University Calculus: Early Transcendentals (3rd Edition)

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