If ` veca , vec b ` all are unit vector, then the greatest value of ` |veca +vecb| + |veca-vecb|` is

Options

A. `2 sqrt2`
B. sqrt2
C. 2
D. 4sqrt2

Answer

To solve the problem of finding the greatest value of \( |\vec{a} + \vec{b}| + |\vec{a} - \vec{b}| \) where \( \vec{a} \) and \( \vec{b} \) are unit vectors, we can follow these steps:

### Step 1: Define the angle between the vectors
Let \( \theta \) be the angle between the unit vectors \( \vec{a} \) and \( \vec{b} \). Since both vectors are unit vectors, we have:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]

### Step 2: Use the dot product to express \( |\vec{a} + \vec{b}|^2 \)
Using the formula for the square of the magnitude of the sum of two vectors, we have:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 |\vec{a}| |\vec{b}| \cos \theta \]
Substituting the magnitudes:
\[ |\vec{a} + \vec{b}|^2 = 1^2 + 1^2 + 2 \cdot 1 \cdot 1 \cdot \cos \theta = 2 + 2 \cos \theta \]
Thus,
\[ |\vec{a} + \vec{b}| = \sqrt{2 + 2 \cos \theta} \]

### Step 3: Use the dot product to express \( |\vec{a} - \vec{b}|^2 \)
Similarly, for the difference of the vectors:
\[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2 |\vec{a}| |\vec{b}| \cos \theta \]
Substituting the magnitudes:
\[ |\vec{a} - \vec{b}|^2 = 1^2 + 1^2 - 2 \cdot 1 \cdot 1 \cdot \cos \theta = 2 - 2 \cos \theta \]
Thus,
\[ |\vec{a} - \vec{b}| = \sqrt{2 - 2 \cos \theta} \]

### Step 4: Combine the two magnitudes
Now we can add the two magnitudes:
\[ |\vec{a} + \vec{b}| + |\vec{a} - \vec{b}| = \sqrt{2 + 2 \cos \theta} + \sqrt{2 - 2 \cos \theta} \]

### Step 5: Simplify the expression
Factor out \( \sqrt{2} \):
\[ = \sqrt{2} \left( \sqrt{1 + \cos \theta} + \sqrt{1 - \cos \theta} \right) \]
Using the identities \( 1 + \cos \theta = 2 \cos^2(\theta/2) \) and \( 1 - \cos \theta = 2 \sin^2(\theta/2) \):
\[ = \sqrt{2} \left( \sqrt{2 \cos^2(\theta/2)} + \sqrt{2 \sin^2(\theta/2)} \right) \]
\[ = \sqrt{2} \cdot \sqrt{2} \left( \cos(\theta/2) + \sin(\theta/2) \right) = 2 \left( \cos(\theta/2) + \sin(\theta/2) \right) \]

### Step 6: Find the maximum value
The expression \( \cos(\theta/2) + \sin(\theta/2) \) achieves its maximum value of \( \sqrt{2} \) when \( \theta/2 = \frac{\pi}{4} \) or \( \theta = \frac{\pi}{2} \). Therefore:
\[ \text{Maximum value} = 2 \cdot \sqrt{2} = 2\sqrt{2} \]

### Final Answer
The greatest value of \( |\vec{a} + \vec{b}| + |\vec{a} - \vec{b}| \) is \( 2\sqrt{2} \).