 # How to Calculate Area of a Triangle with Vectors

Finding the area of a triangle spanned by $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$ is equal to the length of the $\stackrel{\to }{u}×\stackrel{\to }{v}$-vector divided by two. That means it is half the area of a parallelogram. If you know the angle between the two vectors, you can use this formula:

Formula

### TheAreaofaTrianglewithaKnownAngle

$\begin{array}{llll}\hfill & \frac{1}{2}\cdot \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{u}×\stackrel{\to }{v}|=\frac{1}{2}\cdot \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{u}|\cdot \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{v}|\cdot \mathrm{sin}\alpha ,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\alpha =\angle \phantom{\rule{-0.17em}{0ex}}\left(\stackrel{\to }{u},\stackrel{\to }{v}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\frac{1}{2}\cdot \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{u}×\stackrel{\to }{v}|=\frac{1}{2}\cdot \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{u}|\cdot \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{v}|\cdot \mathrm{sin}\alpha ,\phantom{\rule{2em}{0ex}}\alpha =\angle \phantom{\rule{-0.17em}{0ex}}\left(\stackrel{\to }{u},\stackrel{\to }{v}\right)$

If you have the vectors on coordinate form, you can use this formula:

Formula

### TheAreaofaTriangleonVectorCoordinateForm

$\begin{array}{llll}\hfill \frac{1}{2}\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{u}×\stackrel{\to }{v}|& =\frac{1}{2}|\begin{array}{ccc}\hfill \left({x}_{1},\hfill & \hfill {y}_{1},\hfill & \hfill {z}_{1}\right)\hfill \\ \hfill \hfill & \hfill ×\hfill & \hfill \hfill \\ \hfill \left({x}_{2},\hfill & \hfill {y}_{2},\hfill & \hfill {z}_{2}\right)\hfill \end{array}|\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{2}|\left({y}_{1}{z}_{2}-{y}_{2}{z}_{1},{z}_{1}{x}_{2}-{z}_{2}{x}_{1},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}{x}_{1}{y}_{2}-{x}_{2}{y}_{1}\right)|\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \frac{1}{2}\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{u}×\stackrel{\to }{v}|& =\frac{1}{2}|\begin{array}{ccc}\hfill \left({x}_{1},\hfill & \hfill {y}_{1},\hfill & \hfill {z}_{1}\right)\hfill \\ \hfill \hfill & \hfill ×\hfill & \hfill \hfill \\ \hfill \left({x}_{2},\hfill & \hfill {y}_{2},\hfill & \hfill {z}_{2}\right)\hfill \end{array}|\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{2}\phantom{\rule{-0.17em}{0ex}}|\phantom{\rule{-0.17em}{0ex}}\left({y}_{1}{z}_{2}-{y}_{2}{z}_{1},{z}_{1}{x}_{2}-{z}_{2}{x}_{1},{x}_{1}{y}_{2}-{x}_{2}{y}_{1}\right)|\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Example 1

Find the area of the triangle spanned by $\stackrel{\to }{u}=\phantom{\rule{-0.17em}{0ex}}\left(1,3,2\right)$ and $\stackrel{\to }{v}=\phantom{\rule{-0.17em}{0ex}}\left(3,2,4\right)$.

First, you have to find the cross product of the vectors, which turns out to be $\phantom{\rule{-0.17em}{0ex}}\left(16,2,11\right)$. The length of this vector will be equal to the area of the parallelogram $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$ spans. That means you have to divide the length by $2$ to find the area of the triangle. $\begin{array}{llll}\hfill \frac{1}{2}\sqrt{1{6}^{2}+{2}^{2}+1{1}^{2}}& =\frac{1}{2}\sqrt{256+4+121}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{2}\sqrt{381}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 9.8\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The area of the triangle is approximately equal to $9.8$.