What can be the angle between `(vec(P) + vec(Q))` and `(vec(P) - vec(Q))` ?

Options

A. `90^(@)`
B. `0^(@)` only
C. any angle between `0^(@)` and `180^(@)`
D. `180^(@)`

Answer

To determine the angle between the vectors \( \vec{P} + \vec{Q} \) and \( \vec{P} - \vec{Q} \), we can use the concept of the dot product of vectors. Here’s a step-by-step solution:

### Step 1: Define the angle
Let \( \alpha \) be the angle between the vectors \( \vec{P} + \vec{Q} \) and \( \vec{P} - \vec{Q} \).

### Step 2: Use the dot product formula
The dot product of two vectors can be expressed as:
\[ (\vec{A} \cdot \vec{B}) = |\vec{A}| |\vec{B}| \cos(\theta) \]
In our case, we have:
\[ (\vec{P} + \vec{Q}) \cdot (\vec{P} - \vec{Q}) = |\vec{P} + \vec{Q}| |\vec{P} - \vec{Q}| \cos(\alpha) \]

### Step 3: Calculate the left-hand side
Now, we can expand the left-hand side:
\[ (\vec{P} + \vec{Q}) \cdot (\vec{P} - \vec{Q}) = \vec{P} \cdot \vec{P} - \vec{P} \cdot \vec{Q} + \vec{Q} \cdot \vec{P} - \vec{Q} \cdot \vec{Q} \]
This simplifies to:
\[ |\vec{P}|^2 - |\vec{Q}|^2 \]
since \( \vec{P} \cdot \vec{Q} = \vec{Q} \cdot \vec{P} \).

### Step 4: Calculate the magnitudes
Next, we need to find the magnitudes of \( \vec{P} + \vec{Q} \) and \( \vec{P} - \vec{Q} \):
\[ |\vec{P} + \vec{Q}| = \sqrt{|\vec{P}|^2 + |\vec{Q}|^2 + 2(\vec{P} \cdot \vec{Q})} \]
\[ |\vec{P} - \vec{Q}| = \sqrt{|\vec{P}|^2 + |\vec{Q}|^2 - 2(\vec{P} \cdot \vec{Q})} \]

### Step 5: Substitute into the equation
Now substituting these back into our equation gives:
\[ |\vec{P}|^2 - |\vec{Q}|^2 = |\vec{P} + \vec{Q}| |\vec{P} - \vec{Q}| \cos(\alpha) \]

### Step 6: Analyze the cases
1. **Case 1**: If \( |\vec{P}| > |\vec{Q}| \), then \( |\vec{P}|^2 - |\vec{Q}|^2 > 0 \) which implies \( \cos(\alpha) > 0 \) and thus \( \alpha \) is between \( 0^\circ \) and \( 90^\circ \).
2. **Case 2**: If \( |\vec{P}| < |\vec{Q}| \), then \( |\vec{P}|^2 - |\vec{Q}|^2 < 0 \) which implies \( \cos(\alpha) < 0 \) and thus \( \alpha \) is between \( 90^\circ \) and \( 180^\circ \).

### Conclusion
Combining both cases, we conclude that the angle \( \alpha \) can be any angle between \( 0^\circ \) and \( 180^\circ \).

### Final Answer
The angle between \( \vec{P} + \vec{Q} \) and \( \vec{P} - \vec{Q} \) can be any angle between \( 0^\circ \) and \( 180^\circ \).

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