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, [latex]\overrightarrow{u}=\lt x_{1},y_{1}\gt[/latex] and [latex]\overrightarrow{v}=\lt x_{2},y_{2}\gt[/latex] . A [latex]2\times2[/latex] 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 [latex]2\times2[/latex] determinant defined by [latex]\overrightarrow{u}=\lt 1,2\gt[/latex] and [latex]\overrightarrow{v}=\lt 2,-2\gt[/latex].

Definition: 

Given three vectors in 3-dimensional space, [latex]\overrightarrow{u}=\lt x_{1},y_{1},z_{1}\gt[/latex], [latex]\overrightarrow{v}=\lt x_{2},y_{2},z_{2}\gt[/latex] and [latex]\overrightarrow{w}=\lt x_{3},y_{3},z_{3}\gt[/latex].  A [latex]3\times3[/latex] 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 [latex]\overrightarrow{u}=\lt 1,-2,3\gt,\overrightarrow{v}=\lt 2,1,0\gt[/latex], and [latex]\overrightarrow{w}=\lt -2,0,1\gt[/latex]. Find the [latex]3\times3[/latex] determinant is defined by those three vectors using the given order.

Exercise 1: Let [latex]\overrightarrow{u}=\lt -1,2,-3\gt,\overrightarrow{v}=\lt -2,0,1\gt[/latex], and [latex]\overrightarrow{w}=\lt 3,2,-1\gt[/latex]. Find the [latex]3\times3[/latex] determinant is defined by those three vectors using the given order.

We are now ready for the cross product.

Definition: 

Given [latex]\overrightarrow{u}=\lt x_{1},y_{1},z_{1}\gt[/latex] and [latex]\overrightarrow{v}=\lt x_{2},y_{2},z_{2}\gt[/latex]. Then, the cross product, [latex]\overrightarrow{u}\times\overrightarrow{v},[/latex] is vector 

\[\overrightarrow{u}\times\overrightarrow{v} \]

\[=\lt y_{1}z_{2}-y_{2}z_{1} ,-(x_{1}z_{2}-x_{2}z_{1}), x_{1}y_{2}-x_{2}y_{1} \gt \]

\[ =(y_{1}z_{2}-y_{2}z_{1})\overrightarrow{i}-(x_{1}z_{2}-x_{2}z_{1})\overrightarrow{j}+(x_{1}y_{2}-x_{2}y_{1})\overrightarrow{k} \]

\[=\left|\begin{array}{cc} y_{1} & z_{1}\\ y_{2} & z_{2} \end{array}\right|\overrightarrow{i}-\left|\begin{array}{cc} x_{1} & z_{1}\\ x_{2} & z_{2} \end{array}\right|\overrightarrow{j}+\left|\begin{array}{cc} x_{1} & y_{1}\\ x_{2} & y_{2} \end{array}\right|\overrightarrow{k}\]

\[=\left|\begin{array}{ccc} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ x_{1} & y_{1} & z_{1}\\ x_{2} & y_{2} & z_{2} \end{array}\right|.\]

Proof of this vector is orthogonal to both [latex]\overrightarrow{u}[/latex] and [latex]\overrightarrow{v}[/latex].

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 [latex]\overrightarrow{u}[/latex], [latex]\overrightarrow{v}[/latex] and [latex]\overrightarrow{w}[/latex] vectors in 3-dimensional space?

A: [latex](\overrightarrow{u}\times\overrightarrow{v})[/latex] is a vector;

B: [latex](\overrightarrow{u}\cdot\overrightarrow{v})[/latex] is a number;

C: [latex](\overrightarrow{u}\times\overrightarrow{v})\cdot\overrightarrow{w}[/latex] is a number;

D: [latex](\overrightarrow{u}\times\overrightarrow{v})\times\overrightarrow{w}[/latex] is a number.

Example 2: Let [latex]\overrightarrow{u}=\lt 1,-2,3\gt,\overrightarrow{v}=\lt 2,1,0\gt[/latex], and [latex]\overrightarrow{w}=\lt -2,0,1\gt[/latex] be vectors. 

(a) Find [latex](\overrightarrow{u}\times\overrightarrow{v})[/latex]

(b) Find [latex](\overrightarrow{u}\times\overrightarrow{w})[/latex]

(c) Find [latex](\overrightarrow{u}\times\overrightarrow{v})\cdot\overrightarrow{w}[/latex]

(d) Find [latex](\overrightarrow{u}\times\overrightarrow{v})\times\overrightarrow{w}[/latex]

Exercise 2: Let [latex]\overrightarrow{u}=\lt -1,2,-3\gt,\overrightarrow{v}=\lt -2,0,1\gt[/latex], and [latex]\overrightarrow{w}=\lt 3,2,-1\gt[/latex] be vectors. 

(a) Find [latex](\overrightarrow{u}\times\overrightarrow{v})[/latex]

(b) [latex](\overrightarrow{u}\times\overrightarrow{w})[/latex]

(c) [latex](\overrightarrow{u}\times\overrightarrow{v})\cdot\overrightarrow{w}[/latex]

(d) [latex](\overrightarrow{u}\times\overrightarrow{v})\times\overrightarrow{w}[/latex]

Theorem: Properties of cross product 

Let [latex]\overrightarrow{u},\overrightarrow{v},[/latex] and [latex]\overrightarrow{w}[/latex] be vectors. Let [latex]r[/latex] be scalars. 

i. [latex]\overrightarrow{u}\times\overrightarrow{v}=-\overrightarrow{v}\times\overrightarrow{u}[/latex] Anti-commutative property 

ii. [latex]\overrightarrow{u}\times(\overrightarrow{v}+\overrightarrow{w})=\overrightarrow{u}\times\overrightarrow{v}+\overrightarrow{u}\times\overrightarrow{w}[/latex] Distributive property 

iii. [latex]c(\overrightarrow{u}\times\overrightarrow{v})=(c\overrightarrow{u})\times\overrightarrow{v}=\overrightarrow{u}\times(c\overrightarrow{v})[/latex] Associative property

iv. [latex]\overrightarrow{u}\times\overrightarrow{0}=\overrightarrow{0}\times\overrightarrow{u}=\overrightarrow{0}[/latex] Cross product of the zero vector

v. [latex]\overrightarrow{v}\times\overrightarrow{v}=\overrightarrow{0}[/latex] Cross product of a vector with itself

vi. [latex]\overrightarrow{u}\cdot(\overrightarrow{v}\times\overrightarrow{w})=(\overrightarrow{u}\times\overrightarrow{v})\cdot\overrightarrow{w}[/latex] Scalar triple product.

Proof of vi using matrix:

EQ3: Which one is false for given vectors

[latex]\overrightarrow{u}[/latex], [latex]\overrightarrow{v}[/latex] and [latex]\overrightarrow{w}[/latex] vectors in 3-dimensional space?

A: [latex]\overrightarrow{u}\times\overrightarrow{v}=\overrightarrow{v}\times\overrightarrow{u}[/latex];

B: [latex]\overrightarrow{u}\cdot(\overrightarrow{v}\times\overrightarrow{w})=(\overrightarrow{u}\times\overrightarrow{v})\cdot\overrightarrow{w}[/latex];

C: [latex]\overrightarrow{v}\times\overrightarrow{v}=\overrightarrow{0}[/latex];

D: [latex](\overrightarrow{u}+\overrightarrow{v})\times\overrightarrow{w}=(\overrightarrow{u}\times\overrightarrow{w})+(\overrightarrow{v}\times\overrightarrow{w})[/latex]. 

Theorem:

Let [latex]\overrightarrow{u}[/latex] and [latex]\overrightarrow{v}[/latex] be vectors, and let [latex]\theta[/latex] be the angle between them. Then 

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

Proof: Use [latex]||\overrightarrow{u}\times\overrightarrow{v}||^{2}[/latex]

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

a. [latex]\overrightarrow{i}+\overrightarrow{j}[/latex] and [latex]\overrightarrow{j}-\overrightarrow{k}[/latex] 

b. [latex]\lt 1,2,-3\gt[/latex] and [latex]\lt 3,0,1\gt[/latex]

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

a. [latex]\overrightarrow{i}-\overrightarrow{j}+2\overrightarrow{k}[/latex] and [latex]\overrightarrow{i}+2\overrightarrow{j}-\overrightarrow{k}[/latex] 

b. [latex]\lt -1,2,-3\gt[/latex] and [latex]\lt 0,3,2\gt[/latex]

Theorem: Parallel Vectors

The nonzero vectors [latex]\overrightarrow{u}[/latex] and [latex]\overrightarrow{v}[/latex] are parallel vectors if and only if [latex]\overrightarrow{u}\times\overrightarrow{v}=0.[/latex]

Theorem: Area of parallelogram

Given nonzero vectors [latex]\overrightarrow{u}[/latex] and [latex]\overrightarrow{v},[/latex]

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

Example 4: Find the area of parallelogram determined by [latex]\overrightarrow{u}=\lt 1,-2,3\gt,\overrightarrow{v}=\lt 2,1,0\gt[/latex].

Exercise 4: Find the area of parallelogram determined by [latex]\overrightarrow{u}=\lt -1,2,-3\gt,\overrightarrow{v}=\lt -2,0,1\gt[/latex].

Theorem: Volume of parallelepiped Given nonzero vectors [latex]\overrightarrow{u}[/latex], [latex]\overrightarrow{v}[/latex] and [latex]\overrightarrow{w}[/latex] then 

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

is the volume of the parallelepiped determined by [latex]\overrightarrow{u}[/latex], [latex]\overrightarrow{v}[/latex] and [latex]\overrightarrow{w}[/latex].

Proof: Use picture.

Example 5: Find the volume of the parallelepiped determined by [latex]\overrightarrow{u}=\lt 1,-2,3\gt,\overrightarrow{v}=\lt 2,1,0\gt[/latex] and [latex]\overrightarrow{w}=\lt 0,3,-1\gt[/latex].

Exercise 5: Find the volume of the parallelepiped determined by [latex]\overrightarrow{u}=\lt -1,2,-3\gt,\overrightarrow{v}=\lt -2,0,1\gt[/latex] and [latex]\overrightarrow{w}=\lt 2,-2,0\gt[/latex].

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

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

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

Definition: Torque

Torque [latex]\overrightarrow{\tau}[/latex] (the Greek letter tau), measures the tendency of a force to produce rotation about an axis of rotation. Let [latex]\overrightarrow{r}[/latex] 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 [latex]\overrightarrow{F}[/latex] represent the force. Then torque is equal to the cross product of [latex]\overrightarrow{r}[/latex] and [latex]\overrightarrow{F}[/latex]: [latex]\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}[/latex].

Example 7: A bolt is tightened by applying a 30-N force to a 0.15-m wrench at a [latex]90[/latex] degree to the wrench. Find the magnitude of the torque about the center of the bolt.

Example 8: Let [latex]P=(1,0,0),Q=(0,1,0)[/latex], and [latex]R=(0,0,1)[/latex] be the vertices of a triangle. Find its area.

Group work:

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

2. Let [latex]P=(1,1,0),Q=(0,1,1)[/latex], and [latex]R=(1,0,1)[/latex] 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 [latex]P=(9,-3,-2)[/latex],

[latex]Q=(1,3,0)[/latex], and [latex]R=(-2,5,0)[/latex].

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

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