In a ` triangle ABC` if `vecabs(AB)=7 ,vecabs(BC)=5 ,and vecabs(CA)=3.` . If the projection of `vec(BC)` on `vec(CA)` is `n/2`, then the value of n is

Answer

To solve the problem, we need to find the value of \( n \) given the conditions of triangle \( ABC \) and the projection of vector \( \vec{BC} \) onto vector \( \vec{CA} \).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- \( |\vec{AB}| = 7 \)
- \( |\vec{BC}| = 5 \)
- \( |\vec{CA}| = 3 \)

2. **Understand the Projection**:
- The projection of vector \( \vec{BC} \) onto vector \( \vec{CA} \) can be calculated using the formula:
\[ \text{Projection of } \vec{BC} \text{ on } \vec{CA} = |\vec{BC}| \cdot \cos(\theta) \]
where \( \theta \) is the angle between vectors \( \vec{BC} \) and \( \vec{CA} \).

3. **Use the Cosine Rule to Find \( \cos(\theta) \)**:
- According to the cosine rule:
\[ c^2 = a^2 + b^2 - 2ab \cos(\theta) \]
Here, let:
- \( a = |\vec{CA}| = 3 \)
- \( b = |\vec{BC}| = 5 \)
- \( c = |\vec{AB}| = 7 \)

Plugging the values into the cosine rule:
\[ 7^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cos(\theta) \]
This simplifies to:
\[ 49 = 9 + 25 - 30 \cos(\theta) \]
\[ 49 = 34 - 30 \cos(\theta) \]
Rearranging gives:
\[ 30 \cos(\theta) = 34 - 49 \]
\[ 30 \cos(\theta) = -15 \]
\[ \cos(\theta) = -\frac{15}{30} = -\frac{1}{2} \]

4. **Calculate the Projection**:
- Now substituting \( \cos(\theta) \) back into the projection formula:
\[ \text{Projection of } \vec{BC} \text{ on } \vec{CA} = |\vec{BC}| \cdot \cos(\theta) = 5 \cdot \left(-\frac{1}{2}\right) = -\frac{5}{2} \]
Since we are interested in the magnitude, we take the absolute value:
\[ \text{Magnitude of Projection} = \frac{5}{2} \]

5. **Set Up the Equation**:
- According to the problem, this projection is equal to \( \frac{n}{2} \):
\[ \frac{n}{2} = \frac{5}{2} \]

6. **Solve for \( n \)**:
- Multiplying both sides by 2 gives:
\[ n = 5 \]

### Conclusion:
The value of \( n \) is \( 5 \).