If `vecP` and `vecQ` denote the sides of parallelogram and its area is `(1)/(2)`PQ, then the angle between `vecP` and `vecQ `is

Options

A. `0^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Explanation

To solve the problem, we need to find the angle between the vectors \(\vec{P}\) and \(\vec{Q}\) given that the area of the parallelogram formed by these vectors is \(\frac{1}{2} PQ\).

### Step-by-Step Solution:

1. **Understanding the Area of a Parallelogram**:
The area \(A\) of a parallelogram formed by two vectors \(\vec{P}\) and \(\vec{Q}\) is given by the magnitude of the cross product of the two vectors:
\[ A = |\vec{P} \times \vec{Q}| \]

2. **Using the Formula for the Cross Product**:
The magnitude of the cross product can also be expressed in terms of the magnitudes of the vectors and the sine of the angle \(\theta\) between them:
\[ |\vec{P} \times \vec{Q}| = |\vec{P}| |\vec{Q}| \sin \theta \]

3. **Setting Up the Equation**:
According to the problem, the area of the parallelogram is given as:
\[ A = \frac{1}{2} PQ \]
where \(P = |\vec{P}|\) and \(Q = |\vec{Q}|\). Therefore, we can equate the two expressions for the area:
\[ |\vec{P}| |\vec{Q}| \sin \theta = \frac{1}{2} PQ \]

4. **Cancelling Out the Magnitudes**:
Since \(P = |\vec{P}|\) and \(Q = |\vec{Q}|\), we can simplify the equation:
\[ PQ \sin \theta = \frac{1}{2} PQ \]
Dividing both sides by \(PQ\) (assuming \(PQ \neq 0\)):
\[ \sin \theta = \frac{1}{2} \]

5. **Finding the Angle**:
We need to find the angle \(\theta\) such that \(\sin \theta = \frac{1}{2}\). The angle that satisfies this equation is:
\[ \theta = \sin^{-1}\left(\frac{1}{2}\right) \]
From trigonometric values, we know:
\[ \theta = 30^\circ \]

6. **Conclusion**:
Therefore, the angle between the vectors \(\vec{P}\) and \(\vec{Q}\) is:
\[ \theta = 30^\circ \]

### Final Answer:
The angle between \(\vec{P}\) and \(\vec{Q}\) is \(30^\circ\).

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