## Thomas' Calculus 13th Edition

$a.\quad 2\sqrt{6}$ $b.\displaystyle \quad \frac{2}{\sqrt{6}} {\bf i}+ \frac{1}{\sqrt{6}} {\bf j}+ \frac{1}{\sqrt{6}} {\bf k}$
$(a)$ ${\bf u}= \overrightarrow{PQ} = \langle 2-1,0+1,-1-2 \rangle= \langle 1,1,-3 \rangle$ ${\bf v}= \overrightarrow{PR} = \langle 0-1,2+1,1-2 \rangle= \langle-1,3,-1 \rangle$ We find the area of the parallelogram $|{\bf u}\times{\bf v}|$ The area of the triangle is half the area of the parallelogram. $A=\displaystyle \frac{1}{2}\cdot|{\bf u}\times{\bf v}|$ ${\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 1 & 1 & -3\\ -1 & 3 & -1 \end{array}\right|=(-1+9){\bf i}-(-1-3){\bf j}+(3+1){\bf k}$ $=8{\bf i}+4{\bf j}+4{\bf k}$ $|{\bf u}\times{\bf v}|=\sqrt{64+16+16}=\sqrt{96}=4\sqrt{6}$ $A=\displaystyle \frac{1}{2}\cdot 4\sqrt{6}=2\sqrt{6}$ $(b)$ ${\bf w}={\bf u}\times{\bf v}$ is perpendicular to both ${\bf u}$ and ${\bf v}$ (and the plane they belong to). A unit vector has length 1, so we take $\displaystyle \frac{{\bf w}}{|{\bf w}|}= \frac{1}{4\sqrt{6}} (8{\bf i}+4{\bf j}+4{\bf k})$ $= \displaystyle \frac{2}{\sqrt{6}} {\bf i}+ \frac{1}{\sqrt{6}} {\bf j}+ \frac{1}{\sqrt{6}} {\bf k}$