`vec(a)+vec(b)+vec(c)=vec(0)` such that `|vec(a)|=3, |vec(b)|=5 and |vec(c)|=7`. What is cosine of the angle between `vec(b) and vec(c)` ?

Options

A. `11//12`
B. `13//14`
C. `-11//12`
D. `-13//14`

Explanation

To solve the problem, we start with the equation given for the vectors:

1. **Given Equation**:
\[ \vec{a} + \vec{b} + \vec{c} = \vec{0} \]
This implies:
\[ \vec{b} + \vec{c} = -\vec{a} \]

2. **Magnitude of Vectors**:
We know the magnitudes of the vectors:
\[ |\vec{a}| = 3, \quad |\vec{b}| = 5, \quad |\vec{c}| = 7 \]

3. **Squaring Both Sides**:
We square both sides of the equation \(\vec{b} + \vec{c} = -\vec{a}\):
\[ |\vec{b} + \vec{c}|^2 = |\vec{a}|^2 \]

4. **Expanding the Left Side**:
Using the formula for the square of a vector sum:
\[ |\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2 \vec{b} \cdot \vec{c} \]
Therefore, we have:
\[ |\vec{b}|^2 + |\vec{c}|^2 + 2 \vec{b} \cdot \vec{c} = |\vec{a}|^2 \]

5. **Substituting the Magnitudes**:
Substitute the known magnitudes into the equation:
\[ 5^2 + 7^2 + 2 \vec{b} \cdot \vec{c} = 3^2 \]
This simplifies to:
\[ 25 + 49 + 2 \vec{b} \cdot \vec{c} = 9 \]

6. **Combining Terms**:
Combine the constants:
\[ 74 + 2 \vec{b} \cdot \vec{c} = 9 \]

7. **Isolating the Dot Product**:
Rearranging gives:
\[ 2 \vec{b} \cdot \vec{c} = 9 - 74 \]
\[ 2 \vec{b} \cdot \vec{c} = -65 \]
Therefore:
\[ \vec{b} \cdot \vec{c} = -\frac{65}{2} \]

8. **Using the Dot Product Formula**:
The dot product can also be expressed as:
\[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta \]
Substituting the magnitudes:
\[ -\frac{65}{2} = 5 \cdot 7 \cos \theta \]
\[ -\frac{65}{2} = 35 \cos \theta \]

9. **Solving for Cosine**:
Now, isolate \(\cos \theta\):
\[ \cos \theta = -\frac{65}{2 \cdot 35} \]
Simplifying this gives:
\[ \cos \theta = -\frac{65}{70} = -\frac{13}{14} \]

Thus, the cosine of the angle between \(\vec{b}\) and \(\vec{c}\) is:
\[ \cos \theta = -\frac{13}{14} \]