In a right angled triangle hypotenuse AC= p, then `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

Answer

To solve the problem, we need to evaluate the expression:

\[ \vec{AB} \cdot \vec{AC} + \vec{BC} \cdot \vec{BA} + \vec{CA} \cdot \vec{CB} \]

Given that triangle ABC is a right-angled triangle with hypotenuse AC = p, we can follow these steps:

### Step 1: Understand the vectors involved
In a right-angled triangle:
- \(\vec{AB}\) is one leg,
- \(\vec{BC}\) is the other leg,
- \(\vec{AC}\) is the hypotenuse.

### Step 2: Analyze the dot products
Since \(\vec{AB}\) and \(\vec{BC}\) are perpendicular to each other, we know that:

\[ \vec{BC} \cdot \vec{AB} = 0 \]

This means that the term \(\vec{BC} \cdot \vec{BA}\) (which is \(-\vec{BC} \cdot \vec{AB}\)) will also equal 0.

### Step 3: Rewrite the expression
Now we can simplify the expression:

\[ \vec{AB} \cdot \vec{AC} + 0 + \vec{CA} \cdot \vec{CB} \]

### Step 4: Change the direction of vectors
Notice that \(\vec{CA} = -\vec{AC}\) and \(\vec{CB} = -\vec{BC}\). Thus, we can rewrite \(\vec{CA} \cdot \vec{CB}\) as:

\[ \vec{CA} \cdot \vec{CB} = (-\vec{AC}) \cdot (-\vec{BC}) = \vec{AC} \cdot \vec{BC} \]

### Step 5: Combine the terms
Now we can combine the terms:

\[ \vec{AB} \cdot \vec{AC} + \vec{AC} \cdot \vec{BC} \]

### Step 6: Factor out \(\vec{AC}\)
We can factor out \(\vec{AC}\):

\[ \vec{AC} \cdot (\vec{AB} + \vec{BC}) \]

### Step 7: Use the Pythagorean theorem
In a right triangle, the sum of the squares of the legs equals the square of the hypotenuse. Therefore, we can express \(\vec{AB} + \vec{BC}\) in terms of \(\vec{AC}\):

\[ \vec{AB} + \vec{BC} = \vec{AC} \]

Thus, we have:

\[ \vec{AC} \cdot \vec{AC} \]

### Step 8: Calculate the final result
The dot product \(\vec{AC} \cdot \vec{AC}\) is equal to the magnitude squared of \(\vec{AC}\):

\[ \vec{AC} \cdot \vec{AC} = |\vec{AC}|^2 = p^2 \]

### Final Answer
So the value of the expression is:

\[ p^2 \]