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

Options

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

Answer

To solve the problem, we need to find the angle between the vectors \(\vec{P}\) and \(\vec{Q}\) given the equation:

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

### Step 1: Understand the given equation
We start with the equation \( |\vec{P} + \vec{Q}| = |\vec{P}| - |\vec{Q}| \). This implies that the magnitude of the vector sum of \(\vec{P}\) and \(\vec{Q}\) is equal to the difference of their magnitudes.

### Step 2: Square both sides
To eliminate the absolute values, we square both sides of the equation:

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

### Step 3: Expand both sides
Now we expand both sides:

- Left side:
\[ |\vec{P} + \vec{Q}|^2 = |\vec{P}|^2 + |\vec{Q}|^2 + 2|\vec{P}||\vec{Q}|\cos\theta \]

- Right side:
\[ (|\vec{P}| - |\vec{Q}|)^2 = |\vec{P}|^2 - 2|\vec{P}||\vec{Q}| + |\vec{Q}|^2 \]

### Step 4: Set the expanded forms equal to each other
Now we set the left side equal to the right side:

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

### Step 5: Simplify the equation
We can simplify this equation by canceling out \( |\vec{P}|^2 \) and \( |\vec{Q}|^2 \) from both sides:

\[ 2|\vec{P}||\vec{Q}|\cos\theta = -2|\vec{P}||\vec{Q}| \]

### Step 6: Divide by \(2|\vec{P}||\vec{Q}|\) (assuming they are not zero)
Assuming \( |\vec{P}| \) and \( |\vec{Q}| \) are not zero, we can divide both sides by \(2|\vec{P}||\vec{Q}|\):

\[ \cos\theta = -1 \]

### Step 7: Find the angle \(\theta\)
The angle \(\theta\) for which \(\cos\theta = -1\) is:

\[ \theta = \cos^{-1}(-1) = 180^\circ \]

### Conclusion
Thus, the angle between the vectors \(\vec{P}\) and \(\vec{Q}\) is \(180^\circ\).

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