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

Options

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

Explanation

To find the angle between vectors **b** and **c** given that **a + b + c = 0**, and the magnitudes of the vectors are |a| = √37, |b| = 3, and |c| = 4, we can follow these steps:

### Step-by-Step Solution:

1. **Express vector a in terms of b and c:**
Given the equation **a + b + c = 0**, we can rearrange it to express **a** as:
\[ a = -b - c \]

2. **Square the magnitudes:**
We know that the magnitude of vector **a** can be expressed as:
\[ |a|^2 = |b + c|^2 \]
Using the property of magnitudes, we can expand this:
\[ |a|^2 = |b|^2 + |c|^2 + 2(b \cdot c) \]

3. **Substitute the known values:**
We know:
- |a| = √37, so |a|² = 37
- |b| = 3, so |b|² = 9
- |c| = 4, so |c|² = 16

Substituting these values into the equation gives:
\[ 37 = 9 + 16 + 2(b \cdot c) \]

4. **Simplify the equation:**
Combine the constants on the right side:
\[ 37 = 25 + 2(b \cdot c) \]
Now, isolate the dot product:
\[ 37 - 25 = 2(b \cdot c) \]
\[ 12 = 2(b \cdot c) \]
\[ b \cdot c = 6 \]

5. **Use the dot product to find the angle:**
The dot product can also be expressed as:
\[ b \cdot c = |b| |c| \cos \theta \]
Substituting the magnitudes:
\[ 6 = 3 \cdot 4 \cdot \cos \theta \]
\[ 6 = 12 \cos \theta \]
Now, solve for cos θ:
\[ \cos \theta = \frac{6}{12} = \frac{1}{2} \]

6. **Find the angle θ:**
The angle whose cosine is \( \frac{1}{2} \) is:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \text{ radians} = 60^\circ \]

### Conclusion:
The angle between vectors **b** and **c** is \( 60^\circ \).