If `vecA+ vecB = vecC and A+B+C=0`, then the angle between `vecA and vecB` is :

Options

A. 0
B. `(pi)/(4)`
C. `(pi)/(2)`
D. `pi`

Explanation

To solve the problem, we need to find the angle between the vectors \(\vec{A}\) and \(\vec{B}\) given the equations:

1. \(\vec{A} + \vec{B} = \vec{C}\)
2. \(\vec{A} + \vec{B} + \vec{C} = 0\)

### Step-by-Step Solution:

**Step 1: Rewrite the second equation.**
From the second equation, we can express \(\vec{C}\) in terms of \(\vec{A}\) and \(\vec{B}\):
\[ \vec{C} = -(\vec{A} + \vec{B}) \]

**Step 2: Substitute \(\vec{C}\) into the first equation.**
Now, substitute this expression for \(\vec{C}\) into the first equation:
\[ \vec{A} + \vec{B} = -(\vec{A} + \vec{B}) \]

**Step 3: Take the magnitude of both sides.**
Taking the magnitude of both sides gives us:
\[ |\vec{A} + \vec{B}| = |-\vec{C}| = |\vec{C}| \]
Since the magnitude of a vector is always non-negative, we can write:
\[ |\vec{A} + \vec{B}| = |\vec{C}| \]

**Step 4: Use the formula for the magnitude of the sum of two vectors.**
The magnitude of the sum of two vectors can be expressed using the cosine of the angle \(\theta\) between them:
\[ |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta} \]
Let \(A = |\vec{A}|\) and \(B = |\vec{B}|\). Thus:
\[ |\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB\cos\theta} \]

**Step 5: Substitute \(|\vec{C}|\) into the equation.**
From the earlier step, we can also express \(|\vec{C}|\):
\[ |\vec{C}| = |-\vec{C}| = |\vec{A} + \vec{B}| \]
Thus, we have:
\[ \sqrt{A^2 + B^2 + 2AB\cos\theta} = |\vec{C}| \]

**Step 6: Square both sides.**
Squaring both sides gives:
\[ A^2 + B^2 + 2AB\cos\theta = C^2 \]

**Step 7: Substitute for \(C^2\).**
Since \(\vec{C} = -(\vec{A} + \vec{B})\), we can write:
\[ C^2 = |\vec{C}|^2 = |\vec{A} + \vec{B}|^2 = A^2 + B^2 + 2AB\cos\theta \]

**Step 8: Set the equations equal and simplify.**
Setting the two expressions for \(C^2\) equal gives:
\[ A^2 + B^2 + 2AB\cos\theta = A^2 + B^2 + 2AB \]
Subtracting \(A^2 + B^2\) from both sides:
\[ 2AB\cos\theta = 2AB \]

**Step 9: Divide by \(2AB\) (assuming \(AB \neq 0\)).**
\[ \cos\theta = 1 \]

**Step 10: Solve for \(\theta\).**
The angle \(\theta\) that satisfies \(\cos\theta = 1\) is:
\[ \theta = 0^\circ \]

### Final Answer:
The angle between \(\vec{A}\) and \(\vec{B}\) is \(0^\circ\).

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