Question

Let ABCD be a quadrilateral with \(\overline{AB} = \overset{⃗}{a}\), \(\overline{AD} = \overset{⃗}{b}\) and \(\overline{AC} = 3\overset{⃗}{a}+2\overset{⃗}{b}\). If its area is \(α\) times the area of the parallelogram with AB, AD as adjacent sides, then the value of \(α\) is equal to

• A. \(\frac{3}{2}\)
• B. \(\frac{5}{2}\)
• C. \(\frac{1}{2}\)
• D. \(1\)

Hint:

Area of a quadrilateral is half the magnitude of the cross product of its diagonals.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For quadrilateral ABCD, area = \(\frac12|\overrightarrow{AC} \times \overrightarrow{BD}|\). The parallelogram on AB and AD has area \(|\vec a \times \vec b|\).

Step 2: Find BD
\(\overrightarrow{BD} = \overrightarrow{AD} - \overrightarrow{AB} = \vec b - \vec a\).

Step 3: Cross product
\[ \overrightarrow{AC} \times \overrightarrow{BD} = (3\vec a + 2\vec b) \times (\vec b - \vec a) = 3(\vec a \times \vec b) - 2(\vec b \times \vec a) = 3(\vec a \times \vec b) + 2(\vec a \times \vec b) = 5(\vec a \times \vec b) \]
\[ \text{Area} = \frac12\cdot 5|\vec a \times \vec b| = \frac52|\vec a \times \vec b| \]
So \(\alpha = \frac52\), option (B).

Final Answer:
\(\alpha = \frac52\), option (B). \[ \boxed{\frac{5}{2}} \]