The adjacent sides of a parallelogram are `vec(P)=2i-3j+k` and `vec(Q)=-2i+4j-k`. What is the area of the parallelogram ?

Options

A. 4 units
B. 7 units
C. `sqrt(5)` units
D. `sqrt(8)` units

Answer

To find the area of the parallelogram formed by the vectors \(\vec{P} = 2\hat{i} - 3\hat{j} + \hat{k}\) and \(\vec{Q} = -2\hat{i} + 4\hat{j} - \hat{k}\), we will use the formula for the area of a parallelogram, which is given by the magnitude of the cross product of the two vectors.

### Step-by-Step Solution:

1. **Write down the vectors**:
\[ \vec{P} = 2\hat{i} - 3\hat{j} + 1\hat{k} \]
\[ \vec{Q} = -2\hat{i} + 4\hat{j} - 1\hat{k} \]

2. **Set up the cross product using the determinant**:
The cross product \(\vec{P} \times \vec{Q}\) can be calculated using the determinant of a matrix formed by the unit vectors and the components of \(\vec{P}\) and \(\vec{Q}\):
\[ \vec{P} \times \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ -2 & 4 & -1 \end{vmatrix} \]

3. **Calculate the determinant**:
Expanding the determinant:
\[ \vec{P} \times \vec{Q} = \hat{i} \begin{vmatrix} -3 & 1 \\ 4 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ -2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} \]

- For the \(\hat{i}\) component:
\[ \begin{vmatrix} -3 & 1 \\ 4 & -1 \end{vmatrix} = (-3)(-1) - (1)(4) = 3 - 4 = -1 \]

- For the \(\hat{j}\) component:
\[ \begin{vmatrix} 2 & 1 \\ -2 & -1 \end{vmatrix} = (2)(-1) - (1)(-2) = -2 + 2 = 0 \]

- For the \(\hat{k}\) component:
\[ \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} = (2)(4) - (-3)(-2) = 8 - 6 = 2 \]

Combining these results, we have:
\[ \vec{P} \times \vec{Q} = -1\hat{i} - 0\hat{j} + 2\hat{k} = -\hat{i} + 2\hat{k} \]

4. **Find the magnitude of the cross product**:
The magnitude of the vector \(\vec{P} \times \vec{Q}\) is given by:
\[ |\vec{P} \times \vec{Q}| = \sqrt{(-1)^2 + 0^2 + 2^2} = \sqrt{1 + 0 + 4} = \sqrt{5} \]

5. **Conclusion**:
The area of the parallelogram is equal to the magnitude of the cross product:
\[ \text{Area} = |\vec{P} \times \vec{Q}| = \sqrt{5} \]

### Final Answer:
The area of the parallelogram is \(\sqrt{5}\) square units.

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