Question:medium

ABCD is a quadrilateral with $\overline{AB} = \overline{a}, \overline{AD} = \overline{b}$ and $\overline{AC} = 2\overline{a} + 3\overline{b}$. If its area is $\alpha$ times the area of the parallelogram with AB, AD as adjacent sides, then the value of $\alpha$ is

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Area of $\triangle$ with vectors $\vec{u}, \vec{v}$ is $\frac{1}{2}|\vec{u} \times \vec{v}|$.
Updated On: May 14, 2026
  • $\frac{1}{2}$
  • $\frac{5}{2}$
  • $\frac{3}{2}$
  • 2
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
We need to relate the area of a general quadrilateral $ABCD$ to the area of a parallelogram defined by its sides $\vec{a}$ and $\vec{b}$.
Step 2: Key Formula or Approach:
Area of parallelogram with adjacent sides $\vec{a}, \vec{b}$ is $|\vec{a} \times \vec{b}|$.
Area of quadrilateral $ABCD = \frac{1}{2} |\vec{AC} \times \vec{BD}|$.
Step 3: Detailed Explanation:
1. We are given $\vec{AB} = \vec{a}$, $\vec{AD} = \vec{b}$, and $\vec{AC} = 2\vec{a} + 3\vec{b}$.
2. Find vector $\vec{BD}$:
$\vec{BD} = \vec{AD} - \vec{AB} = \vec{b} - \vec{a}$.
3. Calculate the area of quadrilateral $ABCD$:
$Area_{quad} = \frac{1}{2} |(2\vec{a} + 3\vec{b}) \times (\vec{b} - \vec{a})|$
$Area_{quad} = \frac{1}{2} |2(\vec{a} \times \vec{b}) - 2(\vec{a} \times \vec{a}) + 3(\vec{b} \times \vec{b}) - 3(\vec{b} \times \vec{a})|$
Since $\vec{v} \times \vec{v} = \vec{0}$ and $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$:
$Area_{quad} = \frac{1}{2} |2(\vec{a} \times \vec{b}) - 0 + 0 + 3(\vec{a} \times \vec{b})| = \frac{1}{2} |5(\vec{a} \times \vec{b})| = \frac{5}{2} |\vec{a} \times \vec{b}|$.
4. We are given $Area_{quad} = \alpha \times Area_{para}$, and $Area_{para} = |\vec{a} \times \vec{b}|$.
Comparing terms: $\alpha = \frac{5}{2}$.
Step 4: Final Answer:
The value of $\alpha$ is $\frac{5}{2}$.
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