## University Calculus: Early Transcendentals (3rd Edition)

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