Let `veca,vecb` and `vecc` are three vectors such that `|veca|=3, |vecb|=3, |vecc|=2, |veca+vecb+vecc|=4` and `veca` is perpendicular to `vecb,vecc` makes angle `theta` and `phi` with `veca` and `vecb` respectively, then `cos theta+cos theta=`

Options

A. `(3)/(4)`
B. `-(3)/(4)`
C. `(1)/(2)`
D. `-(1)/(2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Set up the equation for the magnitudes
We know that the magnitude of the sum of the vectors is given by:
\[ |\vec{a} + \vec{b} + \vec{c}| = 4 \]
Squaring both sides gives:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 16 \]

### Step 2: Expand the left-hand side
Using the property of dot products, we can expand the left-hand side:
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 16 \]

### Step 3: Substitute the known magnitudes
Given that \(|\vec{a}| = 3\), \(|\vec{b}| = 3\), and \(|\vec{c}| = 2\), we can substitute these values:
\[ 3^2 + 3^2 + 2^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 16 \]
This simplifies to:
\[ 9 + 9 + 4 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 16 \]
\[ 22 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 16 \]

### Step 4: Rearranging the equation
Now, we can rearrange the equation to isolate the dot products:
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 16 - 22 \]
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -6 \]
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -3 \]

### Step 5: Calculate the dot products
Since \(\vec{a}\) is perpendicular to \(\vec{b}\), we have:
\[ \vec{a} \cdot \vec{b} = 0 \]
Thus, we can simplify the equation:
\[ 0 + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -3 \]
\[ \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -3 \]

### Step 6: Express the dot products in terms of angles
Using the definitions of the dot product:
\[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \phi = 3 \cdot 2 \cos \phi = 6 \cos \phi \]
\[ \vec{c} \cdot \vec{a} = |\vec{c}| |\vec{a}| \cos \theta = 2 \cdot 3 \cos \theta = 6 \cos \theta \]
Substituting these into our equation gives:
\[ 6 \cos \phi + 6 \cos \theta = -3 \]
Dividing through by 6:
\[ \cos \phi + \cos \theta = -\frac{1}{2} \]

### Final Result
Thus, the required value is:
\[ \cos \theta + \cos \phi = -\frac{1}{2} \]