The area of a parallelogram constructed on the vectors `a +3b` and `3a +b` where `|a| = |b| =1` and the angle between a and b is `60^(@)`, is 4sq. Units. True or False

Answer

To determine whether the area of the parallelogram constructed on the vectors \( \mathbf{a} + 3\mathbf{b} \) and \( 3\mathbf{a} + \mathbf{b} \) is 4 square units, we will follow these steps:

### Step 1: Identify the vectors
Let:
- \( \mathbf{u} = \mathbf{a} + 3\mathbf{b} \)
- \( \mathbf{v} = 3\mathbf{a} + \mathbf{b} \)

### Step 2: Calculate the cross product \( \mathbf{u} \times \mathbf{v} \)
The area \( A \) of the parallelogram formed by vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by the magnitude of the cross product:
\[ A = |\mathbf{u} \times \mathbf{v}| \]

### Step 3: Expand the cross product
Using the distributive property of the cross product:
\[ \mathbf{u} \times \mathbf{v} = (\mathbf{a} + 3\mathbf{b}) \times (3\mathbf{a} + \mathbf{b}) \]
Expanding this gives:
\[ = \mathbf{a} \times (3\mathbf{a}) + \mathbf{a} \times \mathbf{b} + 3\mathbf{b} \times (3\mathbf{a}) + 3\mathbf{b} \times \mathbf{b} \]
Since \( \mathbf{a} \times \mathbf{a} = \mathbf{0} \) and \( \mathbf{b} \times \mathbf{b} = \mathbf{0} \), we simplify:
\[ = 0 + \mathbf{a} \times \mathbf{b} + 9(\mathbf{b} \times \mathbf{a}) + 0 \]
Using the property \( \mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b}) \):
\[ = \mathbf{a} \times \mathbf{b} - 9(\mathbf{a} \times \mathbf{b}) = -8(\mathbf{a} \times \mathbf{b}) \]

### Step 4: Calculate the magnitude of the cross product
The magnitude of the cross product \( |\mathbf{a} \times \mathbf{b}| \) is given by:
\[ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin(\theta) \]
Where \( \theta = 60^\circ \) and \( |\mathbf{a}| = |\mathbf{b}| = 1 \):
\[ |\mathbf{a} \times \mathbf{b}| = 1 \cdot 1 \cdot \sin(60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

### Step 5: Substitute back to find the area
Now substituting back into the area formula:
\[ A = |-8(\mathbf{a} \times \mathbf{b})| = 8 |\mathbf{a} \times \mathbf{b}| = 8 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3} \]

### Step 6: Compare with the given area
The calculated area \( 4\sqrt{3} \) is not equal to 4 square units. Therefore, the statement is **False**.