# Find the area of triangle PQR, P(0,-2,0), Q(4,1,-2), R(5,3,1)

Find the area of triangle PQR, P(0,-2,0), Q(4,1,-2), R(5,3,1)

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May Dunn
Area of a vector with vertices at P,Q and R is
$$\displaystyle{A}{r}{e}{a}={\frac{{{1}}}{{{2}}}}{\left|\vec{{{P}{Q}}}\times\vec{{{P}{R}}}\right|}$$
$$\displaystyle\vec{{{P}{Q}}}={<}{4},{2},{0}\succ{<}{0},{0},-{3}\ge{<}{4},{2},{3}{>}$$
$$\displaystyle\vec{{{P}{R}}}={<}{3},{3},{1}\succ{<}{0},{0},-{3}\ge{<}{3},{3},{4}{>}$$
$Area=\frac{1}{2}|<4,2,3>\times<3,3,4>|=\frac{1}{2}\begin{vmatrix}i & j&k \\4 & 2&3\\3&3&4 \end{vmatrix}$
$$\displaystyle={\frac{{{\left|{i}{\left({8}-{9}\right)}-{j}{\left({16}-{9}\right)}+{k}{\left({12}-{6}\right)}\right|}}}{{{2}}}}$$
$$\displaystyle={\frac{{{\mid}-{i}-{7}{j}+{6}{k}}}{{{2}}}}$$
$$\displaystyle={\frac{{\sqrt{{{1}^{{2}}+{7}^{{2}}+{6}^{{2}}}}}}{{{2}}}}$$
$$\displaystyle={\frac{{\sqrt{{{1}+{49}+{36}}}}}{{{2}}}}={\frac{{\sqrt{{{86}}}}}{{{2}}}}$$