## Thomas' Calculus 13th Edition

$\sqrt{129}$
The first job is to find pairs of parallel vectors, so we are able to tell which two vectors that originate from a point (say, A) define the parallelogram. The three vectors with initial point A are $\overrightarrow{AB} = \langle 3-0, 2-0,4-0 \rangle= \langle 3, 2, 4 \rangle$ $\overrightarrow{AC} = \langle 5-0, 1-0,4-0 \rangle= \langle 5, 1, 4 \rangle$ $\overrightarrow{AD} = \langle 2, -1, 0 \rangle$ We search for parallel vectors $\overrightarrow{BC} = \langle 5-3, 1-2,4-4 \rangle= \langle 2, -1, 0 \rangle,$ which is parallel to $\overrightarrow{AD}$, and has the same orientation. $\overrightarrow{CD} = \langle 2-5, -1-1,0-4 \rangle= \langle-3, -2, -4 \rangle,$ So, $\overrightarrow{DC}$ is parallel to $\overrightarrow{AB}$. We have our pairs of parallel vectors. The vectors defining the paralelogram are $\overrightarrow{AB}$ and $\overrightarrow{AD}$ ($\overrightarrow{AC}$ is the diagonal). Area =$| \overrightarrow{AB}\times \overrightarrow{AD} |$ $\overrightarrow{AB}\times\overrightarrow{AD}$=$\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 3 & 2 & 4\\ 2 & -1 & 0 \end{array}\right|=\langle 0+4 ,\ -(0-8) ,\ -3-4 \rangle$ $=\langle 4 ,\ 8 ,\ -7 \rangle$ Area =$| \overrightarrow{AB}\times\overrightarrow{AD} |$=$\sqrt{16+64+49}=\sqrt{129}$