Question

The diagonals of a parallelogram are given by \( \mathbf{a} = 2 \hat{i} - \hat{j} + \hat{k} \) and \( \mathbf{b} = \hat{i} + 3 \hat{j} - \hat{k}\) . Find the area of the parallelogram.

Hint:

The area of a parallelogram formed by two vectors is given by the magnitude of their cross product.

Answer 1

The area of the parallelogram is given by the magnitude of the cross product of the diagonals: \[ \text{Area} = |\mathbf{a} \times \mathbf{b}| \] First, compute the cross product \( \mathbf{a} \times \mathbf{b} \). Let \( \mathbf{a} = 2 \hat{i} - \hat{j} + \hat{k} \) and \( \mathbf{b} = \hat{i} + 3 \hat{j} - \hat{k} \), so the determinant form of the cross product is: \[ \mathbf{a} \times \mathbf{b} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{matrix} \right| \] Expanding the determinant: \[ \mathbf{a} \times \mathbf{b} = \hat{i} \left[ (-1)(-1) - (1)(3) \right] - \hat{j} \left[ (2)(-1) - (1)(1) \right] + \hat{k} \left[ (2)(3) - (-1)(1) \right] \] \[ = \hat{i} (1 - 3) - \hat{j} (-2 - 1) + \hat{k} (6 + 1) = -2 \hat{i} + 3 \hat{j} + 7 \hat{k} \] Now, find the magnitude of the cross product: \[ |\mathbf{a} \times \mathbf{b}| = \sqrt{(-2)^2 + 3^2 + 7^2} = \sqrt{4 + 9 + 49} = \sqrt{62} \] So, the area of the parallelogram is \( 7 \).