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For any two vectors A and B, if vec(A).v...

For any two vectors A and B, if `vec(A).vec(B) = |vec(A) xx vec(B)|`, the magnitude of `vec(C ) = vec(A) + vec(B)` is equal to :

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)AB)`

Text Solution

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To solve the problem, we need to find the magnitude of the vector \( \vec{C} = \vec{A} + \vec{B} \) given that \( \vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}| \). ### Step-by-step Solution: 1. **Understanding the Given Condition**: We have the condition \( \vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}| \). This means that the dot product of vectors \( \vec{A} \) and \( \vec{B} \) is equal to the magnitude of their cross product. 2. **Using the Definitions**: The dot product is given by: \[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \] The magnitude of the cross product is given by: \[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \] where \( \theta \) is the angle between the two vectors. 3. **Setting Up the Equation**: From the given condition, we can set up the equation: \[ |\vec{A}| |\vec{B}| \cos \theta = |\vec{A}| |\vec{B}| \sin \theta \] 4. **Dividing Both Sides**: Assuming \( |\vec{A}| \) and \( |\vec{B}| \) are not zero, we can divide both sides by \( |\vec{A}| |\vec{B}| \): \[ \cos \theta = \sin \theta \] 5. **Finding the Angle**: The equation \( \cos \theta = \sin \theta \) implies: \[ \tan \theta = 1 \implies \theta = 45^\circ \text{ or } \frac{\pi}{4} \text{ radians} \] 6. **Calculating the Magnitude of \( \vec{C} \)**: The magnitude of \( \vec{C} = \vec{A} + \vec{B} \) can be calculated using the formula: \[ |\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta \] Substituting \( \theta = 45^\circ \) (where \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)): \[ |\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \left(\frac{1}{\sqrt{2}}\right) \] 7. **Final Expression**: Therefore, we have: \[ |\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + \sqrt{2} |\vec{A}| |\vec{B}| \] Taking the square root gives: \[ |\vec{C}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + \sqrt{2} |\vec{A}| |\vec{B}|} \]
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