If in a right-angled triangle ABC, hypotenuse AC=p, then what is `vec(AB).vec(AC)+vec(BC).vec(BA)+vec(CA).vec(CB)` equal to ?

Options

A. `p^(2)`
B. `2p^(2)`
C. `(p^(2))/(2)`
D. p

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 AC is the hypotenuse of length \( p \).

### Step-by-Step Solution:

1. **Understand the Geometry**:
In triangle ABC, angle B is the right angle, and AC is the hypotenuse. We can denote the sides as follows:
- \( \vec{AB} \) is one leg,
- \( \vec{BC} \) is the other leg,
- \( \vec{AC} \) is the hypotenuse.

2. **Identify the Dot Products**:
We need to evaluate three dot products:
- \( \vec{AB} \cdot \vec{AC} \)
- \( \vec{BC} \cdot \vec{BA} \)
- \( \vec{CA} \cdot \vec{CB} \)

3. **Analyze \( \vec{BC} \cdot \vec{BA} \)**:
Notice that \( \vec{BA} = -\vec{AB} \). Therefore,
\[ \vec{BC} \cdot \vec{BA} = \vec{BC} \cdot (-\vec{AB}) = -\vec{BC} \cdot \vec{AB} \]
Since \( \vec{BC} \) and \( \vec{AB} \) are perpendicular (as they are legs of the right triangle), we have:
\[ \vec{BC} \cdot \vec{AB} = 0 \]
Thus,
\[ \vec{BC} \cdot \vec{BA} = 0 \]

4. **Rewrite the Expression**:
Now we can simplify the original expression:
\[ \vec{AB} \cdot \vec{AC} + 0 + \vec{CA} \cdot \vec{CB} \]
This simplifies to:
\[ \vec{AB} \cdot \vec{AC} + \vec{CA} \cdot \vec{CB} \]

5. **Substituting \( \vec{CA} \) and \( \vec{CB} \)**:
We know that \( \vec{CA} = -\vec{AC} \) and \( \vec{CB} = -\vec{BC} \). Therefore,
\[ \vec{CA} \cdot \vec{CB} = (-\vec{AC}) \cdot (-\vec{BC}) = \vec{AC} \cdot \vec{BC} \]

6. **Combine the Terms**:
Now we have:
\[ \vec{AB} \cdot \vec{AC} + \vec{AC} \cdot \vec{BC} \]
We can factor out \( \vec{AC} \):
\[ \vec{AC} \cdot (\vec{AB} + \vec{BC}) \]

7. **Using Vector Addition**:
Since \( \vec{AB} + \vec{BC} = \vec{AC} \) (by the triangle law of addition), we get:
\[ \vec{AC} \cdot \vec{AC} \]
This is the dot product of a vector with itself, which gives us the square of its magnitude:
\[ |\vec{AC}|^2 = AC^2 = p^2 \]

8. **Final Result**:
Therefore, the value of the expression \( \vec{AB} \cdot \vec{AC} + \vec{BC} \cdot \vec{BA} + \vec{CA} \cdot \vec{CB} \) is:
\[ p^2 \]

### Conclusion:
The answer is \( p^2 \).