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: \[ A_{\textstyle\diamond}=|\vec u\times\vec v| \]

The image shows a parallelogram. The base of the parallelogram is
spanned by a vector, labeled u. A second side is a vector, labeled v
startong at the same point as the vector u. The angle between them is an
acute angle, labeled theta. A line, labeled h, goes from the tip of the
vector v to a point on the vector u and is perpendicular to u. The sides
opposite to u and v are solid lines.

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: \[ A_\Delta=\dfrac{1}{2}|\vec u\times\vec v| \]

The image shows the same a parallelogram as above, but now the sides
opposite to u and v are dashed lines. There is also a line from the tip of
u to the tip of v splitting the parallelogram into 2 triangles with equal
areas.

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

5. Cross Product

c. Area

Parallelogram

Triangle

Heading