The area of parallelogram constructed on the vector `a =m + 2n and b =2m +n` where m and n are unit vectors forming an angle of `30^(@)` is

Options

A. `3//2`
B. `5//2`
C. `7//2`
D. none of these

Answer

To find the area of the parallelogram constructed on the vectors \( \mathbf{a} = \mathbf{m} + 2\mathbf{n} \) and \( \mathbf{b} = 2\mathbf{m} + \mathbf{n} \), we will use the formula for the area of a parallelogram defined by two vectors, which is given by the magnitude of their cross product:

\[ \text{Area} = |\mathbf{a} \times \mathbf{b}| \]

### Step 1: Write down the vectors
We have:
\[ \mathbf{a} = \mathbf{m} + 2\mathbf{n} \]
\[ \mathbf{b} = 2\mathbf{m} + \mathbf{n} \]

### Step 2: Compute the cross product \( \mathbf{a} \times \mathbf{b} \)
Using the distributive property of the cross product:
\[ \mathbf{a} \times \mathbf{b} = (\mathbf{m} + 2\mathbf{n}) \times (2\mathbf{m} + \mathbf{n}) \]

Expanding this:
\[ \mathbf{a} \times \mathbf{b} = \mathbf{m} \times (2\mathbf{m}) + \mathbf{m} \times \mathbf{n} + 2\mathbf{n} \times (2\mathbf{m}) + 2\mathbf{n} \times \mathbf{n} \]

### Step 3: Simplify the cross product
1. \( \mathbf{m} \times (2\mathbf{m}) = 0 \) (since the cross product of any vector with itself is zero)
2. \( 2\mathbf{n} \times \mathbf{n} = 0 \) (same reason)
3. \( \mathbf{m} \times \mathbf{n} = \mathbf{m} \times \mathbf{n} \) (remains as is)
4. \( 2\mathbf{n} \times (2\mathbf{m}) = 4(\mathbf{n} \times \mathbf{m}) = -4(\mathbf{m} \times \mathbf{n}) \) (using the property \( \mathbf{n} \times \mathbf{m} = -(\mathbf{m} \times \mathbf{n}) \))

Putting it all together:
\[ \mathbf{a} \times \mathbf{b} = 0 + \mathbf{m} \times \mathbf{n} - 4(\mathbf{m} \times \mathbf{n}) + 0 = -3(\mathbf{m} \times \mathbf{n}) \]

### Step 4: Find the magnitude of the cross product
The area is given by:
\[ \text{Area} = |\mathbf{a} \times \mathbf{b}| = |-3(\mathbf{m} \times \mathbf{n})| = 3|\mathbf{m} \times \mathbf{n}| \]

### Step 5: Calculate \( |\mathbf{m} \times \mathbf{n}| \)
Using the formula for the magnitude of the cross product:
\[ |\mathbf{m} \times \mathbf{n}| = |\mathbf{m}||\mathbf{n}|\sin(\theta) \]
where \( \theta = 30^\circ \) and both \( \mathbf{m} \) and \( \mathbf{n} \) are unit vectors (\( |\mathbf{m}| = 1 \) and \( |\mathbf{n}| = 1 \)):
\[ |\mathbf{m} \times \mathbf{n}| = 1 \cdot 1 \cdot \sin(30^\circ) = \sin(30^\circ) = \frac{1}{2} \]

### Step 6: Substitute back to find the area
Now substituting back:
\[ \text{Area} = 3 \cdot \frac{1}{2} = \frac{3}{2} \]

### Final Answer
Thus, the area of the parallelogram is:
\[ \frac{3}{2} \]

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