Given that `vecx + vecy = vecz` and `x + Y =z`, Then the angle between `vecx` and `vecy` is

Options

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

Answer

To find the angle between the vectors \(\vec{x}\) and \(\vec{y}\) given that \(\vec{x} + \vec{y} = \vec{z}\) and \(x + y = z\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Vector Addition**:
- We have two vectors \(\vec{x}\) and \(\vec{y}\) that add up to \(\vec{z}\). This can be visualized as forming a triangle where \(\vec{z}\) is the resultant vector.

2. **Using the Triangle Law of Vector Addition**:
- According to the triangle law, the magnitude of the resultant vector \(\vec{z}\) is equal to the magnitude of the sum of the two vectors \(\vec{x}\) and \(\vec{y}\).

3. **Applying the Law of Cosines**:
- The law of cosines states that for any triangle with sides \(a\), \(b\), and \(c\), where \(c\) is the side opposite to angle \(\theta\):
\[ c^2 = a^2 + b^2 - 2ab \cos(\theta) \]
- In our case, we can set \(a = |\vec{x}|\), \(b = |\vec{y}|\), and \(c = |\vec{z}|\).

4. **Setting Up the Equation**:
- Since \(\vec{x} + \vec{y} = \vec{z}\), we have:
\[ |\vec{z}| = |\vec{x} + \vec{y}| \]
- Therefore, we can write:
\[ |\vec{z}|^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2 |\vec{x}| |\vec{y}| \cos(\theta) \]

5. **Substituting the Given Condition**:
- Given that \(x + y = z\), we can also express this in terms of magnitudes:
\[ z^2 = x^2 + y^2 + 2xy \cos(\theta) \]
- Since \(|\vec{z}| = |\vec{x} + \vec{y}|\), we can equate the two expressions:
\[ z^2 = x^2 + y^2 \]

6. **Equating the Two Expressions**:
- From the two equations we have:
\[ x^2 + y^2 + 2xy \cos(\theta) = x^2 + y^2 \]

7. **Simplifying the Equation**:
- By cancelling \(x^2 + y^2\) from both sides, we get:
\[ 2xy \cos(\theta) = 0 \]

8. **Finding the Angle**:
- Since \(xy \neq 0\) (assuming both vectors are non-zero), we can conclude:
\[ \cos(\theta) = 0 \]
- This implies that \(\theta = 0^\circ\).

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