Question

The total area of a parallelogram constructed with adjacent vector sides given by \( \vec{a} = \hat{i} - \hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} - 7\hat{j} + \hat{k} \) is equal to:

• A. \( 15\sqrt{2} \)
• B. \( 15 \)
• C. \( \sqrt{35} \)
• D. \( 0 \)

Hint:

If a question asks for the area of a triangle sharing those same adjacent vector sides instead of a parallelogram, simply evaluate the cross product magnitude and divide it by two (\( \text{Area} = \frac{1}{2}|\vec{a} \times \vec{b}| \)).

Answer 1

The Correct Option is A

Concept: The geometric area of a parallelogram spanned by two adjacent vector sides \( \vec{a} \) and \( \vec{b} \) is equal to the vector magnitude of their cross product: \[ \text{Area} = \left| \vec{a} \times \vec{b} \right| \] This cross product vector is calculated using a standard \( 3 \times 3 \) matrix determinant layout.

Step 1: Compute the vector cross product \( \vec{a} \times \vec{b} \).
Set up the component determinant using unit vectors: \[ \vec{a} \times \vec{b} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} 1 & -1 & 3 2 & -7 & 1 \end{matrix} \right| \] Expanding the determinant along the top row: \[ = \hat{i}\left[(-1)(1) - (3)(-7)\right] - \hat{j}\left[(1)(1) - (3)(2)\right] + \hat{k}\left[(1)(-7) - (-1)(2)\right] \] \[ = \hat{i}(-1 + 21) - \hat{j}(1 - 6) + \hat{k}(-7 + 2) \] \[ = 20\hat{i} + 5\hat{j} - 5\hat{k} \]

Step 2: Calculate the magnitude of this cross product vector.
The magnitude gives the scalar area of the bounded parallelogram region: \[ \text{Area} = \sqrt{(20)^2 + (5)^2 + (-5)^2} = \sqrt{400 + 25 + 25} = \sqrt{450} \] Simplify the radical expression by factoring out perfect squares: \[ \sqrt{450} = \sqrt{225 \times 2} = 15\sqrt{2} \text{ square units} \]