If ` veca , vecb` are unit vectors such that ` |vec a+vecb|=-1 " then " |2veca -3vecb| =`

Options

A. 19
B. `sqrt19`
C. `sqrt13`
D. 4

Explanation

To solve the problem, we need to find the magnitude of \( |2\vec{a} - 3\vec{b}| \) given that \( |\vec{a} + \vec{b}| = 1 \) and both \( \vec{a} \) and \( \vec{b} \) are unit vectors.

### Step-by-step Solution:

1. **Square the Magnitude Equation**:
\[ |\vec{a} + \vec{b}|^2 = 1^2 \]
This gives us:
\[ |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 1 \]

2. **Use the fact that \( \vec{a} \) and \( \vec{b} \) are unit vectors**:
Since \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \), we have:
\[ 1^2 + 1^2 + 2\vec{a} \cdot \vec{b} = 1 \]
Simplifying this gives:
\[ 1 + 1 + 2\vec{a} \cdot \vec{b} = 1 \]
\[ 2 + 2\vec{a} \cdot \vec{b} = 1 \]

3. **Solve for \( \vec{a} \cdot \vec{b} \)**:
Rearranging gives:
\[ 2\vec{a} \cdot \vec{b} = 1 - 2 \]
\[ 2\vec{a} \cdot \vec{b} = -1 \]
\[ \vec{a} \cdot \vec{b} = -\frac{1}{2} \]

4. **Find the magnitude of \( |2\vec{a} - 3\vec{b}| \)**:
We start by squaring the magnitude:
\[ |2\vec{a} - 3\vec{b}|^2 = (2\vec{a} - 3\vec{b}) \cdot (2\vec{a} - 3\vec{b}) \]
Expanding this gives:
\[ = 4|\vec{a}|^2 + 9|\vec{b}|^2 - 2(2\vec{a} \cdot 3\vec{b}) \]
Substituting the values:
\[ = 4(1) + 9(1) - 2(6\vec{a} \cdot \vec{b}) \]
\[ = 4 + 9 - 12(\vec{a} \cdot \vec{b}) \]
We already found \( \vec{a} \cdot \vec{b} = -\frac{1}{2} \):
\[ = 4 + 9 - 12\left(-\frac{1}{2}\right) \]
\[ = 4 + 9 + 6 \]
\[ = 19 \]

5. **Take the square root to find the magnitude**:
\[ |2\vec{a} - 3\vec{b}| = \sqrt{19} \]

### Final Answer:
\[ |2\vec{a} - 3\vec{b}| = \sqrt{19} \]