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

signokodo7h 2021-11-15 Answered
Find the area of triangle PQR, P(0,-2,0), Q(4,1,-2), R(5,3,1)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

May Dunn
Answered 2021-11-16 Author has 456 answers
Area of a vector with vertices at P,Q and R is
\(\displaystyle{A}{r}{e}{a}={\frac{{{1}}}{{{2}}}}{\left|\vec{{{P}{Q}}}\times\vec{{{P}{R}}}\right|}\)
\(\displaystyle\vec{{{P}{Q}}}={<}{4},{2},{0}\succ{<}{0},{0},-{3}\ge{<}{4},{2},{3}{>}\)
\(\displaystyle\vec{{{P}{R}}}={<}{3},{3},{1}\succ{<}{0},{0},-{3}\ge{<}{3},{3},{4}{>}\)
\[Area=\frac{1}{2}|<4,2,3>\times<3,3,4>|=\frac{1}{2}\begin{vmatrix}i & j&k \\4 & 2&3\\3&3&4 \end{vmatrix}\]
\(\displaystyle={\frac{{{\left|{i}{\left({8}-{9}\right)}-{j}{\left({16}-{9}\right)}+{k}{\left({12}-{6}\right)}\right|}}}{{{2}}}}\)
\(\displaystyle={\frac{{{\mid}-{i}-{7}{j}+{6}{k}}}{{{2}}}}\)
\(\displaystyle={\frac{{\sqrt{{{1}^{{2}}+{7}^{{2}}+{6}^{{2}}}}}}{{{2}}}}\)
\(\displaystyle={\frac{{\sqrt{{{1}+{49}+{36}}}}}{{{2}}}}={\frac{{\sqrt{{{86}}}}}{{{2}}}}\)
Have a similar question?
Ask An Expert
0
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more
...