Question

Let \( \vec{a}, \vec{b}, \vec{c} \) be vectors of equal magnitude such that the angle between \( \vec{a} \) and \( \vec{b} \) is \( \alpha \), between \( \vec{b} \) and \( \vec{c} \) is \( \beta \), and 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. \( -\frac{1}{2} \)
• C. \( \frac{3}{2} \)
• D. \( -\frac{3}{2} \)

Hint:

For vector angle sum problems: \begin{itemize} \item Use \( |\vec{a}+\vec{b}+\vec{c}|^2 \ge 0 \). \item Convert dot products to cosines. \item Equality occurs when vector sum is zero. \end{itemize}

Answer 1

The Correct Option is D

Concept: Use identity: \[ |\vec{a}+\vec{b}+\vec{c}|^2 \ge 0 \] Expand using dot products: \[ = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \] Step 1: {\color{red}Use equal magnitudes.} Let each magnitude = 1 (scaling does not affect cosines). Then: \[ |\vec{a}+\vec{b}+\vec{c}|^2 = 3 + 2(\cos\alpha + \cos\beta + \cos\gamma) \] Since square ≥ 0: \[ 3 + 2S \ge 0 \] where \( S = \cos\alpha + \cos\beta + \cos\gamma \). Step 2: {\color{red}Find minimum.} \[ S \ge -\frac{3}{2} \] Step 3: {\color{red}Achieving equality.} Minimum occurs when: \[ \vec{a} + \vec{b} + \vec{c} = 0 \] which is possible for three equal vectors at \( 120^\circ \).