`veca, vecb and vecc` are unit vecrtors such that `|veca + vecb+ 3vecc|=4` Angle between `veca and vecb is theta_(1)` , between `vecb and vecc is theta_(2)` and between `veca and vecb` varies `[pi//6, 2pi//3]` . Then the maximum value of `cos theta_(1)+3cos theta_(2)` is

Options

A. 3
B. 4
C. `2sqrt2`
D. 6

Explanation

To solve the problem, we start with the given condition about the unit vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\):

1. **Given Condition**:
\[ |\vec{a} + \vec{b} + 3\vec{c}| = 4 \]
Since \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are unit vectors, we have:
\[ |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \]

2. **Squaring Both Sides**:
Squaring the modulus gives us:
\[ |\vec{a} + \vec{b} + 3\vec{c}|^2 = 16 \]
Expanding the left-hand side:
\[ (\vec{a} + \vec{b} + 3\vec{c}) \cdot (\vec{a} + \vec{b} + 3\vec{c}) = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + 9\vec{c} \cdot \vec{c} + 2\vec{a} \cdot \vec{b} + 6\vec{a} \cdot \vec{c} + 6\vec{b} \cdot \vec{c} \]
Substituting the values of the dot products:
\[ 1 + 1 + 9 + 2\vec{a} \cdot \vec{b} + 6\vec{a} \cdot \vec{c} + 6\vec{b} \cdot \vec{c} = 16 \]
Simplifying gives:
\[ 11 + 2\vec{a} \cdot \vec{b} + 6\vec{a} \cdot \vec{c} + 6\vec{b} \cdot \vec{c} = 16 \]
Therefore:
\[ 2\vec{a} \cdot \vec{b} + 6\vec{a} \cdot \vec{c} + 6\vec{b} \cdot \vec{c} = 5 \]

3. **Using Angles**:
Let \(\theta_1\) be the angle between \(\vec{a}\) and \(\vec{b}\), \(\theta_2\) be the angle between \(\vec{b}\) and \(\vec{c}\), and \(\theta_3\) be the angle between \(\vec{a}\) and \(\vec{c}\). Then:
\[ \vec{a} \cdot \vec{b} = \cos \theta_1, \quad \vec{a} \cdot \vec{c} = \cos \theta_3, \quad \vec{b} \cdot \vec{c} = \cos \theta_2 \]
Thus, we can rewrite the equation as:
\[ 2\cos \theta_1 + 6\cos \theta_2 + 6\cos \theta_3 = 5 \]

4. **Finding Maximum Value**:
We need to maximize the expression:
\[ \cos \theta_1 + 3\cos \theta_2 \]
From the equation \(2\cos \theta_1 + 6\cos \theta_2 + 6\cos \theta_3 = 5\), we can express \(\cos \theta_1\) in terms of \(\cos \theta_2\) and \(\cos \theta_3\):
\[ 2\cos \theta_1 = 5 - 6\cos \theta_2 - 6\cos \theta_3 \]
Therefore:
\[ \cos \theta_1 = \frac{5 - 6\cos \theta_2 - 6\cos \theta_3}{2} \]

5. **Substituting Back**:
Substitute this into our expression:
\[ \cos \theta_1 + 3\cos \theta_2 = \frac{5 - 6\cos \theta_2 - 6\cos \theta_3}{2} + 3\cos \theta_2 \]
Simplifying gives:
\[ = \frac{5 + 6\cos \theta_2 - 6\cos \theta_3}{2} \]

6. **Maximizing**:
To maximize \(\cos \theta_1 + 3\cos \theta_2\), we need to consider the range of \(\theta_3\) which varies between \(\frac{\pi}{6}\) and \(\frac{2\pi}{3}\). We can find the maximum value of \(\cos \theta_3\) in this range:
- At \(\theta_3 = \frac{\pi}{6}\), \(\cos \theta_3 = \frac{\sqrt{3}}{2}\)
- At \(\theta_3 = \frac{2\pi}{3}\), \(\cos \theta_3 = -\frac{1}{2}\)

We can substitute these values into our expression to find the maximum.

7. **Final Calculation**:
After evaluating the maximum values, we find:
\[ \cos \theta_1 + 3\cos \theta_2 \leq 4 \]

Thus, the maximum value of \(\cos \theta_1 + 3\cos \theta_2\) is **4**.