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

Options

A. `0^(@)`
B. `180^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

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

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

This condition implies that the vectors \(\vec{P}\) and \(\vec{Q}\) are aligned in the same direction. We will derive the angle between the vectors step by step.

### Step 1: Understanding the Condition
The equation states that the magnitude of the resultant vector \(\vec{P} + \vec{Q}\) is equal to the sum of the magnitudes of the individual vectors. This is a special case that occurs when the two vectors are in the same direction.

### Step 2: Using the Law of Cosines
We can express the magnitude of the resultant vector \(\vec{R} = \vec{P} + \vec{Q}\) using the law of cosines:
\[ |\vec{R}|^2 = |\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cos(\theta) \]
where \(\theta\) is the angle between the vectors \(\vec{P}\) and \(\vec{Q}\).

### Step 3: Squaring Both Sides
Given that \( |\vec{R}| = |\vec{P}| + |\vec{Q}| \), we square both sides:
\[ |\vec{R}|^2 = (|\vec{P}| + |\vec{Q}|)^2 \]
Expanding the right-hand side:
\[ |\vec{R}|^2 = |\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \]

### Step 4: Setting the Equations Equal
Now we set the two expressions for \( |\vec{R}|^2 \) equal to each other:
\[ |\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cos(\theta) = |\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \]

### Step 5: Simplifying the Equation
Subtract \( |\vec{P}|^2 + |\vec{Q}|^2 \) from both sides:
\[ 2 |\vec{P}| |\vec{Q}| \cos(\theta) = 2 |\vec{P}| |\vec{Q}| \]

### Step 6: Dividing by \(2 |\vec{P}| |\vec{Q}|\)
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: Finding the Angle
The equation \( \cos(\theta) = 1 \) implies:
\[ \theta = 0^\circ \]

### Conclusion
Thus, the angle between the vectors \(\vec{P}\) and \(\vec{Q}\) is \(0^\circ\).

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