Show that the scalar `vec(A)*(vec(B)+vec(C)) times (vec(A)+vec(B)+vec(C))=0`.

Answer

To show that the scalar \( \vec{A} \cdot (\vec{B} + \vec{C}) \times (\vec{A} + \vec{B} + \vec{C}) = 0 \), we will follow the steps below:

### Step 1: Expand the Expression
We start with the expression:
\[ \vec{A} \cdot (\vec{B} + \vec{C}) \times (\vec{A} + \vec{B} + \vec{C}) \]
Using the distributive property of the cross product, we can rewrite this as:
\[ \vec{A} \cdot \left( \vec{B} \times (\vec{A} + \vec{B} + \vec{C}) + \vec{C} \times (\vec{A} + \vec{B} + \vec{C}) \right) \]

### Step 2: Calculate the Cross Products
Now, we compute the cross products:
1. \( \vec{B} \times (\vec{A} + \vec{B} + \vec{C}) = \vec{B} \times \vec{A} + \vec{B} \times \vec{B} + \vec{B} \times \vec{C} \)
- Note that \( \vec{B} \times \vec{B} = 0 \) (the cross product of any vector with itself is zero).
- Thus, this simplifies to:
\[ \vec{B} \times \vec{A} + \vec{B} \times \vec{C} \]

2. Similarly, for \( \vec{C} \times (\vec{A} + \vec{B} + \vec{C}) \):
\[ \vec{C} \times \vec{A} + \vec{C} \times \vec{B} + \vec{C} \times \vec{C} \]
- Again, \( \vec{C} \times \vec{C} = 0 \), so this simplifies to:
\[ \vec{C} \times \vec{A} + \vec{C} \times \vec{B} \]

### Step 3: Combine the Results
Now we can combine the results of the cross products:
\[ \vec{A} \cdot \left( \vec{B} \times \vec{A} + \vec{B} \times \vec{C} + \vec{C} \times \vec{A} + \vec{C} \times \vec{B} \right) \]

### Step 4: Apply Properties of the Dot and Cross Products
Using the property that \( \vec{A} \cdot (\vec{B} \times \vec{C}) \) is equal to the scalar triple product, we can analyze each term:
1. \( \vec{A} \cdot (\vec{B} \times \vec{A}) = 0 \) (since the angle between \( \vec{A} \) and \( \vec{B} \times \vec{A} \) is 0).
2. \( \vec{A} \cdot (\vec{C} \times \vec{A}) = 0 \) (for the same reason).

Thus, we are left with:
\[ \vec{A} \cdot (\vec{B} \times \vec{C}) + \vec{A} \cdot (\vec{C} \times \vec{B}) \]
Using the property of the scalar triple product, we know that:
\[ \vec{A} \cdot (\vec{C} \times \vec{B}) = -\vec{A} \cdot (\vec{B} \times \vec{C}) \]
This means that:
\[ \vec{A} \cdot (\vec{B} \times \vec{C}) + (-\vec{A} \cdot (\vec{B} \times \vec{C})) = 0 \]

### Conclusion
Thus, we have shown that:
\[ \vec{A} \cdot (\vec{B} + \vec{C}) \times (\vec{A} + \vec{B} + \vec{C}) = 0 \]
Hence, the statement is proved.