If $\vec a = 2\hat j - \hat k, \vec b = 2\hat i-3\hat j +\hat k$ and $\vec c = -\hat i+\hat k$ are three vectors, then the area (in sq. units) of the parallelogram whose diagonals are $(\vec b+\vec c)$ and $(\vec a + \vec c)$ is

$\frac{1}{2}\sqrt{21}$

The correct answer is Option (2) → $\frac{1}{2}\sqrt{21}$

Given:

$\vec{a} = 2\hat{j} - \hat{k}$

$\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$

$\vec{c} = -\hat{i} + \hat{k}$

Diagonals of the parallelogram:

$\vec{d_1} = \vec{b} + \vec{c} = (2 - 1)\hat{i} + (-3)\hat{j} + (1 + 1)\hat{k} = \hat{i} - 3\hat{j} + 2\hat{k}$

$\vec{d_2} = \vec{a} + \vec{c} = (-1)\hat{i} + 2\hat{j} + (1 - 1)\hat{k} = -\hat{i} + 2\hat{j}$

The area of the parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is:

$A = \frac{1}{2} \left| \vec{d_1} \times \vec{d_2} \right|$

Compute $\vec{d_1} \times \vec{d_2}$:

$\vec{d_1} = \hat{i} - 3\hat{j} + 2\hat{k}$

$\vec{d_2} = -\hat{i} + 2\hat{j}$

$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0 \\ \end{vmatrix} = \hat{i}( -3 \cdot 0 - 2 \cdot 2 ) - \hat{j}( 1 \cdot 0 - 2 \cdot (-1) ) + \hat{k}( 1 \cdot 2 - (-3) \cdot (-1) )$

$= \hat{i}(-4) - \hat{j}(2) + \hat{k}(2 - 3)$

$= -4\hat{i} - 2\hat{j} - \hat{k}$

Magnitude of the cross product:

$\left| \vec{d_1} \times \vec{d_2} \right| = \sqrt{(-4)^2 + (-2)^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$

Area = $\frac{1}{2} \sqrt{21}$ sq. units