The angle between `vec(A)+vec(B)` and `vec(A)xxvec(B)` is

Options

A. 0
B. `pi//4`
C. `pi//2`
D. `pi`

Explanation

To find the angle between the vectors \(\vec{A} + \vec{B}\) and \(\vec{A} \times \vec{B}\), we can follow these steps:

### Step 1: Understand the vectors involved
We have two vectors:
- \(\vec{A}\)
- \(\vec{B}\)

We need to find the angle between the vector sum \(\vec{A} + \vec{B}\) and the vector cross product \(\vec{A} \times \vec{B}\).

### Step 2: Define the cross product
The cross product \(\vec{A} \times \vec{B}\) results in a vector that is perpendicular to both \(\vec{A}\) and \(\vec{B}\). This means that:
\[ \vec{A} \times \vec{B} \perp \vec{A} \quad \text{and} \quad \vec{A} \times \vec{B} \perp \vec{B} \]

### Step 3: Use the dot product to find the angle
The angle \(\theta\) between two vectors \(\vec{X}\) and \(\vec{Y}\) can be found using the dot product formula:
\[ \cos \theta = \frac{\vec{X} \cdot \vec{Y}}{|\vec{X}| |\vec{Y}|} \]
In our case, we want to find:
\[ \cos \theta = \frac{(\vec{A} \times \vec{B}) \cdot (\vec{A} + \vec{B})}{|\vec{A} \times \vec{B}| |\vec{A} + \vec{B}|} \]

### Step 4: Expand the dot product
Using the distributive property of the dot product, we can expand the numerator:
\[ (\vec{A} \times \vec{B}) \cdot (\vec{A} + \vec{B}) = (\vec{A} \times \vec{B}) \cdot \vec{A} + (\vec{A} \times \vec{B}) \cdot \vec{B} \]
Since \(\vec{A} \times \vec{B}\) is perpendicular to both \(\vec{A}\) and \(\vec{B}\), we have:
\[ (\vec{A} \times \vec{B}) \cdot \vec{A} = 0 \quad \text{and} \quad (\vec{A} \times \vec{B}) \cdot \vec{B} = 0 \]
Thus,
\[ (\vec{A} \times \vec{B}) \cdot (\vec{A} + \vec{B}) = 0 + 0 = 0 \]

### Step 5: Conclude the angle
Since the dot product is zero, we have:
\[ \cos \theta = \frac{0}{|\vec{A} \times \vec{B}| |\vec{A} + \vec{B}|} = 0 \]
This implies that:
\[ \theta = \frac{\pi}{2} \text{ radians} \quad \text{or} \quad 90^\circ \]

### Final Answer
The angle between \(\vec{A} + \vec{B}\) and \(\vec{A} \times \vec{B}\) is \(90^\circ\).