Find the area $A$ of the parallelogram shown in Figure 49, which has vertices at the points $P_{1}=(0, 0, 0)$, $P_{2}=(-1, 2, 4)$, $P_{3}=(1, 1, 8)$, and $P_{4}=(2, -1, 4)$.

Figure 49

Solution The area $A$ of a parallelogram is $\Vert \mathbf{v} \times \mathbf{w}\Vert$, where $\mathbf{v}$ and $\mathbf{w}$ are two adjacent sides of the parallelogram. Be careful! Not all pairs of vertices give a side. For example, $\overrightarrow{P_{1}~P_{3}}$ is not a side; it is a diagonal of the parallelogram. It is helpful to sketch the parallelogram before choosing $\mathbf{v}$ and $\mathbf{w}$. Choose $$\begin{equation*} \mathbf{v}=\overrightarrow{P_{1}~P_{2}}= \left\langle -1,\,2,4\right\rangle =-\mathbf{i}+2\mathbf{j}+4\mathbf{k} \end{equation*}$$

729

and $$\begin{equation*} \hbox{}\mathbf{w}=\overrightarrow{P_{1}~P_{4}} =\left\langle 2,-1,4\right\rangle =2\mathbf{i}-\mathbf{j}+4\mathbf{k} \end{equation*}$$

Then $$\begin{equation*} \mathbf{v}\times \mathbf{w}= \left|\begin{array}{r@{\quad}r@{\quad}r} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] -1 & 2 & 4 \\[3pt] 2 & -1 & 4 \end{array}\right| =12\mathbf{i}+12\mathbf{j}-3\mathbf{k} \end{equation*}$$

The area of the parallelogram is $$\left\Vert \mathbf{v}\times \mathbf{w}\right\Vert = \sqrt{144+144+9}= \sqrt{297} \approx 17.234~\hbox{square units}$$