Cross Product and Applications: Learn It 3

Using the Cross Product

The cross product is a powerful tool for solving geometric problems in three dimensions. Its key applications include finding vectors orthogonal to two given vectors, calculating areas of parallelograms and triangles, and determining volumes of three-dimensional shapes called parallelepipeds.

Finding Orthogonal Vectors

Since [latex]\mathbf{u} \times \mathbf{v}[/latex] is always orthogonal to both [latex]\mathbf{u}[/latex] and [latex]\mathbf{v}[/latex], cross products provide an easy way to find perpendicular vectors.

Let [latex]\mathbf{a} = \langle 5,2,-1\rangle[/latex] and [latex]\mathbf{b} = \langle 0,-1,4\rangle[/latex]. Find a unit vector orthogonal to both [latex]\textbf a[/latex] and [latex]\textbf b[/latex].

Remember that [latex]\mathbf{u} \times \mathbf{v}[/latex] and [latex]\mathbf{v} \times \mathbf{u}[/latex] point in opposite directions. Both are orthogonal to [latex]\mathbf{u}[/latex] and [latex]\mathbf{v}[/latex], but choose the one appropriate for your application.

Calculating Areas

The cross product provides an elegant way to calculate areas of parallelograms and triangles formed by vectors.

area of a parallelogram

If vectors [latex]\mathbf{u}[/latex] and [latex]\mathbf{v}[/latex] form adjacent sides of a parallelogram, then the area of the parallelogram is:

[latex]\text{Parallelogram Area } = ||\mathbf{u} \times \mathbf{v}||[/latex]

This figure is a parallelogram. One side is represented with a vector labeled “v.” The second side, the base, has the same initial point as vector v and is labeled “u.” The angle between u and v is theta. Also, a perpendicular line segment is drawn from the terminal point of v to vector u. It is labeled “|v|sin(theta).”
Figure 1. The parallelogram with adjacent sides [latex]{\bf{u}}[/latex] and [latex]{\bf{v}}[/latex] has base [latex]∥{\bf{u}}∥[/latex] and height [latex]∥{\bf{v}}∥\sin\theta[/latex].

Proof

From basic geometry, the area of a parallelogram equals base times height. Using [latex]\mathbf{u}[/latex] as the base:

  • Base = [latex]||\mathbf{u}||[/latex]
  • Height = [latex]||\mathbf{v}|| \sin \theta[/latex] (perpendicular distance from [latex]\mathbf{v}[/latex] to [latex]\mathbf{u}[/latex])

Therefore:

Area = [latex]||\mathbf{u}|| \cdot ||\mathbf{v}|| \sin \theta = ||\mathbf{u} \times \mathbf{v}||[/latex]

[latex]_\blacksquare[/latex]

area of a triangle

Since a triangle has half the area of the parallelogram formed by the same two vectors:

[latex]\text{Triangle Area }= \frac{1}{2}||\mathbf{u} \times \mathbf{v}||[/latex]

This geometric interpretation makes cross products invaluable for area calculations in computer graphics, engineering, and physics applications where you need to find areas from vector data.

Let [latex]P=(1, 0, 0)[/latex], [latex]Q=(0, 1, 0)[/latex], and [latex]R=(0, 0, 1)[/latex] be the vertices of a triangle (Figure 2). Find its area.

This figure is the 3-dimensional coordinate system. It has a triangle drawn in the first octant. The vertices of the triangle are points P(1, 0, 0); Q(0, 1, 0); and R(0, 0, 1).
Figure 2. Finding the area of a triangle by using the cross product.