Question

Let \(\vec{u}\) and \(\vec{v}\) be unit vectors such that the angle between them is \(\theta =\frac{\pi }{6}\). If the adjacent sides of a parallelogram are represented by the vectors \(\vec{a}=2\vec{u}+3\vec{v}\) and \(\vec{b}=\vec{u}-\vec{v}\), then the area of the parallelogram is

• A. \(5\)
• B. \(\frac{5\sqrt{3}}{2}\)
• C. \(2.5\)
• D. \(16\sqrt{3}\)

Hint:

Expand the cross product of the two sides and use the magnitude of u cross v.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The area of a parallelogram with adjacent sides \(\vec{a}\) and \(\vec{b}\) is \(|\vec{a}\times\vec{b}|\).

Step 2: Expand the cross product
\(\vec{a}\times\vec{b}=(2\vec{u}+3\vec{v})\times(\vec{u}-\vec{v})\).
\(=-2(\vec{u}\times\vec{v})+3(\vec{v}\times\vec{u})=-2(\vec{u}\times\vec{v})-3(\vec{u}\times\vec{v})=-5(\vec{u}\times\vec{v})\).

Step 3: Find the magnitude
\(|\vec{u}\times\vec{v}|=\sin\frac{\pi}{6}=\frac{1}{2}\) since both are unit vectors.

Step 4: Area
\(\text{Area}=5\times\frac{1}{2}=2.5\) square units.

Final Answer:
The area of the parallelogram is 2.5 square units. \[ \boxed{\text{(C) }2.5} \]