Question

Find the area of the parallelogram whose adjacent sides are represented by the vectors \( \vec{a} = 3\hat{i} + \hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 2\hat{k} \).

Hint:

The area of a parallelogram can be found by calculating the magnitude of the cross product of the vectors representing its adjacent sides.

Answer 1

Step 1: The area of the parallelogram is given by the magnitude of the cross product of the vectors \( \vec{a} \) and \( \vec{b} \): \[ \text{Area} = |\vec{a} \times \vec{b}| \] 

Step 2: Compute the cross product \( \vec{a} \times \vec{b} \): \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 
3 & 1 & 2 
1 & 2 & -2 \end{vmatrix} \] 

Step 3: Expand the determinant: \[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 1 & 2 
2 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 2 
1 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 1 
1 & 2 \end{vmatrix} \] \[ = \hat{i} (-2 - 4) - \hat{j} (-6 - 2) + \hat{k} (6 - 1) \] \[ = -6 \hat{i} + 8 \hat{j} + 5 \hat{k} \] Step 4: Find the magnitude of the cross product: \[ |\vec{a} \times \vec{b}| = \sqrt{(-6)^2 + 8^2 + 5^2} = \sqrt{36 + 64 + 25} = \sqrt{125} = 10 \] \bigskip