If ` veca, vecb` are unit vectors such that ` |veca +vecb|=1 and |veca -vecb|=sqrt3`, then |3`veca+2vecb` |=`

Options

A. 7
B. 4
C. ` sqrt7`
D. `sqrt19`

Explanation

To solve the problem step by step, we will use the given conditions and properties of vectors.

### Step 1: Understand the given conditions
We know that:
1. \( |\vec{a} + \vec{b}| = 1 \)
2. \( |\vec{a} - \vec{b}| = \sqrt{3} \)

### Step 2: Use the formula for the magnitude of vectors
We can square both sides of the equations to eliminate the square roots:
- From \( |\vec{a} + \vec{b}| = 1 \):
\[ |\vec{a} + \vec{b}|^2 = 1^2 \implies |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 1 \]
- From \( |\vec{a} - \vec{b}| = \sqrt{3} \):
\[ |\vec{a} - \vec{b}|^2 = (\sqrt{3})^2 \implies |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} = 3 \]

### Step 3: Substitute the values of unit vectors
Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, we have:
- \( |\vec{a}|^2 = 1 \)
- \( |\vec{b}|^2 = 1 \)

Substituting these into the equations:
1. \( 1 + 1 + 2\vec{a} \cdot \vec{b} = 1 \)
\[ 2 + 2\vec{a} \cdot \vec{b} = 1 \implies 2\vec{a} \cdot \vec{b} = 1 - 2 \implies \vec{a} \cdot \vec{b} = -\frac{1}{2} \]

2. \( 1 + 1 - 2\vec{a} \cdot \vec{b} = 3 \)
\[ 2 - 2\vec{a} \cdot \vec{b} = 3 \implies -2\vec{a} \cdot \vec{b} = 3 - 2 \implies \vec{a} \cdot \vec{b} = -\frac{1}{2} \]

### Step 4: Find the angle between the vectors
The dot product \( \vec{a} \cdot \vec{b} \) can also be expressed in terms of the angle \( \theta \) between them:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = 1 \cdot 1 \cdot \cos \theta = \cos \theta \]
Thus, we have:
\[ \cos \theta = -\frac{1}{2} \]
This implies that \( \theta = 120^\circ \) or \( \theta = \frac{2\pi}{3} \).

### Step 5: Calculate \( |3\vec{a} + 2\vec{b}| \)
Now we need to find \( |3\vec{a} + 2\vec{b}| \). We will square this magnitude:
\[ |3\vec{a} + 2\vec{b}|^2 = (3\vec{a} + 2\vec{b}) \cdot (3\vec{a} + 2\vec{b}) \]
Expanding this:
\[ = 9|\vec{a}|^2 + 4|\vec{b}|^2 + 2(3)(2)(\vec{a} \cdot \vec{b}) \]
Substituting the known values:
\[ = 9(1) + 4(1) + 12(\vec{a} \cdot \vec{b}) = 9 + 4 + 12(-\frac{1}{2}) \]
Calculating this:
\[ = 13 - 6 = 7 \]

### Step 6: Final result
Thus, we have:
\[ |3\vec{a} + 2\vec{b}| = \sqrt{7} \]

### Final Answer
\[ |3\vec{a} + 2\vec{b}| = \sqrt{7} \]

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