Consider the following for the next two items that follow
Let `vec(a).vec(b) and vec(c )` be three vectors such that `vec(a) +vec(b) + vec(c )= vec(0) and |vec(a)|= 10 |vec(b)| =6 and |vec(c )|=14`.
What is the angle between `vec(a) and vec(b)`?

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `75^(@)`

Answer

To find the angle between the vectors \(\vec{a}\) and \(\vec{b}\), we start with the given information:

1. \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\)
2. \(|\vec{a}| = 10\), \(|\vec{b}| = 6\), and \(|\vec{c}| = 14\)

### Step 1: Rearranging the equation
From the equation \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\), we can rearrange it to express \(\vec{c}\):
\[ \vec{c} = -(\vec{a} + \vec{b}) \]

### Step 2: Taking the modulus
Taking the modulus of both sides gives us:
\[ |\vec{c}| = |-(\vec{a} + \vec{b})| = |\vec{a} + \vec{b}| \]
Since the modulus of a vector is always non-negative, we can write:
\[ |\vec{c}| = |\vec{a} + \vec{b}| \]

### Step 3: Squaring both sides
Now, squaring both sides yields:
\[ |\vec{c}|^2 = |\vec{a} + \vec{b}|^2 \]
Using the property of modulus, we have:
\[ |\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b} \]

### Step 4: Substituting known values
Substituting the known magnitudes:
\[ 14^2 = 10^2 + 6^2 + 2 |\vec{a}| |\vec{b}| \cos \theta \]
Calculating the squares:
\[ 196 = 100 + 36 + 2 \cdot 10 \cdot 6 \cos \theta \]
This simplifies to:
\[ 196 = 136 + 120 \cos \theta \]

### Step 5: Isolating \(\cos \theta\)
Now, we isolate \(\cos \theta\):
\[ 196 - 136 = 120 \cos \theta \]
\[ 60 = 120 \cos \theta \]
\[ \cos \theta = \frac{60}{120} = \frac{1}{2} \]

### Step 6: Finding the angle
The angle \(\theta\) whose cosine is \(\frac{1}{2}\) is:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

Thus, the angle between \(\vec{a}\) and \(\vec{b}\) is \(60^\circ\).

### Summary of Steps:
1. Rearranged the vector equation.
2. Took the modulus of both sides.
3. Squared both sides to eliminate the modulus.
4. Substituted the known magnitudes into the equation.
5. Isolated \(\cos \theta\).
6. Found the angle using the inverse cosine function.