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

Concept:
If \(\vec{d}_1\) and \(\vec{d}_2\) are the diagonals of a parallelogram, then: \[ \text{Area}=\frac12 |\vec{d}_1 \times \vec{d}_2| \] Also, for unit vectors \(\bar{a}\) and \(\bar{b}\) making angle \(45^\circ\), \[ |\bar{a}\times \bar{b}| = |\bar{a}||\bar{b}|\sin 45^\circ = \frac{1}{\sqrt{2}} \] ip

Step 1: Write the diagonals.
\[ \vec{d}_1 = 2\bar{a}-\bar{b} \] \[ \vec{d}_2 = 4\bar{a}-5\bar{b} \] ip

Step 2: Find the cross product.
\[ \vec{d}_1 \times \vec{d}_2 = (2\bar{a}-\bar{b})\times(4\bar{a}-5\bar{b}) \] Expand: \[ = 8(\bar{a}\times \bar{a}) -10(\bar{a}\times \bar{b}) -4(\bar{b}\times \bar{a}) +5(\bar{b}\times \bar{b}) \] Since \[ \bar{a}\times\bar{a}=0,\qquad \bar{b}\times\bar{b}=0,\qquad \bar{b}\times\bar{a}=-(\bar{a}\times\bar{b}) \] we get: \[ \vec{d}_1 \times \vec{d}_2 = -10(\bar{a}\times\bar{b}) +4(\bar{a}\times\bar{b}) = -6(\bar{a}\times\bar{b}) \] So, \[ |\vec{d}_1 \times \vec{d}_2| = 6|\bar{a}\times\bar{b}| \] ip

Step 3: Use the angle between \(\bar{a}\) and \(\bar{b}\).
\[ |\bar{a}\times\bar{b}|=\sin45^\circ=\frac{1}{\sqrt{2}} \] Hence, \[ |\vec{d}_1 \times \vec{d}_2| = 6\cdot \frac{1}{\sqrt{2}} = \frac{6}{\sqrt{2}} \] ip

Step 4: Find the area of the parallelogram.
\[ \text{Area}=\frac12 |\vec{d}_1\times \vec{d}_2| = \frac12 \cdot \frac{6}{\sqrt{2}} = \frac{3}{\sqrt{2}} \] ip Hence, the correct answer is:
\[ \boxed{(B)\ \frac{3}{\sqrt{2}} \text{ sq. units}} \]