If `|vec(a)|=sqrt(2), |vec(b)|=sqrt(3) and |vec(a)+vec(b)|=sqrt(6)`, then what is `|vec(a)-vec(b)|` equal to ?

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to find the magnitude of the vector \(\vec{a} - \vec{b}\) given the magnitudes of \(\vec{a}\), \(\vec{b}\), and \(\vec{a} + \vec{b}\).

### Step-by-Step Solution:

1. **Given Values**:
- \(|\vec{a}| = \sqrt{2}\)
- \(|\vec{b}| = \sqrt{3}\)
- \(|\vec{a} + \vec{b}| = \sqrt{6}\)

2. **Using the Formula for Magnitude of Sum of Vectors**:
We know that:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 |\vec{a}| |\vec{b}| \cos \theta \]
where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\).

3. **Substituting the Known Values**:
\[ |\vec{a} + \vec{b}|^2 = (\sqrt{6})^2 = 6 \]
\[ |\vec{a}|^2 = (\sqrt{2})^2 = 2 \]
\[ |\vec{b}|^2 = (\sqrt{3})^2 = 3 \]
Thus, we have:
\[ 6 = 2 + 3 + 2 \cdot \sqrt{2} \cdot \sqrt{3} \cdot \cos \theta \]

4. **Simplifying the Equation**:
\[ 6 = 5 + 2 \cdot \sqrt{6} \cdot \cos \theta \]
\[ 1 = 2 \cdot \sqrt{6} \cdot \cos \theta \]
\[ \cos \theta = \frac{1}{2\sqrt{6}} \]

5. **Using the Formula for Magnitude of Difference of Vectors**:
We also have:
\[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2 |\vec{a}| |\vec{b}| \cos \theta \]

6. **Substituting the Known Values**:
\[ |\vec{a} - \vec{b}|^2 = 2 + 3 - 2 \cdot \sqrt{2} \cdot \sqrt{3} \cdot \frac{1}{2\sqrt{6}} \]
\[ = 5 - \frac{2 \cdot \sqrt{6}}{2\sqrt{6}} = 5 - 1 = 4 \]

7. **Finding the Magnitude**:
\[ |\vec{a} - \vec{b}| = \sqrt{4} = 2 \]

Thus, the magnitude \(|\vec{a} - \vec{b}|\) is equal to \(2\).