## Thomas' Calculus 13th Edition

$a.\displaystyle \quad \frac{\sqrt{14}}{2}$ $b.\displaystyle \quad \frac{\sqrt{14}}{7} {\bf i}+ \frac{3\sqrt{14}}{14} {\bf j}+ \frac{\sqrt{14}}{14} {\bf k}$
$(a)$ ${\bf u}= \overrightarrow{PQ} = \langle 0+2, 1-2, -1-0 \rangle= \langle 2, -1, -1 \rangle$ ${\bf v}= \overrightarrow{PR} = \langle-1+2, 2-2, -2-0 \rangle= \langle 1, 0, -2 \rangle$ We find the area of the parallelogram $|{\bf u}\times{\bf v}|$ The area of the triangle is half the area of the parallelogram. $A=\displaystyle \frac{1}{2}\cdot|{\bf u}\times{\bf v}|$ ${\bf u}\times{\bf v}=\left|\begin{array}{rrr} {\bf i} & {\bf j} & {\bf k}\\ 2 & -1 & -1\\ 1 & 0 & -2 \end{array}\right|=(2-0){\bf i}-(-4+1){\bf j}+(0+1){\bf k}$ $= 2{\bf i}+3{\bf j}+{\bf k}$ $|{\bf u}\times{\bf v}|=\sqrt{4+9+1}=\sqrt{14}$ $A=\displaystyle \frac{1}{2}\cdot\sqrt{14}= \frac{\sqrt{14}}{2}$ $(b)$ ${\bf w}={\bf u}\times{\bf v}$ is perpendicular to both ${\bf u}$ and ${\bf v}$ (and the plane they belong to) A unit vector has length 1, so we take $\displaystyle \frac{{\bf w}}{|{\bf w}|}= \frac{1}{\sqrt{14}} (2{\bf i}+3{\bf j}+{\bf k} )$ $= \displaystyle \frac{2}{\sqrt{14}} {\bf i}+ \frac{3}{\sqrt{14}} {\bf j}+ \frac{1}{\sqrt{14}} {\bf k}$ $= \displaystyle \frac{\sqrt{14}}{7} {\bf i}+ \frac{3\sqrt{14}}{14} {\bf j}+ \frac{\sqrt{14}}{14} {\bf k}$