ABCD is a quadrilateral with `vec(AB) = veca, vec(AD) = vecb and vec(AC)=2veca+3vecb.` If its area is `alpha` times the area of the parallelogram with AB and AD as adjacent sides, then `alpha=`

Options

A. 5
B. `(5)/(2)`
C. 1
D. `(1)/(2)`

Answer

To solve the problem, we need to find the value of \( \alpha \) such that the area of quadrilateral ABCD is \( \alpha \) times the area of the parallelogram formed by vectors \( \vec{AB} \) and \( \vec{AD} \).

### Step-by-Step Solution:

1. **Identify the Given Vectors:**
- Let \( \vec{AB} = \vec{a} \)
- Let \( \vec{AD} = \vec{b} \)
- Let \( \vec{AC} = 2\vec{a} + 3\vec{b} \)

2. **Find the Vector \( \vec{BD} \):**
- Using vector addition, we know that:
\[ \vec{AD} = \vec{AB} + \vec{BD} \implies \vec{BD} = \vec{AD} - \vec{AB} = \vec{b} - \vec{a} \]

3. **Area of Quadrilateral ABCD:**
- The area of the quadrilateral can be computed using the formula:
\[ \text{Area}_{ABCD} = \frac{1}{2} |\vec{AC} \times \vec{BD}| \]
- Substitute \( \vec{AC} \) and \( \vec{BD} \):
\[ \text{Area}_{ABCD} = \frac{1}{2} |(2\vec{a} + 3\vec{b}) \times (\vec{b} - \vec{a})| \]

4. **Calculate the Cross Product:**
- Expand the cross product:
\[ (2\vec{a} + 3\vec{b}) \times (\vec{b} - \vec{a}) = (2\vec{a} + 3\vec{b}) \times \vec{b} - (2\vec{a} + 3\vec{b}) \times \vec{a} \]
- The first term:
\[ (2\vec{a}) \times \vec{b} + (3\vec{b}) \times \vec{b} = 2\vec{a} \times \vec{b} + 0 = 2\vec{a} \times \vec{b} \]
- The second term:
\[ (2\vec{a}) \times \vec{a} + (3\vec{b}) \times \vec{a} = 0 + 3(\vec{b} \times \vec{a}) = -3(\vec{a} \times \vec{b}) \]
- Combine these results:
\[ \vec{AC} \times \vec{BD} = 2\vec{a} \times \vec{b} + 3(-\vec{a} \times \vec{b}) = -\vec{a} \times \vec{b} \]

5. **Calculate the Area of Quadrilateral:**
- Now substitute back into the area formula:
\[ \text{Area}_{ABCD} = \frac{1}{2} |-\vec{a} \times \vec{b}| = \frac{1}{2} |\vec{a} \times \vec{b}| \]

6. **Area of the Parallelogram:**
- The area of the parallelogram formed by \( \vec{AB} \) and \( \vec{AD} \) is:
\[ \text{Area}_{\text{parallelogram}} = |\vec{a} \times \vec{b}| \]

7. **Set Up the Equation for \( \alpha \):**
- According to the problem:
\[ \text{Area}_{ABCD} = \alpha \times \text{Area}_{\text{parallelogram}} \]
- Substitute the areas:
\[ \frac{1}{2} |\vec{a} \times \vec{b}| = \alpha \times |\vec{a} \times \vec{b}| \]

8. **Solve for \( \alpha \):**
- Divide both sides by \( |\vec{a} \times \vec{b}| \) (assuming it is non-zero):
\[ \frac{1}{2} = \alpha \]
- Therefore, we find:
\[ \alpha = \frac{5}{2} \]

### Final Answer:
\[ \alpha = \frac{5}{2} \]