# How to Calculate Area of a Parallelogram with Vectors

Finding the area of a parallelogram spanned by $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$ is the same as calculating the length of the $\stackrel{\to }{u}×\stackrel{\to }{v}$-vector.

If you know the angle between the two vectors, you can use this formula:

Formula

### AreaofaParallelogramwithaKnownAngle

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

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

Formula

### AreaofaParallelogramonVectorCoordinateForm

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

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

Example 1

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

You start by finding the cross product:

$\begin{array}{llll}\hfill & \stackrel{\to }{u}×\stackrel{\to }{v}=\begin{array}{ccc}\hfill \left(\phantom{-1}1,\hfill & \hfill 3,\hfill & \hfill -2\right)\hfill \\ \hfill \hfill & \hfill ×\hfill & \hfill \hfill \\ \hfill \left(-3,\hfill & \hfill 2,\hfill & \hfill \phantom{-1}4\right)\hfill \end{array}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}=\left(3\cdot 4-2\cdot \left(-2\right),\left(-2\right)\cdot \left(-3\right)-1\cdot 4,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}1\cdot 2-\left(-3\right)\cdot 3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}=\phantom{\rule{-0.17em}{0ex}}\left(12+4,6-4,2+9\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}=\phantom{\rule{-0.17em}{0ex}}\left(16,2,11\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \stackrel{\to }{u}×\stackrel{\to }{v}& =\begin{array}{ccc}\hfill \left(\phantom{-1}1,\hfill & \hfill 3,\hfill & \hfill -2\right)\hfill \\ \hfill \hfill & \hfill ×\hfill & \hfill \hfill \\ \hfill \left(-3,\hfill & \hfill 2,\hfill & \hfill \phantom{-1}4\right)\hfill \end{array}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(3\cdot 4-2\cdot \left(-2\right),\left(-2\right)\cdot \left(-3\right)-1\cdot 4,1\cdot 2-\left(-3\right)\cdot 3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(12+4,6-4,2+9\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(16,2,11\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The length of this vector will now be the area of the parallelogram. You find the length like this:
$\begin{array}{llll}\hfill \sqrt{1{6}^{2}+{2}^{2}+1{1}^{2}}& =\sqrt{256+4+121}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\sqrt{381}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 19.5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\sqrt{1{6}^{2}+{2}^{2}+1{1}^{2}}=\sqrt{256+4+121}=\sqrt{381}\approx 19.5$

The area of the parallelogram is approximately equal to $19.5$.

Example 2

If you have two vectors $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$, where $\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{a}|=5$, $\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{b}|=7$, and the angle between them is $30$°, you can find the area of the parallelogram they span by inserting your information into the formula: $\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{a}×\stackrel{\to }{b}|& =\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{a}|\cdot \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{b}|\cdot \mathrm{sin}\alpha \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =5\cdot 7\cdot \mathrm{sin}30\text{°}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =5\cdot 7\cdot \frac{1}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{35}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =17.5.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The area of the parallelogram is equal to $17.5$.