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 solve the problem step by step, we will analyze the given information and apply vector operations.

### Step 1: Understand the Given Condition
We are given that for two vectors \( \vec{A} \) and \( \vec{B} \):
\[ \vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}| \]
This means that the dot product of the vectors is equal to the magnitude of their cross product.

### Step 2: Write the Formulas for Dot Product and Cross Product
The dot product \( \vec{A} \cdot \vec{B} \) can be expressed as:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]
where \( \theta \) is the angle between the two vectors.

The magnitude of the cross product \( |\vec{A} \times \vec{B}| \) can be expressed as:
\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \]

### Step 3: Set Up the Equation
From the given condition, we can equate the two expressions:
\[ |\vec{A}| |\vec{B}| \cos \theta = |\vec{A}| |\vec{B}| \sin \theta \]

### Step 4: Simplify the Equation
Assuming \( |\vec{A}| \) and \( |\vec{B}| \) are not zero, we can divide both sides by \( |\vec{A}| |\vec{B}| \):
\[ \cos \theta = \sin \theta \]

### Step 5: Solve for \( \theta \)
The equation \( \cos \theta = \sin \theta \) implies:
\[ \tan \theta = 1 \]
This means:
\[ \theta = 45^\circ \]

### Step 6: Find the Magnitude of \( \vec{C} = \vec{A} + \vec{B} \)
The magnitude of the resultant vector \( \vec{C} \) can be calculated using the formula:
\[ |\vec{C}| = |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta} \]

### Step 7: Substitute \( \theta \) into the Formula
Since \( \theta = 45^\circ \), we have:
\[ \cos 45^\circ = \frac{1}{\sqrt{2}} \]
Substituting this into the equation gives:
\[ |\vec{C}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cdot \frac{1}{\sqrt{2}}} \]

### Step 8: Simplify the Expression
This simplifies to:
\[ |\vec{C}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + \sqrt{2} |\vec{A}| |\vec{B}|} \]

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