If in a right-angled triangle `A B C ,` the hypotenuse `A B=p ,t h e n vec A BdotA C+ vec B Cdot vec B A+ vec C Adot vec C B` is equal to `2p^2` b. `(p^2)/2` c. `p^2` d. none of these

Options

A. `2p^(2)`
B. `p^(2)/2`
C. `p^(2)`
D. none of these

Explanation

To solve the problem, we need to evaluate the expression \( \vec{AB} \cdot \vec{AC} + \vec{BC} \cdot \vec{BA} + \vec{CA} \cdot \vec{CB} \) in the context of a right-angled triangle \( ABC \) where \( AB = p \).

### Step-by-step Solution:

1. **Identify the Vectors**:
- Let \( \vec{AB} \) be the hypotenuse, which is given as \( p \).
- The vectors can be represented as:
- \( \vec{AB} \) from point \( A \) to point \( B \)
- \( \vec{AC} \) from point \( A \) to point \( C \)
- \( \vec{BC} \) from point \( B \) to point \( C \)
- \( \vec{BA} \) from point \( B \) to point \( A \)
- \( \vec{CA} \) from point \( C \) to point \( A \)
- \( \vec{CB} \) from point \( C \) to point \( B \)

2. **Use the Right-Angle Property**:
- In triangle \( ABC \), since \( C \) is the right angle, the vectors \( \vec{CA} \) and \( \vec{CB} \) are perpendicular.
- Therefore, \( \vec{CA} \cdot \vec{CB} = 0 \).

3. **Rewrite the Expression**:
- The expression simplifies to:
\[ \vec{AB} \cdot \vec{AC} + \vec{BC} \cdot \vec{BA} + 0 \]
- This can be rewritten as:
\[ \vec{AB} \cdot \vec{AC} + \vec{BC} \cdot (-\vec{AB}) \]
- This gives:
\[ \vec{AB} \cdot \vec{AC} - \vec{BC} \cdot \vec{AB} \]

4. **Substitute \( \vec{BC} \)**:
- We know \( \vec{BC} = -\vec{CB} \), so we can substitute:
\[ \vec{AB} \cdot \vec{AC} + \vec{AB} \cdot \vec{CB} \]

5. **Factor Out \( \vec{AB} \)**:
- Factor out \( \vec{AB} \):
\[ \vec{AB} \cdot (\vec{AC} + \vec{CB}) \]

6. **Use Triangle Properties**:
- Since \( \vec{AC} + \vec{CB} = \vec{AB} \), we can substitute:
\[ \vec{AB} \cdot \vec{AB} \]

7. **Calculate the Dot Product**:
- The dot product \( \vec{AB} \cdot \vec{AB} \) is equal to the magnitude squared of \( \vec{AB} \):
\[ |\vec{AB}|^2 = p^2 \]

### Final Result:
Thus, the value of the expression \( \vec{AB} \cdot \vec{AC} + \vec{BC} \cdot \vec{BA} + \vec{CA} \cdot \vec{CB} \) is equal to \( p^2 \).

### Answer:
c. \( p^2 \)