Question

Find the area of a parallelogram whose adjacent sides are \( \vec{a} = 3\hat{i} + \hat{j} + 4\hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \).

Hint:

The magnitude of the cross product of two vectors gives the area of the parallelogram they form.

Answer 1

The area of a parallelogram is given by: \[ |\vec{a} \times \vec{b}|. \] Computing the cross product: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & 1 & 4
1 & -1 & 1 \end{vmatrix} \] Expanding along the first row: \[ = \hat{i} \begin{vmatrix} 1 & 4
-1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 4
1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 1
1 & -1 \end{vmatrix}. \] \[ = \hat{i} (1 \times 1 - (-1) \times 4) - \hat{j} (3 \times 1 - 4 \times 1) + \hat{k} (3 \times (-1) - 1 \times 1). \] \[ = \hat{i} (1 + 4) - \hat{j} (3 - 4) + \hat{k} (-3 -1). \] \[ = 5\hat{i} + \hat{j} - 4\hat{k}. \] \[ |\vec{a} \times \vec{b}| = \sqrt{(5)^2 + (1)^2 + (-4)^2} = \sqrt{25 + 1 + 16} = \sqrt{42}. \]