5. Cross Product

c. Area

Parallelogram

Consider a parallelogram whose sides are the vectors \(\vec u\) and \(\vec v\). Take \(\vec u\) as the base. Let \(\theta\) be the angle between these vectors.

The area of the parallelogram is the product of its base and its height \[ A_{\textstyle\diamond}=bh \] where the base is \(b=|\vec u|\) and the height is \(h=|\vec v|\sin\theta\). Thus \[ A_{\textstyle\diamond}=bh=|\vec u|\,|\vec v|\sin\theta \] However this is is just the length of the cross product \(|\vec u\times\vec v|\). Therefore:

The Area of a Parallelogram with adjacent edges \(\vec u\) and \(\vec v\) is: \[ A_{\textstyle\diamond}=|\vec u\times\vec v| \]

DefAreaPara

Find the area of the parallelogram with adjacent edges \(\vec u=\left\langle 2,3,1\right\rangle\) and \(\vec v=\left\langle 2,0,-1\right\rangle\).

The cross product is \[\begin{aligned} \vec u\times\vec v &=\left\langle 2,3,1\right\rangle\times\left\langle 2,0,-1\right\rangle \\ &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 2 & 3 & 1 \\ 2 & 0 & -1 \end{vmatrix} =\hat\imath(-3)-\hat\jmath(-2-2)+\hat k(-6) \\ &=\left\langle -3,4,-6\right\rangle \end{aligned}\] So the area is \[ A_{\textstyle\diamond}=|\vec u\times\vec v|=\sqrt{9+16+36}=\sqrt{61} \]

Find the area of the parallelogram with adjacent edges \(\vec a=\left\langle 4,3,2\right\rangle\) and \(\vec b=\left\langle 1,2,3\right\rangle\).

\(A_{\textstyle\diamond}=5\sqrt{6}\)

The cross product is \[\begin{aligned} \vec a\times\vec b &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 4 & 3 & 2 \\ 1 & 2 & 3 \end{vmatrix} =\hat\imath(5)-\hat\jmath(10)+\hat k(5) \\ &=\left\langle 5,-10,5\right\rangle \end{aligned}\] So the area is \[ A_{\textstyle\diamond}=|\vec a\times\vec b|=\sqrt{25+100+25}=5\sqrt{6} \]

Triangle

A triangle is just half of a parallelogram.

The Area of a Triangle with adjacent edges \(\vec u\) and \(\vec v\) is: \[ A_\Delta=\dfrac{1}{2}|\vec u\times\vec v| \]

DefAreaTri

Find the area of the triangle with vertices \(A=\left\langle 3,1,0\right\rangle\), \(B=\left\langle 3,0,2\right\rangle\) and \(C=\left\langle 4,1,3\right\rangle\).

\(A_\Delta=\dfrac{1}{2}\sqrt{14}\)

Two adjacent edges are \[\begin{aligned} \overrightarrow{AB}&=B-A =\left\langle 3,0,2\right\rangle-\left\langle 3,1,0\right\rangle =\left\langle 0,-1,2\right\rangle \\ \overrightarrow{AC}&=C-A =\left\langle 4,1,3\right\rangle-\left\langle 3,1,0\right\rangle =\left\langle 1,0,3\right\rangle \end{aligned}\] Their cross product is \[ \overrightarrow{AB}\times\overrightarrow{AC} =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 0 & -1 & 2 \\ 1 & 0 & 3 \end{vmatrix} =\hat\imath(-3)-\hat\jmath(-2)+\hat k(1) \] So the area is \[ A_\Delta =\dfrac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right| =\dfrac{1}{2}\sqrt{14} \]

You can also practice computing the area of a triangle using vectors and cross products in 3D by using the following Maplet (requires Maple on the computer where this is executed):

Areas of Triangles in 3D Space   Rate It

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Supported in part by NSF Grant #1123255