Problem Statement
Find the area of a parallelogram whose diagonals are determined by the vectors $\vec{d}_1 = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{d}_2 = \hat{i} - 3\hat{j} + 4\hat{k}$.
Verified Solution & Marking Scheme
Compute Cross Product d₁ × d₂
$\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix} = \hat{i}(4 - 6) - \hat{j}(12 - (-2)) + \hat{k}(-9 - 1) = -2\hat{i} - 14\hat{j} - 10\hat{k}$
Calculate Magnitude |d₁ × d₂|
$|\vec{d}_1 \times \vec{d}_2| = \sqrt{(-2)^2 + (-14)^2 + (-10)^2} = \sqrt{4 + 196 + 100} = \sqrt{300} = 10\sqrt{3}$
Apply Area Formula Area = 1/2 |d₁ × d₂|
$\text{Area} = \frac{1}{2} |\vec{d}_1 \times \vec{d}_2| = \frac{1}{2}(10\sqrt{3}) = 5\sqrt{3} \text{ sq. units}$