Question

The area of a parallelogram whose diagonals are given by $ \vec{u} + \vec{v} $ and $ \vec{v} + \vec{w} $, where: 
$ \vec{u} = 2\hat{i} - 3\hat{j} + \hat{k}, \quad \vec{v} = -\hat{i} + \hat{k}, \quad \vec{w} = 2\hat{j} - \hat{k} $ is:

• A. \( \sqrt{14} \) sq. unit
• B. \( \sqrt{21} \) sq. unit
• C. \( \frac{1}{2} \sqrt{21} \) sq. unit
• D. \( \frac{1}{2} \sqrt{14} \) sq. unit

Hint:

To find the area of a parallelogram, compute the cross product of its diagonals, then take half of the magnitude of the cross product.

Answer 1

The Correct Option is B

We are given that the area of a parallelogram is half the magnitude of the cross product of its diagonals. Here, the diagonals are \( \vec{u} + \vec{v} \) and \( \vec{v} + \vec{w} \).
Step 1:
First, compute the diagonals: \[ \vec{u} + \vec{v} = (2\hat{i} - 3\hat{j} + \hat{k}) + (-\hat{i} + \hat{k}) = \hat{i} - 3\hat{j} + 2\hat{k} \] \[ \vec{v} + \vec{w} = (-\hat{i} + \hat{k}) + (2\hat{j} - \hat{k}) = -\hat{i} + 2\hat{j} \] 
Step 2:
Now, find the cross product of these two vectors: \[ (\vec{u} + \vec{v}) \times (\vec{v} + \vec{w}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 1 & -3 & 2 \\ -1 & 2 & 0 \end{vmatrix} \] Expanding the determinant: \[ = \hat{i} \begin{vmatrix} -3 & 2 2 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ -1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -3 \\ -1 & 2 \end{vmatrix} \] \[ = \hat{i}((-3)(0) - (2)(2)) - \hat{j}((1)(0) - (2)(-1)) + \hat{k}((1)(2) - (-3)(-1)) \] \[ = \hat{i}(-4) - \hat{j}(2) + \hat{k}(-1) \] \[ = (-4, 2, -1) \] 
Step 3:
Find the magnitude of the cross product: \[ |\vec{u} + \vec{v} \times \vec{v} + \vec{w}| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21} \] 
Step 4:
The area of the parallelogram is half of the magnitude of the cross product: \[ \text{Area} = \frac{1}{2} \times \sqrt{21} = \sqrt{21} \text{ sq. units.} \]