Vector Cross Product

Cross Product

Definition

The cross product of two vectors is defined as follows

(1)
\begin{align} \vec{u}\times\vec{v}=(|\vec{u}||\vec{v}|\sin \theta)\hat{n} \end{align}

where the vector $\hat{n}$ is a vector of unit length that is perpendicular to both the vector $\vec{u}$ and $\vec{v}$. The direction of the vector is determined by the right-hand rule. Point your fingers in the direction of $\vec{u}$ and curl them in the direction of $\vec{v}$ and your thumb will point in the direction of $\hat{n}$

right-hand-rule.gif

Because of the right-hand rule, if you switch the order of the vectors, $\hat{n}$ switches to the opposite direction. Therefore, $\vec{u} \times \vec{v}=-(\vec{u} \times \vec{v})$

Since $\hat{n}$ is a unit vector, $\vec{u} \times \vec{v}$ is written as the product of its magnitude and direction. Where the direction of $\vec{u} \times \vec{v}$ is $\hat{n}$ and $|\vec{u} \times \vec{v}|=|\vec{u}||\vec{v}|\sin \theta$

Example
Find the magnitude of the cross product of $\vec{u}=15@45^\circ$ and $\vec{v}=4@100^\circ$

Notice that if $\vec{u}$ and $\vec{v}$ are parallel then,

(2)
\begin{align} \vec{u}\times\vec{v}=(|u||v|\sin 0^\circ)\hat{n}=0\hat{n}=\vec{0} \end{align}

or

(3)
\begin{align} \vec{u}\times\vec{v}=(|u||v|\sin 180^\circ)\hat{n}=0\hat{n}=\vec{0} \end{align}

Calculating the Cross Product

For the purposes of calculation, this definition is not usually useful. It does not explicitly tell us how to find the vector $\hat{n}$. Instead, if we are given two vectors in component form

(4)
\begin{eqnarray} \vec{u}&=&<u_x,u_y,u_z>\\ \vec{v}&=&<v_x,v_y,v_z>\\ \end{eqnarray}

then

(5)
\begin{align} \vec{u}\times\vec{v}= \left | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ u_x& u_y & u_z\\ v_x & v_y & v_z \\ \end{array} \right | =(u_yv_z-v_yu_z)\hat{i}-(u_xv_z-v_xu_z)\hat{j}+(u_xv_y-v_xu_y)\hat{k} \end{align}

Unlike other operations the cross product is not commutative. That is to say $\vec{u}\times\vec{v}\neq\vec{v}\times \vec{u}$.

(6)
\begin{align} \vec{v}\times\vec{u}= \left | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ v_x& v_y & v_z\\ u_x & u_y & u_z \\ \end{array} \right | =(v_yu_z-u_yv_z)\hat{i}-(v_xu_z-u_xv_z)\hat{j}+(v_xu_y-u_xv_y)\hat{k}=-(\vec{u}\times\vec{v}) \end{align}

Finding the Equation of a Plane

We describe a plane by a point on the plane, $P(x_0,y_0,z_0)$, and a vector perpendicular to the plane called a normal vector, $\vec{N}=<A,B,C>$. Looking at the image below, notice that any vector that we create in the plane will be perpendicular to $\vec{N}$. So the vector $\vec{PQ}=<x-x_0,y-y_0,z-z_0>$ is perpendicular to $\vec{N}=<A,B,C>$. We know, that for perpendicular vectors, the dot product is zero. Therefore we have,

(7)
\begin{eqnarray} \vec{N} \cdot \vec{PQ} = 0\\ <A,B,C>\cdot <x-x_0,y-y_0,z-z_0>=0\\ A(x-x_0)+B(y-y_0)+C(z-z_0)=0 \end{eqnarray}

The standard equation of a plane containing a point $P(x_0,y_0,z_0)$ that has a normal of $\vec{N}=<A,B,C>$ is therefore,

(8)
\begin{equation} A(x-x_0)+B(y-y_0)+C(z-z_0)=0 \end{equation}
plane.jpg

In two dimensions, we know that two unique points will define a line. In three dimensions, the analog is that three non-collinear points, not on a common line, will define a plane. Given those three points, we can construct two nonparallel vectors that lie in the plane. If we take the cross product of those two vectors, we will get a vector perpendicular to the plane. We can use that cross product as the $\vec{N}$ for our plane.

Example
Find the equation of the plane containing the points $P(5,1,1)$ , $Q(0,2,2)$ , and $R(6,2,0)$

Finding the Area of a Parallelogram

We know from geometry that the area of a parallelogram is given by $A=bh$. In the image below we see that the area of the parallelogram would be,

areaparallelogram.jpg
(9)
\begin{eqnarray} A=|\vec{u}|h\\ \end{eqnarray}

But from right triangle trigonometry we have that,

(10)
\begin{eqnarray} \sin \theta = \frac{h}{|\vec{v}|}\\ h=|\vec{v}|\sin \theta \end{eqnarray}

Combining the facts from 9 and 10, we have that

(11)
\begin{eqnarray} A=|\vec{u}|h\\ A=|\vec{u}||\vec{v}|\sin \theta\\ A=|\vec{u} \times \vec{v}| \end{eqnarray}

So the area of the parallelogram whose nonparallel sides are the vectors $\vec{u}$ and $\vec{v}$ can be calculated by the formula

(12)
\begin{align} \mbox{Area}=|\vec{u} \times \vec{v}| \end{align}

Example
Find the area of the parallelogram with vertices $P(1,1)$ , $Q(5,1)$ , $R(3,3)$ and $S(7,3)$

Finding the Area of a Triangle

Since a triangle can be thought of as half of a parallelogram, the area of the triangle with two sides defined by $\vec{u}$ and $\vec{v}$ can be calculated by the formula

(13)
\begin{align} \mbox{Area}=\frac{1}{2}|\vec{u} \times \vec{v}| \end{align}
trianglefrompara.jpg

Example
Find the area of the triangle with vertices $P(1,1,1)$ , $Q(2,1,0)$ and $R(-1,4,2)$

Exercises

  1. Find the magnitude of the cross product of $11@15^\circ$ and $99@-15^\circ$
  2. Find the cross product of $\vec{u}=<1,-1,2>$ and $\vec{v}=<0,2,-1>$.
  3. The points (1,1,3), (2, -3, 3), and (-1, 0, 4) determine a plane. Find the equation of the plane in both standard and dot product form.
  4. Use cross product to show that $\vec{u}=<1,4,-5>$ and $\vec{v}=<-3,-12,15>$ are parallel.
  5. Find the area of the parallelogram with vertices (0,0) , (3,7) , (2,5) and (5,12)
  6. Explain why $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$
  7. Find the area of the triangle with vertices (1,2,-1) , (4,1,0) and (3,8,1)
  8. Find a vector perpendicular to the plane $5x-3y+6z=8$
  9. If the vectors $\vec{u}$ and $\vec{v}$ are parallel to the xy-plane, what do we know about their cross product?
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