Let ` veca , vecb,vecc` be vectors of equal magnitude such that the angle between ` veca and vecb " is " alpha , vecb and vecc " is " beta and vecc and veca " is " gamma` .then minimum value of ` cos alpha + cos beta + cos gamma` is

Options

A. ` 1/2`
B. ` - 1/2`
C. ` 3/2`
D. ` -3/2`

Explanation

To find the minimum value of \( \cos \alpha + \cos \beta + \cos \gamma \), where \( \vec{a}, \vec{b}, \vec{c} \) are vectors of equal magnitude and the angles between them are \( \alpha, \beta, \gamma \) respectively, we can follow these steps:

### Step 1: Define the Magnitude of Vectors
Let the magnitude of each vector \( \vec{a}, \vec{b}, \vec{c} \) be \( d \). Thus, we have:
\[ |\vec{a}| = |\vec{b}| = |\vec{c}| = d \]

### Step 2: Express Dot Products
Using the definition of the dot product, we can express the dot products in terms of the angles:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \alpha = d^2 \cos \alpha \]
\[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \beta = d^2 \cos \beta \]
\[ \vec{c} \cdot \vec{a} = |\vec{c}| |\vec{a}| \cos \gamma = d^2 \cos \gamma \]

### Step 3: Apply the Cauchy-Schwarz Inequality
The Cauchy-Schwarz inequality states that:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 \geq 0 \]
Expanding this, we have:
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq 0 \]
Substituting the magnitudes and dot products:
\[ 3d^2 + 2(d^2 \cos \alpha + d^2 \cos \beta + d^2 \cos \gamma) \geq 0 \]

### Step 4: Simplify the Inequality
Dividing the entire inequality by \( d^2 \) (since \( d \neq 0 \)):
\[ 3 + 2(\cos \alpha + \cos \beta + \cos \gamma) \geq 0 \]
Rearranging gives:
\[ \cos \alpha + \cos \beta + \cos \gamma \geq -\frac{3}{2} \]

### Step 5: Conclusion
Thus, the minimum value of \( \cos \alpha + \cos \beta + \cos \gamma \) is:
\[ \boxed{-\frac{3}{2}} \]