If `F_(1) and F_(2)` are two vectors of equal magnitudes F such that `|F_(1).F_(2)|= |F_(1)xxF_(2)|, "then"|F_(1) +F_(2)|` equals to

Options

A. `sqrt((2+sqrt(2)))F`
B. `2F`
C. `Fsqrt(2)`
D. None of these

Explanation

To solve the problem, we need to find the magnitude of the vector sum \( |F_1 + F_2| \) given that the magnitudes of the dot product and cross product of two vectors \( F_1 \) and \( F_2 \) are equal, and both vectors have the same magnitude \( F \).

### Step-by-Step Solution:

1. **Given Information**:
- \( |F_1| = |F_2| = F \)
- \( |F_1 \cdot F_2| = |F_1 \times F_2| \)

2. **Dot Product**:
- The dot product of two vectors can be expressed as:
\[ |F_1 \cdot F_2| = |F_1| |F_2| \cos \theta \]
- Since \( |F_1| = |F_2| = F \), we can write:
\[ |F_1 \cdot F_2| = F^2 \cos \theta \tag{1} \]

3. **Cross Product**:
- The magnitude of the cross product is given by:
\[ |F_1 \times F_2| = |F_1| |F_2| \sin \theta \]
- Again, using \( |F_1| = |F_2| = F \):
\[ |F_1 \times F_2| = F^2 \sin \theta \tag{2} \]

4. **Equating Dot and Cross Products**:
- From the problem statement, we have:
\[ |F_1 \cdot F_2| = |F_1 \times F_2| \]
- Substituting equations (1) and (2) into this gives:
\[ F^2 \cos \theta = F^2 \sin \theta \]
- Dividing both sides by \( F^2 \) (assuming \( F \neq 0 \)):
\[ \cos \theta = \sin \theta \]
- This implies:
\[ \tan \theta = 1 \implies \theta = 45^\circ \]

5. **Finding the Magnitude of \( F_1 + F_2 \)**:
- The magnitude of the sum of two vectors is given by:
\[ |F_1 + F_2| = \sqrt{|F_1|^2 + |F_2|^2 + 2 |F_1| |F_2| \cos \theta} \]
- Substituting \( |F_1| = |F_2| = F \) and \( \theta = 45^\circ \):
\[ |F_1 + F_2| = \sqrt{F^2 + F^2 + 2F^2 \cos 45^\circ} \]
- Since \( \cos 45^\circ = \frac{1}{\sqrt{2}} \):
\[ |F_1 + F_2| = \sqrt{F^2 + F^2 + 2F^2 \cdot \frac{1}{\sqrt{2}}} \]
- Simplifying further:
\[ |F_1 + F_2| = \sqrt{2F^2 + \frac{2F^2}{\sqrt{2}}} \]
- Factoring out \( F^2 \):
\[ |F_1 + F_2| = F \sqrt{2 + \sqrt{2}} \]

6. **Final Result**:
- Thus, the magnitude of \( |F_1 + F_2| \) is:
\[ |F_1 + F_2| = F \sqrt{2 + \sqrt{2}} \]