If `|vec(a) | = 3, |vec(b)|= 4 and |vec(a)- vec(b)|=5`, then what si the value of `|vec(a) + vec(b)|`=?

Options

A. 8
B. 6
C. `5 sqrt2`
D. 5

Answer

To solve the problem, we need to find the value of \( |\vec{a} + \vec{b}| \) given the magnitudes of the vectors and the magnitude of their difference.

1. **Given Information**:
- \( |\vec{a}| = 3 \)
- \( |\vec{b}| = 4 \)
- \( |\vec{a} - \vec{b}| = 5 \)

2. **Using the formula for the magnitude of the difference of two vectors**:
\[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2 |\vec{a}| |\vec{b}| \cos \theta \]
where \( \theta \) is the angle between the vectors \( \vec{a} \) and \( \vec{b} \).

3. **Substituting the known values**:
\[ 5^2 = 3^2 + 4^2 - 2 \cdot 3 \cdot 4 \cos \theta \]
\[ 25 = 9 + 16 - 24 \cos \theta \]

4. **Simplifying the equation**:
\[ 25 = 25 - 24 \cos \theta \]
\[ 0 = -24 \cos \theta \]
This implies:
\[ \cos \theta = 0 \]
Therefore, \( \theta = 90^\circ \), meaning \( \vec{a} \) and \( \vec{b} \) are perpendicular.

5. **Using the formula for the magnitude of the sum of two vectors**:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 |\vec{a}| |\vec{b}| \cos \theta \]

6. **Substituting the known values**:
Since \( \cos \theta = 0 \):
\[ |\vec{a} + \vec{b}|^2 = 3^2 + 4^2 + 2 \cdot 3 \cdot 4 \cdot 0 \]
\[ |\vec{a} + \vec{b}|^2 = 9 + 16 + 0 \]
\[ |\vec{a} + \vec{b}|^2 = 25 \]

7. **Taking the square root**:
\[ |\vec{a} + \vec{b}| = \sqrt{25} = 5 \]

Thus, the value of \( |\vec{a} + \vec{b}| \) is \( 5 \).