Question

Let \( \vec{a}, \vec{b}, \vec{c} \) be vectors of equal magnitude such that the angle between \( \vec{a} \) and \( \vec{b} \) is \( \alpha \), the angle between \( \vec{b} \) and \( \vec{c} \) is \( \beta \), and the angle between \( \vec{c} \) and \( \vec{a} \) is \( \gamma \). Then the minimum value of \( \cos \alpha + \cos \beta + \cos \gamma \) is:

• A. \( \frac{1}{2} \)
• B. \( \frac{3}{2} \)
• C. \( \frac{1}{3} \)
• D. \( -\frac{3}{2} \)

Hint:

For problems involving minimizing the sum of cosines of angles between vectors, consider using symmetry, especially when vectors are of equal magnitude. The minimum value is often achieved when the vectors are symmetrically arranged.

Answer 1

The Correct Option is D

Given three vectors \( \vec{a}, \vec{b}, \vec{c } \) of equal magnitude, with angles \( \alpha, \beta, \gamma \) between them, find the minimum value of: \[\n\cos \alpha + \cos \beta + \cos \gamma\n\]
Solution:
The sum \( \cos \alpha + \cos \beta + \cos \gamma \) is minimized when the vectors are symmetrically arranged. The simplest symmetrical configuration is when \( \alpha = \beta = \gamma = 120^\circ \).
Step 1: Angle calculation
For \( \alpha = \beta = \gamma = 120^\circ \): \[\n\cos 120^\circ = -\frac{1}{2}\n\]
Step 2: Sum Calculation
\[\n\cos \alpha + \cos \beta + \cos \gamma = 3 \times \left( -\frac{1}{2} \right) = -\frac{3}{2}\n\]
Step 3: Justification
The most symmetrical arrangement with \( 120^\circ \) angles minimizes the sum of cosines.
Step 4: Answer
The minimum value is: \[\n\boxed{-\frac{3}{2}}\n\]
Final Answer: \( \boxed{(D) -\frac{3}{2}} \)