For any two vectors `barA` and `barB` if `barA.barB=|bar AxxbarB|`, the magnitude of `barC=barA+barB` is equal to :

Options

A. `sqrt(A^(2)+B^(2))`
B. `A+B`
C. `sqrt(A^(2)+B^(2)+(AB)/sqrt(2)`
D. `sqrt(A^(2)+B^(2)+sqrt(2)xxAB`

Explanation

To find the magnitude of the vector \( \bar{C} = \bar{A} + \bar{B} \) given that \( \bar{A} \cdot \bar{B} = |\bar{A} \times \bar{B}| \), we can follow these steps:

### Step 1: Understand the given condition
We know that:
\[ \bar{A} \cdot \bar{B} = |\bar{A} \times \bar{B}| \]
This condition implies a specific relationship between the vectors \( \bar{A} \) and \( \bar{B} \).

### Step 2: Use the formulas for dot and cross products
The dot product is given by:
\[ \bar{A} \cdot \bar{B} = |\bar{A}| |\bar{B}| \cos \theta \]
The magnitude of the cross product is given by:
\[ |\bar{A} \times \bar{B}| = |\bar{A}| |\bar{B}| \sin \theta \]
Setting these equal gives us:
\[ |\bar{A}| |\bar{B}| \cos \theta = |\bar{A}| |\bar{B}| \sin \theta \]

### Step 3: Simplify the equation
Assuming \( |\bar{A}| \) and \( |\bar{B}| \) are not zero, we can divide both sides by \( |\bar{A}| |\bar{B}| \):
\[ \cos \theta = \sin \theta \]
This implies:
\[ \tan \theta = 1 \quad \Rightarrow \quad \theta = 45^\circ \]

### Step 4: Substitute \( \theta \) into the formulas
Now, substituting \( \theta = 45^\circ \) into the formulas for dot and cross products:
\[ \bar{A} \cdot \bar{B} = |\bar{A}| |\bar{B}| \cos 45^\circ = |\bar{A}| |\bar{B}| \cdot \frac{1}{\sqrt{2}} \]
\[ |\bar{A} \times \bar{B}| = |\bar{A}| |\bar{B}| \sin 45^\circ = |\bar{A}| |\bar{B}| \cdot \frac{1}{\sqrt{2}} \]

### Step 5: Find the magnitude of \( \bar{C} \)
The magnitude of \( \bar{C} = \bar{A} + \bar{B} \) is given by:
\[ |\bar{C}| = \sqrt{|\bar{A}|^2 + |\bar{B}|^2 + 2(\bar{A} \cdot \bar{B})} \]
Substituting \( \bar{A} \cdot \bar{B} \):
\[ |\bar{C}| = \sqrt{|\bar{A}|^2 + |\bar{B}|^2 + 2\left(|\bar{A}| |\bar{B}| \cdot \frac{1}{\sqrt{2}}\right)} \]
This simplifies to:
\[ |\bar{C}| = \sqrt{|\bar{A}|^2 + |\bar{B}|^2 + \sqrt{2} |\bar{A}| |\bar{B}|} \]

### Conclusion
Thus, the magnitude of \( \bar{C} \) is:
\[ |\bar{C}| = \sqrt{|\bar{A}|^2 + |\bar{B}|^2 + \sqrt{2} |\bar{A}| |\bar{B}|} \]