If ` |vecP + vecQ| = |vecP -vecQ|` the angle between ` vecP and vecQ` is

Options

A. ` 0^(@)`
B. `90^(@)`
C. `180^(@)`
D. `45^(@)`

Answer

To solve the problem, we need to analyze the equation given:

\[ |\vec{P} + \vec{Q}| = |\vec{P} - \vec{Q}| \]

### Step 1: Square both sides of the equation

We start by squaring both sides to eliminate the modulus:

\[ |\vec{P} + \vec{Q}|^2 = |\vec{P} - \vec{Q}|^2 \]

### Step 2: Expand both sides

Using the property of dot products, we can expand both sides:

\[ (\vec{P} + \vec{Q}) \cdot (\vec{P} + \vec{Q}) = (\vec{P} - \vec{Q}) \cdot (\vec{P} - \vec{Q}) \]

This expands to:

\[ \vec{P} \cdot \vec{P} + 2 \vec{P} \cdot \vec{Q} + \vec{Q} \cdot \vec{Q} = \vec{P} \cdot \vec{P} - 2 \vec{P} \cdot \vec{Q} + \vec{Q} \cdot \vec{Q} \]

### Step 3: Simplify the equation

Now, we can simplify the equation by canceling out the common terms:

\[ \vec{P} \cdot \vec{P} + \vec{Q} \cdot \vec{Q} + 2 \vec{P} \cdot \vec{Q} = \vec{P} \cdot \vec{P} + \vec{Q} \cdot \vec{Q} - 2 \vec{P} \cdot \vec{Q} \]

Subtracting \(\vec{P} \cdot \vec{P} + \vec{Q} \cdot \vec{Q}\) from both sides gives:

\[ 2 \vec{P} \cdot \vec{Q} + 2 \vec{P} \cdot \vec{Q} = 0 \]

### Step 4: Solve for the dot product

This simplifies to:

\[ 4 \vec{P} \cdot \vec{Q} = 0 \]

Thus, we have:

\[ \vec{P} \cdot \vec{Q} = 0 \]

### Step 5: Determine the angle between the vectors

The dot product of two vectors is given by:

\[ \vec{P} \cdot \vec{Q} = |\vec{P}| |\vec{Q}| \cos \theta \]

Since \(\vec{P} \cdot \vec{Q} = 0\), we can conclude that:

\[ |\vec{P}| |\vec{Q}| \cos \theta = 0 \]

This implies that \(\cos \theta = 0\), which occurs when:

\[ \theta = 90^\circ \]

### Conclusion

The angle between the vectors \(\vec{P}\) and \(\vec{Q}\) is:

\[ \theta = 90^\circ \]

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