Let a and b be two unit vectors such that `|(a+b)+2(a xx b)|=2 . " if " theta in (0,pi)` is the angle between `hat(a) and hat(b)` then among statements :
`(S1) : 2 |hat(a)xxhat(b)|=|hat(a)-hat(b)|`
(S2) : The projection of `hat(a) ` on `(hat(a) +hat(b))" is "1/2`

Options

A. Only (S1) is true
B. Only (S2) is true
C. Both (S1) and (S2) are true
D. Both (S1) and (S2) are False

Answer

To solve the problem, we need to analyze the given equation involving two unit vectors \( \hat{a} \) and \( \hat{b} \). The equation is:

\[ |(\hat{a} + \hat{b}) + 2(\hat{a} \times \hat{b})| = 2 \]

### Step 1: Understanding the Magnitude
Since \( \hat{a} \) and \( \hat{b} \) are unit vectors, we know that \( |\hat{a}| = 1 \) and \( |\hat{b}| = 1 \). The magnitude of the left-hand side can be expanded using the properties of vectors.

### Step 2: Squaring Both Sides
To simplify the expression, we square both sides:

\[ |(\hat{a} + \hat{b}) + 2(\hat{a} \times \hat{b})|^2 = 4 \]

### Step 3: Expanding the Left Side
Using the formula for the magnitude of a vector, we expand the left side:

\[ |(\hat{a} + \hat{b})|^2 + |2(\hat{a} \times \hat{b})|^2 + 2(\hat{a} + \hat{b}) \cdot (2(\hat{a} \times \hat{b})) \]

### Step 4: Calculating Each Term
1. **Calculating \( |(\hat{a} + \hat{b})|^2 \)**:
\[ |(\hat{a} + \hat{b})|^2 = |\hat{a}|^2 + |\hat{b}|^2 + 2(\hat{a} \cdot \hat{b}) = 1 + 1 + 2(\hat{a} \cdot \hat{b}) = 2 + 2(\hat{a} \cdot \hat{b}) \]

2. **Calculating \( |2(\hat{a} \times \hat{b})|^2 \)**:
\[ |2(\hat{a} \times \hat{b})|^2 = 4|\hat{a} \times \hat{b}|^2 = 4|\hat{a}|^2 |\hat{b}|^2 \sin^2(\theta) = 4 \sin^2(\theta) \]

3. **Calculating the dot product term**:
Since \( \hat{a} \cdot (\hat{a} \times \hat{b}) = 0 \) and \( \hat{b} \cdot (\hat{a} \times \hat{b}) = 0 \), we have:
\[ 2(\hat{a} + \hat{b}) \cdot (2(\hat{a} \times \hat{b})) = 0 \]

### Step 5: Combining Results
Putting it all together, we have:

\[ (2 + 2(\hat{a} \cdot \hat{b})) + 4 \sin^2(\theta) = 4 \]

### Step 6: Simplifying the Equation
This simplifies to:

\[ 2 + 2(\hat{a} \cdot \hat{b}) + 4 \sin^2(\theta) = 4 \]

Subtracting 2 from both sides gives:

\[ 2(\hat{a} \cdot \hat{b}) + 4 \sin^2(\theta) = 2 \]

Dividing through by 2:

\[ \hat{a} \cdot \hat{b} + 2 \sin^2(\theta) = 1 \]

### Step 7: Using the Identity
Using the identity \( \hat{a} \cdot \hat{b} = \cos(\theta) \) and \( \sin^2(\theta) = 1 - \cos^2(\theta) \):

\[ \cos(\theta) + 2(1 - \cos^2(\theta)) = 1 \]

This simplifies to:

\[ \cos(\theta) + 2 - 2\cos^2(\theta) = 1 \]

Rearranging gives:

\[ 2\cos^2(\theta) - \cos(\theta) = 1 \]

### Step 8: Forming a Quadratic Equation
This can be rearranged to form a quadratic equation:

\[ 2\cos^2(\theta) - \cos(\theta) - 1 = 0 \]

### Step 9: Solving the Quadratic
Using the quadratic formula \( \cos(\theta) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ \cos(\theta) = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4} \]

This gives:

\[ \cos(\theta) = 1 \quad \text{or} \quad \cos(\theta) = -\frac{1}{2} \]

### Step 10: Finding the Angles
From \( \cos(\theta) = 1 \), we have \( \theta = 0 \) (not in the interval). From \( \cos(\theta) = -\frac{1}{2} \), we have:

\[ \theta = \frac{2\pi}{3} \quad \text{or} \quad 120^\circ \]

### Step 11: Verifying Statements
Now we need to check the statements:

1. **Statement (S1)**: \( 2 |\hat{a} \times \hat{b}| = |\hat{a} - \hat{b}| \)
2. **Statement (S2)**: The projection of \( \hat{a} \) on \( \hat{a} + \hat{b} \) is \( \frac{1}{2} \)

### Step 12: Evaluating (S1)
Calculating \( |\hat{a} - \hat{b}| \):

\[ |\hat{a} - \hat{b}|^2 = |\hat{a}|^2 + |\hat{b}|^2 - 2(\hat{a} \cdot \hat{b}) = 1 + 1 - 2\cos(\theta) = 2 - 2(-\frac{1}{2}) = 3 \]

Thus, \( |\hat{a} - \hat{b}| = \sqrt{3} \) and \( |\hat{a} \times \hat{b}| = \sin(\theta) = \sin(120^\circ) = \frac{\sqrt{3}}{2} \).

So, \( 2 |\hat{a} \times \hat{b}| = \sqrt{3} \), which confirms that (S1) is true.

### Step 13: Evaluating (S2)
The projection of \( \hat{a} \) on \( \hat{a} + \hat{b} \):

\[ \text{Projection} = \frac{\hat{a} \cdot (\hat{a} + \hat{b})}{|\hat{a} + \hat{b}|} \]

Calculating \( |\hat{a} + \hat{b}| \):

\[ |\hat{a} + \hat{b}|^2 = 2 + 2\cos(\theta) = 2 + 2(-\frac{1}{2}) = 1 \implies |\hat{a} + \hat{b}| = 1 \]

Thus, the projection simplifies to:

\[ \hat{a} \cdot (\hat{a} + \hat{b}) = 1 + \cos(\theta) = 1 - \frac{1}{2} = \frac{1}{2} \]

### Conclusion
Both statements (S1) and (S2) are true.