Section 1.4 Cross Product

1.4 Cross Product

In this section, we introduce the cross product operation of vectors. The idea of dot product is to decide if a pair of vectors are orthogonal to each others. The motivation of cross product is finding a vector that is orthogonal to two given vectors. Cross product helps us to find a non-standard basis of a 3-dimension space from a given set of two vectors. It also gives us the equation of a plane in a 3-dimensional space. Before we define the cross product, we introduce the determinant.

Definition: 

Given two vectors in 2-dimensional space, $\overrightarrow{u}=<x_1,y_1>$ and $\overrightarrow{v}=<x_2,y_2>$ . A $2\times 2$ determinant is defined by

$$\left|\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\end{array}\right|=x_1{y}_2-x_2{y}_1.$$

EQ1: Find the $2\times 2$ determinant defined by $\overrightarrow{u}=<1,2>$ and $\overrightarrow{v}=<2,-2>$.

Definition: 

Given three vectors in 3-dimensional space, $\overrightarrow{u}=<x_1,y_1,z_1>$, $\overrightarrow{v}=<x_2,y_2,z_2>$ and $\overrightarrow{w}=<x_3,y_3,z_3>$.  A $3\times 3$ determinant is defined by

$$\left|\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ x_3& y_3& z_3\end{array}\right|=x_1(y_2{z}_3-y_3{z}_2)-y_1(x_2{z}_3-x_3{z}_2)+z_1(x_2{y}_3-x_3{y}_2).$$

Example 1: Let $\overrightarrow{u}=<1,-2,3>,\overrightarrow{v}=<2,1,0>$, and $\overrightarrow{w}=<-2,0,1>$. Find the $3\times 3$ determinant is defined by those three vectors using the given order.

Exercise 1: Let $\overrightarrow{u}=<-1,2,-3>,\overrightarrow{v}=<-2,0,1>$, and $\overrightarrow{w}=<3,2,-1>$. Find the $3\times 3$ determinant is defined by those three vectors using the given order.

We are now ready for the cross product.

Definition: 

Given $\overrightarrow{u}=<x_1,y_1,z_1>$ and $\overrightarrow{v}=<x_2,y_2,z_2>$. Then, the cross product, $\overrightarrow{u}\times \overrightarrow{v},$ is vector 

$$\overrightarrow{u}\times \overrightarrow{v}$$

$$=<y_1{z}_2-y_2{z}_1,-(x_1{z}_2-x_2{z}_1),x_1{y}_2-x_2{y}_1>$$

$$=(y_1{z}_2-y_2{z}_1)\overrightarrow{i}-(x_1{z}_2-x_2{z}_1)\overrightarrow{j}+(x_1{y}_2-x_2{y}_1)\overrightarrow{k}$$

$$=\left|\begin{array}{cc}{y}_1& z_1\\ y_2& z_2\end{array}\right|\overrightarrow{i}-\left|\begin{array}{cc}{x}_1& z_1\\ x_2& z_2\end{array}\right|\overrightarrow{j}+\left|\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\end{array}\right|\overrightarrow{k}$$

$$=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|.$$

Proof of this vector is orthogonal to both $\overrightarrow{u}$ and $\overrightarrow{v}$.

Notice: Cross product is only defined in 3-dimensional space. Dot product is defined in any dimensional space.

EQ2: Which one is false for given vectors $\overrightarrow{u}$, $\overrightarrow{v}$ and $\overrightarrow{w}$ vectors in 3-dimensional space?

A: $(\overrightarrow{u}\times \overrightarrow{v})$ is a vector;

B: $(\overrightarrow{u}\cdot \overrightarrow{v})$ is a number;

C: $(\overrightarrow{u}\times \overrightarrow{v})\cdot \overrightarrow{w}$ is a number;

D: $(\overrightarrow{u}\times \overrightarrow{v})\times \overrightarrow{w}$ is a number.

Example 2: Let $\overrightarrow{u}=<1,-2,3>,\overrightarrow{v}=<2,1,0>$, and $\overrightarrow{w}=<-2,0,1>$ be vectors. 

(a) Find $(\overrightarrow{u}\times \overrightarrow{v})$

(b) Find $(\overrightarrow{u}\times \overrightarrow{w})$

(c) Find $(\overrightarrow{u}\times \overrightarrow{v})\cdot \overrightarrow{w}$

(d) Find $(\overrightarrow{u}\times \overrightarrow{v})\times \overrightarrow{w}$

Exercise 2: Let $\overrightarrow{u}=<-1,2,-3>,\overrightarrow{v}=<-2,0,1>$, and $\overrightarrow{w}=<3,2,-1>$ be vectors. 

(a) Find $(\overrightarrow{u}\times \overrightarrow{v})$

(b) $(\overrightarrow{u}\times \overrightarrow{w})$

(c) $(\overrightarrow{u}\times \overrightarrow{v})\cdot \overrightarrow{w}$

(d) $(\overrightarrow{u}\times \overrightarrow{v})\times \overrightarrow{w}$

Theorem: Properties of cross product 

Let $\overrightarrow{u},\overrightarrow{v},$ and $\overrightarrow{w}$ be vectors. Let $r$ be scalars. 

i. $\overrightarrow{u}\times \overrightarrow{v}=-\overrightarrow{v}\times \overrightarrow{u}$ Anti-commutative property 

ii. $\overrightarrow{u}\times (\overrightarrow{v}+\overrightarrow{w})=\overrightarrow{u}\times \overrightarrow{v}+\overrightarrow{u}\times \overrightarrow{w}$ Distributive property 

iii. $c(\overrightarrow{u}\times \overrightarrow{v})=(c\overrightarrow{u})\times \overrightarrow{v}=\overrightarrow{u}\times (c\overrightarrow{v})$ Associative property

iv. $\overrightarrow{u}\times \overrightarrow{0}=\overrightarrow{0}\times \overrightarrow{u}=\overrightarrow{0}$ Cross product of the zero vector

v. $\overrightarrow{v}\times \overrightarrow{v}=\overrightarrow{0}$ Cross product of a vector with itself

vi. $\overrightarrow{u}\cdot (\overrightarrow{v}\times \overrightarrow{w})=(\overrightarrow{u}\times \overrightarrow{v})\cdot \overrightarrow{w}$ Scalar triple product.

Proof of vi using matrix:

EQ3: Which one is false for given vectors

$\overrightarrow{u}$, $\overrightarrow{v}$ and $\overrightarrow{w}$ vectors in 3-dimensional space?

A: $\overrightarrow{u}\times \overrightarrow{v}=\overrightarrow{v}\times \overrightarrow{u}$;

B: $\overrightarrow{u}\cdot (\overrightarrow{v}\times \overrightarrow{w})=(\overrightarrow{u}\times \overrightarrow{v})\cdot \overrightarrow{w}$;

C: $\overrightarrow{v}\times \overrightarrow{v}=\overrightarrow{0}$;

D: $(\overrightarrow{u}+\overrightarrow{v})\times \overrightarrow{w}=(\overrightarrow{u}\times \overrightarrow{w})+(\overrightarrow{v}\times \overrightarrow{w})$. 

Theorem:

Let $\overrightarrow{u}$ and $\overrightarrow{v}$ be vectors, and let $\theta$ be the angle between them. Then 

$$||\overrightarrow{u}\times \overrightarrow{v}||=||\overrightarrow{u}||||\overrightarrow{v}||\text{sin}(\theta ).$$

Proof: Use $||\overrightarrow{u}\times \overrightarrow{v}|{|}^2$

Example 3: Find $\text{sin}(\theta )$ where $\theta$ is the angle between each pair of vectors. 

a. $\overrightarrow{i}+\overrightarrow{j}$ and $\overrightarrow{j}-\overrightarrow{k}$ 

b. $<1,2,-3>$ and $<3,0,1>$

Exercise 3: Find $\text{sin}(\theta )$ where $\theta$ is the angle between each pair of vectors. 

a. $\overrightarrow{i}-\overrightarrow{j}+2\overrightarrow{k}$ and $\overrightarrow{i}+2\overrightarrow{j}-\overrightarrow{k}$ 

b. $<-1,2,-3>$ and $<0,3,2>$

Theorem: Parallel Vectors

The nonzero vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ are parallel vectors if and only if $\overrightarrow{u}\times \overrightarrow{v}=0.$

Theorem: Area of parallelogram

Given nonzero vectors $\overrightarrow{u}$ and $\overrightarrow{v},$

$$||\overrightarrow{u}\times \overrightarrow{v}||=||\overrightarrow{u}||||\overrightarrow{v}||\text{sin}(\theta )$$  is the area of the parallelogram determined by $\overrightarrow{u}$ and $\overrightarrow{v}$.

Example 4: Find the area of parallelogram determined by $\overrightarrow{u}=<1,-2,3>,\overrightarrow{v}=<2,1,0>$.

Exercise 4: Find the area of parallelogram determined by $\overrightarrow{u}=<-1,2,-3>,\overrightarrow{v}=<-2,0,1>$.

Theorem: Volume of parallelepiped Given nonzero vectors $\overrightarrow{u}$, $\overrightarrow{v}$ and $\overrightarrow{w}$ then 

$$|\overrightarrow{u}\cdot (\overrightarrow{v}\times \overrightarrow{w})|$$

is the volume of the parallelepiped determined by $\overrightarrow{u}$, $\overrightarrow{v}$ and $\overrightarrow{w}$.

Proof: Use picture.

Example 5: Find the volume of the parallelepiped determined by $\overrightarrow{u}=<1,-2,3>,\overrightarrow{v}=<2,1,0>$ and $\overrightarrow{w}=<0,3,-1>$.

Exercise 5: Find the volume of the parallelepiped determined by $\overrightarrow{u}=<-1,2,-3>,\overrightarrow{v}=<-2,0,1>$ and $\overrightarrow{w}=<2,-2,0>$.

Notice: If the volume of the the parallelepiped determined by $\overrightarrow{u}$, $\overrightarrow{v}$ and $\overrightarrow{w}$ is 0 then $\overrightarrow{u}$, $\overrightarrow{v}$ and $\overrightarrow{w}$ are coplanar which means they lie on the same plane. Hence $\overrightarrow{u}\cdot (\overrightarrow{v}\times \overrightarrow{w})=0$ if and only if $\overrightarrow{u}$, $\overrightarrow{v}$ and $\overrightarrow{w}$ are coplanar.

Example 6: Decide if given four points are on the same plane. $A=(1,0,2)$, $B=(2,1,0)$, $C=(0,2,-1)$, $D=(-3,0,1)$.

Exercise 6: Decide if given four points are on the same plane. $A=(1,3,2)$, $B=(3,-1,6)$, $C=(5,2,0)$, $D=(3,6,-4)$.

Definition: Torque

Torque $\overrightarrow{\tau }$ (the Greek letter tau), measures the tendency of a force to produce rotation about an axis of rotation. Let $\overrightarrow{r}$ be a vector with an initial point located on the axis of rotation and with a terminal point located at the point where the force is applied, and let vector $\overrightarrow{F}$ represent the force. Then torque is equal to the cross product of $\overrightarrow{r}$ and $\overrightarrow{F}$: $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$.

Group work:

1. A bolt is tightened by applying a 40-N force to a 0.1-m wrench at a $60$ degree to the wrench. Find the magnitude of the torque about the center of the bolt.

2. Let $P=(1,1,0),Q=(0,1,1)$, and $R=(1,0,1)$ be the vertices of a triangle. Find its area.

3. Only a single plane can pass through any set of three non-colinear points. Find a vector orthogonal to the plane containing points $P=(9,-3,-2)$,

$Q=(1,3,0)$, and $R=(-2,5,0)$.

4. Nonzero vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ are called collinear if there exists a nonzero scalar $c$ such that $\overrightarrow{v}=c\overrightarrow{u}$. Show that vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are collinear, where $A(4,1,0),B(6,5,-2)$, and $C(5,3,-1)$.

5. Determine a vector of magnitude $10$ perpendicular to the plane passing through the $x$-axis and point $P(1,2,4)$.

6. A bolt is tightened by applying a 30-N force to a 0.15-m wrench at a $90$ degree to the wrench. Find the magnitude of the torque about the center of the bolt.

7. Let $P=(1,0,0),Q=(0,1,0)$, and $R=(0,0,1)$ be the vertices of a triangle. Find its area.