Question

Let \( \mathbf{a} \) and \( \mathbf{b} \) be two unit vectors such that the angle between them is \( \frac{\pi}{3} \). If \( \lambda \mathbf{a} + 2 \mathbf{b} \) and \( 3 \mathbf{a} - \lambda \mathbf{b} \) are perpendicular to each other, then the number of values of \( \lambda \) in \( [-1, 3] \) is:

• A. \( 2 \)
• B. \( 0 \)
• C. \( 3 \)
• D. \( 1 \)

Hint:

When solving for the number of values of \( \lambda \) in vector problems: - Focus on vector dot products to find perpendicularity conditions (i.e., dot product equals zero). - Set up equations using the given conditions (like the angle between vectors and perpendicularity), and solve the resulting quadratic.

Answer 1

The Correct Option is B

Since \( \mathbf{a} \) and \( \mathbf{b} \) are unit vectors and the angle between them is \( \frac{\pi}{3} \), we know that: \[ \mathbf{a} \cdot \mathbf{b} = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}. \] For the vectors \( \lambda \mathbf{a} + 2 \mathbf{b} \) and \( 3 \mathbf{a} - \lambda \mathbf{b} \) to be perpendicular, their dot product must be zero: \[ (\lambda \mathbf{a} + 2 \mathbf{b}) \cdot (3 \mathbf{a} - \lambda \mathbf{b}) = 0. \] Expanding this: \[ \lambda \cdot 3 + 2 \cdot (-\lambda) \cdot \frac{1}{2} = 0. \] Solving this gives \( \lambda = 0 \). Therefore, there is only 1 value of \( \lambda \).