Question

Adjacent sides of a parallelogram are given by \[ \mathbf{a} = 6 \hat{i} - \hat{j} + 5 \hat{k}, \quad \mathbf{b} = \hat{i} + 5 \hat{j} - 2 \hat{k} \] Find the area of the parallelogram.

Hint:

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

Answer 1

Step 1: Formula for the area of the parallelogram.

The area \( A \) of a parallelogram with adjacent sides \( \mathbf{a} \) and \( \mathbf{b} \) is given by the magnitude of the cross product of the two vectors:
\[ A = |\mathbf{a} \times \mathbf{b}| \]
Step 2: Finding the cross product.

We are given:
\[ \mathbf{a} = 6 \hat{i} - \hat{j} + 5 \hat{k}, \quad \mathbf{b} = \hat{i} + 5 \hat{j} - 2 \hat{k} \]
The cross product \( \mathbf{a} \times \mathbf{b} \) is calculated as the determinant:
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & -1 & 5 \\ 1 & 5 & -2 \end{vmatrix} \]
Expanding the determinant:
\[ \mathbf{a} \times \mathbf{b} = \hat{i} \begin{vmatrix} -1 & 5 \\ 5 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 6 & 5 \\ 1 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 6 & -1 \\ 1 & 5 \end{vmatrix} \]
Now, calculate each \( 2 \times 2 \) determinant:

\[ \begin{vmatrix} -1 & 5 \\ 5 & -2 \end{vmatrix} = (-1)(-2) - (5)(5) = 2 - 25 = -23 \]
\[ \begin{vmatrix} 6 & 5 \\ 1 & -2 \end{vmatrix} = (6)(-2) - (5)(1) = -12 - 5 = -17 \]
\[ \begin{vmatrix} 6 & -1 \\ 1 & 5 \end{vmatrix} = (6)(5) - (-1)(1) = 30 + 1 = 31 \]
Step 3: Calculating the cross product.

Substituting these values:
\[ \mathbf{a} \times \mathbf{b} = -23 \hat{i} + 17 \hat{j} + 31 \hat{k} \]
Step 4: Finding the magnitude of the cross product.

The magnitude is:
\[ |\mathbf{a} \times \mathbf{b}| = \sqrt{(-23)^2 + 17^2 + 31^2} \]
\[ |\mathbf{a} \times \mathbf{b}| = \sqrt{529 + 289 + 961} = \sqrt{1779} \]
Thus, the area of the parallelogram is:
\[ A = \sqrt{1779} \approx 42.2 \, \text{square units} \]
Step 5: Conclusion.

Thus, the area of the parallelogram is approximately \( 42.2 \, \text{square units} \).