If `bara and barb` are two unit vectors at `120^@` and c is any vector inclined to `bara and barb` at an angle `theta`. The complete set of values of `cos theta ` is

Options

A. `[-1/2,1/2]`
B. `[0,1/2]`
C. `[1/2,sqrt3/2]`
D. `[-sqrt3/2,sqrt3/2]`

Answer

To solve the problem, we need to find the complete set of values of \( \cos \theta \) where \( \theta \) is the angle between the vector \( C \) and the unit vectors \( \mathbf{a} \) and \( \mathbf{b} \) which are at an angle of \( 120^\circ \) to each other.

### Step 1: Understand the Geometry of the Problem
Given two unit vectors \( \mathbf{a} \) and \( \mathbf{b} \) that are at an angle of \( 120^\circ \), we can represent them in a coordinate system.

Let:
\[ \mathbf{a} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \]
\[ \mathbf{b} = \begin{pmatrix} \cos 120^\circ \\ \sin 120^\circ \end{pmatrix} = \begin{pmatrix} -\frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{pmatrix} \]

### Step 2: Find the Angle Between Vector \( C \) and the Unit Vectors
The vector \( C \) can be expressed in terms of its angle \( \theta \) with respect to \( \mathbf{a} \) and \( \mathbf{b} \). The cosine of the angle \( \theta \) can be expressed using the dot product formula:
\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{C}}{|\mathbf{a}| |\mathbf{C}|} \]
\[ \cos \theta = \mathbf{a} \cdot \mathbf{C} \]
since \( |\mathbf{a}| = 1 \).

Similarly, for \( \mathbf{b} \):
\[ \cos \theta' = \frac{\mathbf{b} \cdot \mathbf{C}}{|\mathbf{b}| |\mathbf{C}|} \]
\[ \cos \theta' = \mathbf{b} \cdot \mathbf{C} \]
since \( |\mathbf{b}| = 1 \).

### Step 3: Set Up the Cosine Relationship
Since \( \mathbf{a} \) and \( \mathbf{b} \) are at an angle of \( 120^\circ \), we know:
\[ \cos(120^\circ) = -\frac{1}{2} \]
This means that the relationship between the angles can be expressed as:
\[ \cos \theta + \cos \theta' = -\frac{1}{2} \]

### Step 4: Solve for \( \cos \theta \)
Let \( x = \cos \theta \) and \( y = \cos \theta' \). From the previous step, we have:
\[ x + y = -\frac{1}{2} \]

### Step 5: Use the Cauchy-Schwarz Inequality
Using the Cauchy-Schwarz inequality, we know that:
\[ (x^2 + y^2)(1 + 1) \geq (x + y)^2 \]
This simplifies to:
\[ 2(x^2 + y^2) \geq \left(-\frac{1}{2}\right)^2 \]
\[ 2(x^2 + y^2) \geq \frac{1}{4} \]
\[ x^2 + y^2 \geq \frac{1}{8} \]

### Step 6: Substitute for \( y \)
Substituting \( y = -\frac{1}{2} - x \) into the inequality:
\[ x^2 + \left(-\frac{1}{2} - x\right)^2 \geq \frac{1}{8} \]
Expanding this gives:
\[ x^2 + \left(\frac{1}{4} + x + x^2\right) \geq \frac{1}{8} \]
\[ 2x^2 + x + \frac{1}{4} \geq \frac{1}{8} \]
\[ 2x^2 + x + \frac{1}{4} - \frac{1}{8} \geq 0 \]
\[ 2x^2 + x + \frac{1}{8} \geq 0 \]

### Step 7: Solve the Quadratic Inequality
Now we can solve the quadratic inequality \( 2x^2 + x + \frac{1}{8} \geq 0 \). The discriminant \( D \) of this quadratic is:
\[ D = b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot \frac{1}{8} = 1 - 1 = 0 \]
Since the discriminant is zero, the quadratic has a double root:
\[ x = \frac{-b}{2a} = \frac{-1}{2 \cdot 2} = -\frac{1}{4} \]

### Step 8: Determine the Range of \( \cos \theta \)
Since the quadratic opens upwards (as the coefficient of \( x^2 \) is positive), the inequality \( 2x^2 + x + \frac{1}{8} \geq 0 \) holds for all \( x \) except at the double root \( x = -\frac{1}{4} \). Thus, the complete set of values for \( \cos \theta \) is:
\[ \cos \theta \leq -\frac{1}{4} \quad \text{or} \quad \cos \theta \geq -\frac{1}{4} \]

### Final Answer
The complete set of values of \( \cos \theta \) is:
\[ \cos \theta \in \left(-1, -\frac{1}{4}\right) \cup \left(-\frac{1}{4}, 1\right) \]