Question

Find the area of a parallelogram whose adjacent sides are given by the vectors \( \mathbf{a} = \hat{i} - \hat{j} + 3 \hat{k} \) and \( \mathbf{b} = 2 \hat{i} - 7 \hat{j} + \hat{k} \).

Hint:

The area of a parallelogram can be found using the magnitude of the cross product of the adjacent vectors.

Answer 1

Step 1: Formula for the area of a parallelogram. 

The area of a parallelogram formed by two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by the magnitude of their cross product: \[ \text{Area} = |\mathbf{a} \times \mathbf{b}|. \]

Step 2: Compute the cross product \( \mathbf{a} \times \mathbf{b} \). 

The vectors \( \mathbf{a} = \hat{i} - \hat{j} + 3 \hat{k} \) and \( \mathbf{b} = 2 \hat{i} - 7 \hat{j} + \hat{k} \) are: \[ \mathbf{a} = \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix}, \mathbf{b} = \begin{pmatrix} 2 \\ -7 \\ 1 \end{pmatrix}. \] Using the cross product formula: \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix}. \] Expanding the determinant: \[ \mathbf{a} \times \mathbf{b} = \hat{i} \begin{vmatrix} -1 & 3 \\ -7 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 2 & -7 \end{vmatrix}. \] Calculating the 2x2 determinants: \[ \mathbf{a} \times \mathbf{b} = \hat{i}((-1)(1) - (3)(-7)) - \hat{j}((1)(1) - (3)(2)) + \hat{k}((1)(-7) - (-1)(2)), \] \[ \mathbf{a} \times \mathbf{b} = \hat{i}( -1 + 21) - \hat{j}(1 - 6) + \hat{k}(-7 + 2), \] \[ \mathbf{a} \times \mathbf{b} = \hat{i}(20) - \hat{j}(-5) + \hat{k}(-5), \] \[ \mathbf{a} \times \mathbf{b} = 20\hat{i} + 5\hat{j} - 5\hat{k}. \]

Step 3: Find the magnitude. 

Now, calculate the magnitude of the cross product: \[ |\mathbf{a} \times \mathbf{b}| = \sqrt{(20)^2 + (5)^2 + (-5)^2} = \sqrt{400 + 25 + 25} = \sqrt{450}. \]

Step 4: Conclusion. 

Thus, the area of the parallelogram is \( \sqrt{450} \), which simplifies to: \[ \boxed{15 \sqrt{2}}. \]