda\) as a linear combination of those dot products.
Step 2: Key Formula or Approach:
Dot product expansion:
\(\hat{u} \cdot \hat{u} = 1\), \(\hat{v} \cdot \hat{v} = 1\).
\(|\hat{u} \times \hat{v}| = |\hat{u}| |\hat{v}| \sin \theta \).
\(\hat{u} \cdot \hat{v} = |\hat{u}| |\hat{v}| \cos \theta \).
Step 3: Detailed Explanation:
Given \(|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2} \implies \sin \theta = \frac{\sqrt{3}}{2}\).
Since the angle is acute, \(\theta = 60^\circ\). Therefore, \(\hat{u} \cdot \hat{v} = \cos 60^\circ = \frac{1}{2}\).
We have \(\vec{A} = \lambda\hat{u} + \hat{v} + (\hat{u} \times \hat{v})\).
Dot product with \(\hat{u}\):
\[ \vec{A} \cdot \hat{u} = \lambda(\hat{u} \cdot \hat{u}) + (\hat{v} \cdot \hat{u}) + (\hat{u} \times \hat{v}) \cdot \hat{u} \] Since \(\hat{u} \times \hat{v}\) is orthogonal to \(\hat{u}\), the third term is zero.
\[ \vec{A} \cdot \hat{u} = \lambda(1) + \frac{1}{2} = \lambda + \frac{1}{2} \quad \text{(Eq 1)} \] Dot product with \(\hat{v}\):
\[ \vec{A} \cdot \hat{v} = \lambda(\hat{u} \cdot \hat{v}) + (\hat{v} \cdot \hat{v}) + (\hat{u} \times \hat{v}) \cdot \hat{v} = \frac{\lambda}{2} + 1 \quad \text{(Eq 2)} \] We need \(\lambda\). From Eq 1, \(\lambda = \vec{A} \cdot \hat{u} - \frac{1}{2}\).
Test the options using these identities.
Option (A):
\[ \frac{4}{3}(\vec{A} \cdot \hat{u}) - \frac{2}{3}(\vec{A} \cdot \hat{v}) = \frac{4}{3}\left(\lambda + \frac{1}{2}\right) - \frac{2}{3}\left(\frac{\lambda}{2} + 1\right) \] \[ = \frac{4\lambda}{3} + \frac{2}{3} - \frac{\lambda}{3} - \frac{2}{3} = \frac{3\lambda}{3} = \lambda \] This perfectly matches \(\lambda\).
Step 4: Final Answer:
Option (A) is the correct expression.