Question

Show that the area of a parallelogram whose diagonals are represented by \( \vec{a} \) and \( \vec{b} \) is given by \[ \text{Area} = \frac{1}{2} | \vec{a} \times \vec{b} |. \] Also, find the area of a parallelogram whose diagonals are \( 2\hat{i} - \hat{j} + \hat{k} \) and \( \hat{i} + 3\hat{j} - \hat{k} \).

Hint:

The area of a parallelogram formed by vectors can be computed using the magnitude of the cross product of the two vectors.

Answer 1

To find the area of a parallelogram, we use the formula for the magnitude of the cross product of two vectors: \[ \text{Area} = \frac{1}{2} | \vec{a} \times \vec{b} |. \] Now, given the diagonals \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 3\hat{j} - \hat{k} \), we compute the cross product: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{vmatrix} = \hat{i} \begin{vmatrix} -1 & 1
3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1
1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1
1 & 3 \end{vmatrix}. \] After calculating the determinants, we get: \[ \vec{a} \times \vec{b} = 2\hat{i} + 3\hat{j} + 7\hat{k}. \] Thus, the magnitude of the cross product is: \[ |\vec{a} \times \vec{b}| = \sqrt{2^2 + 3^2 + 7^2} = \sqrt{4 + 9 + 49} = \sqrt{62}. \] Therefore, the area of the parallelogram is: \[ \text{Area} = \frac{1}{2} \sqrt{62}. \]