`veca and vecb are bot` vector `absveca=absvecb=1` then angle between `veca+vecb+(vecaxxvecb) and veca`

Answer

To solve the problem, we need to find the angle between the vector \( \vec{A} + \vec{B} + (\vec{A} \times \vec{B}) \) and the vector \( \vec{A} \). Given that \( |\vec{A}| = |\vec{B}| = 1 \) and that \( \vec{A} \) and \( \vec{B} \) are perpendicular, we can follow these steps:

### Step 1: Understand the vectors
We know that:
- \( |\vec{A}| = 1 \)
- \( |\vec{B}| = 1 \)
- \( \vec{A} \cdot \vec{B} = 0 \) (since they are perpendicular)

### Step 2: Calculate the dot product
We want to find the angle \( \theta \) between the vector \( \vec{A} + \vec{B} + (\vec{A} \times \vec{B}) \) and \( \vec{A} \). We use the formula for the cosine of the angle between two vectors:
\[ \cos \theta = \frac{(\vec{A} + \vec{B} + (\vec{A} \times \vec{B})) \cdot \vec{A}}{|\vec{A} + \vec{B} + (\vec{A} \times \vec{B})| |\vec{A}|} \]

### Step 3: Calculate the numerator
Now, we calculate the dot product in the numerator:
\[ (\vec{A} + \vec{B} + (\vec{A} \times \vec{B})) \cdot \vec{A} = \vec{A} \cdot \vec{A} + \vec{B} \cdot \vec{A} + (\vec{A} \times \vec{B}) \cdot \vec{A} \]
Substituting the known values:
- \( \vec{A} \cdot \vec{A} = 1 \)
- \( \vec{B} \cdot \vec{A} = 0 \)
- \( (\vec{A} \times \vec{B}) \cdot \vec{A} = 0 \) (since the cross product is perpendicular to both vectors)

Thus, the numerator simplifies to:
\[ 1 + 0 + 0 = 1 \]

### Step 4: Calculate the denominator
Next, we need to calculate the magnitude of the vector \( \vec{A} + \vec{B} + (\vec{A} \times \vec{B}) \):
\[ |\vec{A} + \vec{B} + (\vec{A} \times \vec{B})|^2 = |\vec{A}|^2 + |\vec{B}|^2 + |\vec{A} \times \vec{B}|^2 + 2(\vec{A} \cdot \vec{B}) + 2(\vec{A} \times \vec{B}) \cdot \vec{A} + 2(\vec{A} \times \vec{B}) \cdot \vec{B} \]
Since \( |\vec{A}| = 1 \) and \( |\vec{B}| = 1 \), and \( \vec{A} \cdot \vec{B} = 0 \):
\[ = 1^2 + 1^2 + |\vec{A} \times \vec{B}|^2 + 0 + 0 + 0 \]
Now, \( |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin(90^\circ) = 1 \), so:
\[ = 1 + 1 + 1 = 3 \]
Thus, \( |\vec{A} + \vec{B} + (\vec{A} \times \vec{B})| = \sqrt{3} \).

### Step 5: Substitute into the cosine formula
Now we substitute back into the cosine formula:
\[ \cos \theta = \frac{1}{\sqrt{3} \cdot 1} = \frac{1}{\sqrt{3}} \]

### Step 6: Find the angle
Finally, we find the angle \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

### Final Answer
The angle between \( \vec{A} + \vec{B} + (\vec{A} \times \vec{B}) \) and \( \vec{A} \) is \( \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \).

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