Discussion
The vectors\(-\hat{i} + \frac{1}{4}\hat{j} + 2\hat{k}\) and\(\hat{i} + \frac{1}{4}\hat{j} + 2\hat{k}\),are the adjacent sides of a parallelogram. The area of the parallelogram is
Answer: D
The area is \(\frac{\sqrt{65}}{2}\).
The area of a parallelogram with adjacent sides \(\vec{a}\) and \(\vec{b}\) is \(|\vec{a} \times \vec{b}|\). With \(\vec{a} = (-1, \frac{1}{4}, 2)\) and \(\vec{b} = (1, \frac{1}{4}, 2)\):
\[ \begin{aligned} \vec{a} \times \vec{b} &= \left(\tfrac{1}{2} - \tfrac{1}{2}\right)\hat{i} - (-2 - 2)\hat{j} \\ &\quad + \left(-\tfrac{1}{4} - \tfrac{1}{4}\right)\hat{k} \\ &= 0\hat{i} + 4\hat{j} - \tfrac{1}{2}\hat{k} \\ |\vec{a} \times \vec{b}| &= \sqrt{16 + \tfrac{1}{4}} \\ &= \frac{\sqrt{65}}{2} \end{aligned} \]
\(\frac{\sqrt{65}}{4}\) is the area of the triangle on those sides, half the parallelogram. \(\sqrt{65}\) forgets that \(16 + \frac{1}{4} = \frac{65}{4}\).
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