## Calculus (3rd Edition)

$a\sqrt {c^{2}+b^{2}}$
The area of the parallelogram is given as: $||⟨a,0,0⟩\times ⟨0,b,c⟩||$ Thus we have: $⟨a,0,0⟩\times ⟨0,b,c⟩=\begin{vmatrix}\textbf{i}&\textbf{j}&\textbf{k}\\a&0&0\\0&b&c\end{vmatrix}$ $=\textbf{i}(0-0)-\textbf{j}(ac-0)+\textbf{k}(ab-0)$ $=-ac\textbf{j}+ab\textbf{k}$ $||⟨a,0,0⟩\times ⟨0,b,c⟩||=$ $\sqrt {(-ac)^{2}+(ab)^{2}}=\sqrt {a^{2}c^{2}+a^{2}b^{2}}$ $=\sqrt {a^{2}(c^{2}+b^{2})}=a\sqrt {c^{2}+b^{2}}$