Question

The following two vectors are adjacent sides of a parallelogram: \(\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}\) and \(\vec{B} = 5\hat{i} - 4\hat{k}\).
The magnitude of area of the parallelogram is ________ (Rounded off to two decimal places)

Hint:

Find the cross product \(\vec{A} \times \vec{B}\) and take its magnitude.

Answer 1

Correct Answer: 19.44

Step 1: Set up the cross product.
The area of a parallelogram with adjacent sides \(\vec{A}\) and \(\vec{B}\) equals \(|\vec{A} \times \vec{B}|\).
Here \(\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}\) and \(\vec{B} = 5\hat{i} + 0\hat{j} - 4\hat{k}\).
Write the determinant form:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 5 & 0 & -4 \end{vmatrix} \]

Step 2: Expand the determinant.
The \(\hat{i}\) component is \((3)(-4) - (-1)(0) = -12 - 0 = -12\).
The \(\hat{j}\) component is \(-[(2)(-4) - (-1)(5)] = -[-8 + 5] = 3\).
The \(\hat{k}\) component is \((2)(0) - (3)(5) = 0 - 15 = -15\).
So \(\vec{A} \times \vec{B} = -12\hat{i} + 3\hat{j} - 15\hat{k}\).

Step 3: Find the magnitude.
\[ |\vec{A} \times \vec{B}| = \sqrt{(-12)^2 + 3^2 + (-15)^2} = \sqrt{144 + 9 + 225} = \sqrt{378} \] Working this out gives \(\sqrt{378} = 19.44\).

Final Answer:
The magnitude of the area of the parallelogram is \[ \boxed{19.44} \]