Question

The area of a parallelogram whose diagonals are the vectors \(2\bar{a} - \bar{b}\) and \(4\bar{a} - 5\bar{b}\), where \(\bar{a}\) and \(\bar{b}\) are unit vectors forming an angle of \(45^\circ\) is

• A. \(3\sqrt{2}\) sq. units
• B. \(\frac{3}{\sqrt{2}}\) sq. units
• C. \(\sqrt{2}\) sq. units
• D. \(\frac{\sqrt{2}}{3}\) sq. units

Hint:

For parallelograms, if diagonals are given instead of sides, use: \[ \text{Area}=\frac12 |\vec{d}_1\times \vec{d}_2| \] This is much faster than finding side vectors first.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The area of a parallelogram can be calculated if its diagonals \(\bar{d_1}\) and \(\bar{d_2}\) are known.
The formula involves the cross product of these diagonal vectors.
Step 2: Key Formula or Approach:
1. Area of parallelogram \(= \frac{1}{2} |\bar{d_1} \times \bar{d_2}|\).
2. \(|\bar{a} \times \bar{b}| = |\bar{a}| |\bar{b}| \sin \theta\).
3. Vector cross product properties: \(\bar{a} \times \bar{a} = 0\) and \(\bar{b} \times \bar{a} = -(\bar{a} \times \bar{b})\).
Step 3: Detailed Explanation:
Let \(\bar{d_1} = 2\bar{a} - \bar{b}\) and \(\bar{d_2} = 4\bar{a} - 5\bar{b}\).
Calculate the cross product \(\bar{d_1} \times \bar{d_2}\):
\[ (2\bar{a} - \bar{b}) \times (4\bar{a} - 5\bar{b}) = 8(\bar{a} \times \bar{a}) - 10(\bar{a} \times \bar{b}) - 4(\bar{b} \times \bar{a}) + 5(\bar{b} \times \bar{b}) \] Using the properties \(\bar{a} \times \bar{a} = 0, \bar{b} \times \bar{b} = 0\) and \(\bar{b} \times \bar{a} = - \bar{a} \times \bar{b}\):
\[ = 0 - 10(\bar{a} \times \bar{b}) + 4(\bar{a} \times \bar{b}) + 0 = -6(\bar{a} \times \bar{b}) \] The magnitude is \(|-6| |\bar{a} \times \bar{b}| = 6 |\bar{a} \times \bar{b}|\).
Area \(= \frac{1}{2} \cdot 6 |\bar{a} \times \bar{b}| = 3 |\bar{a} \times \bar{b}|\).
Since \(\bar{a}\) and \(\bar{b}\) are unit vectors, \(|\bar{a}| = 1, |\bar{b}| = 1\). The angle is \(45^\circ\).
\[ \text{Area} = 3 \cdot 1 \cdot 1 \cdot \sin 45^\circ = 3 \cdot \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}} \] Step 4: Final Answer:
The area is \(\frac{3}{\sqrt{2}}\) sq. units.