12.4.9 - Area of a triangle given points in 3D space

[karush]

$\textsf{Find the area of the triangle determined by the points }$
\begin{align*}\displaystyle
&P(1,1,1), \, Q(-2,-7,-1), \, R(-7,-1,4)\\
\end{align*}

\begin{align*}\displaystyle
\vec{PQ}&=(-2-1)i&+(-7-1)J&+(-1-1)k&=-3i-8j-2k\\
\vec{PR}&=(-7-1)i&+(-1-1)j&+(4-1) k&=-8i-2j-3k
\end{align*}
\begin{align*}\displaystyle
\vec{PQ} \times \vec{PR}&=
\begin{vmatrix}
i&j&k\\-3&-8&-2\\-8&-2&3
\end{vmatrix}\\
&=
\begin{vmatrix}
-8&-2\\-2&3
\end{vmatrix}i-
\begin{vmatrix}
-3&-2\\-8&3
\end{vmatrix}j-
\begin{vmatrix}
-3&-8\\-8&-2
\end{vmatrix}k\\
&=(-24-4)i-(-9-16)j+(6-64)k\\
&=-28i+25j-58k\\
&=\sqrt{28^2 + 25^2 + 58^2}\\
&=\sqrt{4773}
\end{align*}
$\textit{divide by half to obtain area of triangle}$
$\displaystyle \frac{\sqrt{4773}}{2}$
I just followed an example but was unsure about the signs in between matrix
and suggestions ?

View attachment 7309

 

[Prove It]

Re: 12.4.9 area of a triangle given points in 3D space

$\textsf{Find the area of the triangle determined by the points }$
\begin{align*}\displaystyle
&P(1,1,1), \, Q(-2,-7,-1), \, R(-7,-1,4)\\
\end{align*}

\begin{align*}\displaystyle
\vec{PQ}&=(-2-1)i&+(-7-1)J&+(-1-1)k&=-3i-8j-2k\\
\vec{PR}&=(-7-1)i&+(-1-1)j&+(4-1) k&=-8i-2j-3k
\end{align*}
\begin{align*}\displaystyle
\vec{PQ} \times \vec{PR}&=
\begin{vmatrix}
i&j&k\\-3&-8&-2\\-8&-2&3
\end{vmatrix}\\
&=
\begin{vmatrix}
-8&-2\\-2&3
\end{vmatrix}i-
\begin{vmatrix}
-3&-2\\-8&3
\end{vmatrix}j-
\begin{vmatrix}
-3&-8\\-8&-2
\end{vmatrix}k\\
&=(-24-4)i-(-9-16)j+(6-64)k\\
&=-28i+25j-58k\\
&=\sqrt{28^2 + 25^2 + 58^2}\\
&=\sqrt{4773}
\end{align*}
$\textit{divide by half to obtain area of triangle}$
$\displaystyle \frac{\sqrt{4773}}{2}$
I just followed an example but was unsure about the signs in between matrix
and suggestions ?

$\displaystyle \begin{align*} \vec{PR} = \left( -8, -2, 3 \right) \end{align*}$, not $\displaystyle \begin{align*} \left( -8, -2, -3 \right) \end{align*}$ as you wrote, but since you have used the correct vector in your calculation of the cross product, I expect this is just a typo.

Also, the area is $\displaystyle \begin{align*} \frac{1}{2} \left| \vec{PQ} \times \vec{PR} \right| \end{align*}$, which you actually calculated, but you wrote $\displaystyle \begin{align*} \frac{1}{2} \left( \vec{PQ} \times \vec{PR} \right) \end{align*}$. PLEASE be careful with your notation!

 

[karush]

Re: 12.4.9 area of a triangle given points in 3D space

thanks for the catch

yes easy sign error typos on these

Still got 5 more vector probs to do

 

[karush]

Re: 12.4.9 area of a triangle given points in 3D space

but you wrote $\displaystyle \begin{align*} \frac{1}{2} \left( \vec{PQ} \times \vec{PR} \right) \end{align*}$. PLEASE be careful with your notation!

where did I write this?

$\displaystyle \begin{align*} \frac{1}{2} \left( \vec{PQ} \times \vec{PR} \right) \end{align*}$

 

[HallsofIvy]

Re: 12.4.9 area of a triangle given points in 3D space

It might be simpler to do this using "Heron's formula": If the lengths of the three sides of a triangle are a, b, and c, then the area of the triangle is [tex]\sqrt{s(s- a)(s- b)(s- c)}[/tex] where "s" is the "half perimeter"- s= (a+ b+ c)/2.

 

[karush]

Re: 12.4.9 area of a triangle given points in 3D space

probably so but it was a homework assignment on this method.