If `vec(a) + vec(b) + vec( c ) = vec(0), |vec(a) | = sqrt( 37), | vec(b)| =3` and `| vec( c)| =4`, then the angle between `vec(b)` and `vec(c )` is

Options

A. `90^(@)`
B. `60^(@)`
C. `45^(@)`
D. `30^(@)`

Answer

To solve the problem, we start with the given equation and the magnitudes of the vectors:

1. **Given Equation**:
\[ \vec{a} + \vec{b} + \vec{c} = \vec{0} \]
This implies:
\[ \vec{a} = -(\vec{b} + \vec{c}) \]

2. **Magnitudes**:
- \(|\vec{a}| = \sqrt{37}\)
- \(|\vec{b}| = 3\)
- \(|\vec{c}| = 4\)

3. **Squaring the Magnitudes**:
We can square both sides of the equation:
\[ |\vec{a}|^2 = |\vec{b} + \vec{c}|^2 \]
Expanding the right side using the formula \(|\vec{x} + \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2\vec{x} \cdot \vec{y}\):
\[ |\vec{a}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2 \vec{b} \cdot \vec{c} \]

4. **Substituting the Values**:
Now substituting the known values:
\[ 37 = 3^2 + 4^2 + 2 \vec{b} \cdot \vec{c} \]
Calculating the squares:
\[ 37 = 9 + 16 + 2 \vec{b} \cdot \vec{c} \]
Simplifying:
\[ 37 = 25 + 2 \vec{b} \cdot \vec{c} \]
Rearranging gives:
\[ 2 \vec{b} \cdot \vec{c} = 37 - 25 \]
\[ 2 \vec{b} \cdot \vec{c} = 12 \]
Thus:
\[ \vec{b} \cdot \vec{c} = 6 \]

5. **Using the Dot Product**:
The dot product can also be expressed in terms of the magnitudes and the cosine of the angle \(\theta\) between the vectors:
\[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta \]
Substituting the magnitudes:
\[ 6 = 3 \cdot 4 \cos \theta \]
Simplifying:
\[ 6 = 12 \cos \theta \]
Thus:
\[ \cos \theta = \frac{6}{12} = \frac{1}{2} \]

6. **Finding the Angle**:
The angle \(\theta\) can be found using the inverse cosine:
\[ \theta = \cos^{-1} \left( \frac{1}{2} \right) \]
This gives:
\[ \theta = 60^\circ \]

**Final Answer**: The angle between \(\vec{b}\) and \(\vec{c}\) is \(60^\circ\).

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