Math 243: Calculus II

12.4 Cross Product

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12.4.1 The Determinant

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Before we can understand cross product we need to understand

• duality which we studied in the last section concerning the dot product

• determinants

Definition

The determinant of an $n\times n$ matrix is the scaling factor of the change in volume(area) in $n$-dimensional space taking orientation into account.

The determinant is best understood in terms of this geometrical notion and by understanding its properties.

Let's start with another video by Grant Sanderson, The Determinant.

We denote determinant of $n\times n$ matrix $A$ by

$$ \det(A) = |A| $$

Notice that we use the absolute value bars even though the determinant of a matrix may have a negative value indicating that the orientation of space has been changed by the linear transformation encoded by the matrix.

For example if $A$ is the matrix

$$A = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix},$$

then we denote the determinant of $A$ by either

$$ \det\left(A\right) = \left|\,A\,\right| = \det\left(\begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}\right) = \begin{vmatrix} 3 & 0 \\ 0 & 2 \end{vmatrix} = (3)(2) - (0)(0) = 6. $$

This indicates that the area of any region $D$ in the domain $\mathbb{R}^2$ will be stretched to a region six times the original area of $D$ and orientation will be preserved.

Example 1

Figure 1

Consider the matrix

$$ A = \begin{bmatrix}\ \ 2\ & -\frac{4}{\sqrt{5}}\ \\ \ \ 0\ &\ \ \frac{2}{\sqrt{5}}\ \end{bmatrix} $$

As one can see in Figure 1, applying matrix $A$ to $\ihat$ yields the vector $\begin{bmatrix}\ 2\ \\ \ 0\ \end{bmatrix}$, while applying matrix $A$ to $\jhat$ results in the vector $\begin{bmatrix} -\frac{4}{\sqrt{5}}\ \\ \ \ \frac{2}{\sqrt{5}}\ \end{bmatrix}$. Every vector on the unit circle is mapped by matrix $A$ onto a new vector. The graph of all images of vectors from the unit circle is an ellipse. The equation for the ellipse can be determined.

$$ x^2 + 4xy + 9y^2 = 4 $$

The area of the ellipse can be computed if we write the equation of the ellipse in normal form, $Ax^2 + Bxy + Cy^2 = 1$,

$$ \frac{1}{4}x^2 + xy + \frac{9}{4}y^2 = 1 $$

The area is determined to be

$$ \text{Area} = \frac{2\pi}{\sqrt{4AC-B^2}} = \frac{2\pi}{\sqrt{4\left(\frac{1}{4}\right)\left(\frac{9}{4}\right) - \left(1\right)^2}} = \frac{2\pi}{\sqrt{\frac{5}{4}}} = \frac{4\pi}{\sqrt{5}} $$

The area of the unit circle is $\pi(1)^2 = \pi$, so the matrix increased the area by a factor of

$$ \frac{\frac{4\pi}{\sqrt{5}}}{\pi} = \frac{4}{\sqrt{5}} $$

The determinant of matrix $A$

$$ \left|A\right| = \begin{vmatrix}\ \ 2\ & -\frac{4}{\sqrt{5}}\ \\ \ \ 0\ &\ \ \frac{2}{\sqrt{5}}\ \end{vmatrix} = (2)\left(\frac{2}{\sqrt{5}}\right) - (0)\left(-\frac{4}{\sqrt{5}}\right) = \frac{4}{\sqrt{5}}\ {\color{green}\Large{\checkmark}} $$

Exercise 1

Compute the determinant of matrix $A = \begin{bmatrix} -\sqrt{2}\ &\ \ 3\ \\ -\sqrt{3}\ &\ \ 1\ \end{bmatrix}$

View Solution

$$ |A| = \begin{vmatrix} -\sqrt{2}\ &\ \ 3\ \\ -\sqrt{3}\ &\ \ 1\ \end{vmatrix} = \left(-\sqrt{2}\right)(1) - \left(3\right)\left(-\sqrt{3}\right) = 3\sqrt{3} - \sqrt{2} $$

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12.4.2 Determinants of $3\times 3$ Matrices

One computes the determinant of and $n\times n$ matrix using the Laplace Expansion. This course requires only the determinant of a $3\times 3$ matrix. The Laplace Expansion for $3\times 3$ matrices uses any row or column and expands the determinant into three $2\times 2$ determinants.

Definition

Minor matrix and Cofactor

Given a $3\times 3$ matrix $A = \begin{bmatrix} a_{ij} \end{bmatrix}$, first one selects an element $a_{pq}$ called the factor. Remove row $p$ and column $q$ to obtain a $2\times 2$ matrix called the minor or minor matrix $M_{pq}$. The determinant of the minor is usually denoted $m_{pq} = \left|M_{pq}\right|$.

The cofactor is defined by

$$ \text{cof}(a_{pq}) := (-1)^{p+q}m_{pq} $$

Example 2

Consider the matrix $A = \begin{bmatrix}\ 4\ &\ \ 2\ & -3\ \\ \ 2\ & -1\ &\ \ 2\ \\ \ 2\ &\ \ 1\ & -4\ \end{bmatrix}$

We can identify nine $2\times 2$ minors of $3\times 3$ matrix $A$. For example consider $a_{\color{#CC0099}{1}\color{darkblue}{1}} = 4$. If one removes row one and column one, from matrix $A$,

Figure 2

The result is a $2\times 2$ minor for element $a_{11}$

$$ M_{11} = \begin{bmatrix} -1\ &\ \ 2\ \\ \ \ 1\ & -4 \end{bmatrix} $$

Notice that because the minor $M_{11}$ is a matrix, we use a capital letter. However there is a minor associated with every element of the original $n\times n$ matrix. As a result we use the denote the minor with the same subscripts that indicate the element of the original matrix.

In this way an $n\times n$ matrix will have $n^2$ minors, one for each element of the matrix. If we choose an element $a_{ij}$ of an $n\times n$ matrix, then the cofactor for element $a_{ij}$ requires the determinant of the minor, $m_{ij} = \det(M_{ij})$,

$$ \text{cof}\left(a_{ij}\right) = (-1)^{i+j}\left|M_{ij}\right| = (-1)^{i+j}m_{ij} $$

The sgn$\left(a_{ij}\right)$ depends only on the position of the element of the matrix. It is defined

$$ \text{sgn}\left(a_{ij}\right) := (-1)^{i+j} = \left\{\begin{array}{lcl}\ \ 1 & \ &\ \text{if } i+j\text{ is even} \\ -1 & \ &\ \text{if }i+j\text{ is odd} \end{array}\right. $$

This creates a pattern to for the sgn function in an $n\times n$ matrix

$$ \begin{bmatrix} \ +\ &\ -\ &\ +\ &\ -\ &\ \ &\ \cdots\ \\ \ -\ &\ +\ &\ -\ &\ +\ &\ \ &\ \cdots\ \\ \ +\ &\ -\ &\ +\ &\ -\ &\ \ &\ \cdots\ \\ \ -\ &\ +\ &\ -\ &\ +\ &\ \ &\ \cdots \ \\ \ \vdots\ &\ \vdots\ &\ \vdots\ &\ \vdots\ &\ \ddots\ &\ \vdots\ \\ \ \cdots\ &\ \cdots\ &\ \cdots\ &\ \cdots\ &\ \ &\ +\ \end{bmatrix} $$

Definition

The Laplace Expansion of the determinant of an $n\times n$ matrix $A$ is computed by choosing a column $j$ so that $1\le j\le n$ and expanding the sum down the $j^{\text{th}}$ column:

$$ \det(A) = \displaystyle\sum_{i=1}^n a_{ij}(-1)^{i+j}m_{ij} $$

Likewise one may choose a row $i$ so that $1\le i\le n$ and expand the sum across the $i^{\text{th}}$ row:

$$ \det(A) = \displaystyle\sum_{j=1}^n a_{ij}(-1)^{i+j}m_{ij} $$

Notice that while this method is very handy for computing the determinant of a $3\times 3$ matrix, it can be very cumbersome for computing determinants of larger matrices.

To compute the determinant of $10\times 10$ matrix using the Laplace Expansion, one would need to compute $10$, $9\times 9$ determinants, $90$, $8\times 8$ determinants, \dots, $10!$ (FACTORIAL!) $2\times 2$ determinants!

Everyone will learn a MUCH better way to compute the determinants of large matrices in Linear Algebra class.

Let's get back to Example 2

$$ \begin{align*} |A| &= \begin{vmatrix}\ 4\ &\ \ 2\ & -3\ \\ \ 2\ & -1\ &\ \ 2\ \\ \ 2\ &\ \ 1\ & -4\ \end{vmatrix} \\ \\ &= 4(-1)^{1+1}\begin{vmatrix} -1\ &\ \ 2\ \\ \ \ 1\ & -4\ \end{vmatrix} + 2(-1)^{2+1}\begin{vmatrix}\ \ 2\ & -3\ \\ \ \ 1\ & -4\ \end{vmatrix} + 2(-1)^{3+1}\begin{vmatrix}\ \ 2\ & -3\ \\ -1\ &\ \ 2\ \end{vmatrix} \\ \\ &= 4\begin{vmatrix} -1\ &\ \ 2\ \\ \ \ 1\ & -4\ \end{vmatrix} - 2\begin{vmatrix}\ \ 2\ & -3\ \\ \ \ 1\ & -4\ \end{vmatrix} + 2\begin{vmatrix}\ \ 2\ & -3\ \\ -1\ &\ \ 2\ \end{vmatrix} \\ \\ &= 4(4 - 2) - 2(-8+3) + 2(4 - 3) = 4(2) + 2(5) + 2(1) = 20 \end{align*} $$

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12.4.3 Cross Product

Just as in dot product and determinant, everyone should start with a good video by Grant Sanderson, Cross Products.

There is a second video on cross products that for anyone who wants to dig deeper into the geometry of cross products.

Definition

The cross product of two vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^3$ is a vector $\mathbf{u}\times\mathbf{v}$ that is perpendicular to both $\mathbf{u}$ and $\mathbf{v}$ whose direction is determined by the right-hand rule, and with length equal to the area of the parallelogram spanned by the two vectors,

$$ \left|\mathbf{u}\times\mathbf{v}\right| = |\mathbf{u}|\,|\mathbf{v}|\,|\sin(\theta)| $$

where $\theta$ is the angle between the two vectors in the interval $[0,\pi]$.

$$ \cos(\theta) = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|} $$

The formula for cross product is derived from the Laplace Expansion of the $3\times 3$ determinant

$$ \begin{align*} \mathbf{u} \times \mathbf{v} &= \begin{vmatrix}\ \ihat\ &\ u_1\ &\ v_1\ \\ \ \jhat\ &\ u_2\ &\ v_2\ \\ \ \khat\ &\ u_3\ &\ v_3\ \end{vmatrix} \\ \\ &= \ihat\begin{vmatrix}\ u_2\ &\ v_2\ \\ \ u_3\ &\ v_3\ \end{vmatrix} - \jhat\begin{vmatrix}\ u_1\ &\ v_1\ \\ \ u_3\ &\ v_3\ \end{vmatrix} + \khat\begin{vmatrix}\ u_1\ &\ v_1\ \\ \ u_2\ &\ v_2\ \end{vmatrix} \\ \\ &= \ihat(u_2v_3 - u_3v_2) - \jhat(u_1v_3 - u_3v_1) + \khat(u_1v_2 - u_2v_1) \end{align*} $$

Example 3

Compute the cross products of vectors $\mathbf{u} = \langle 0, 2, 1 \rangle$ and $\mathbf{v} = \langle 3, -1, 0 \rangle$, $\mathbf{u}\times\mathbf{v}$ and $\mathbf{v}\times\mathbf{u}$.

$$ \begin{align*} \mathbf{u} \times \mathbf{v} &= \begin{vmatrix}\ \ihat\ &\ \ 0\ &\ \ 3\ \\ \ \jhat\ &\ \ 2\ & -1\ \\ \ \khat\ &\ \ 1\ &\ \ 0\ \end{vmatrix} \\ \\ &= \ihat\begin{vmatrix}\ \ 2\ & -1\ \\ \ \ 1\ &\ \ 0\ \end{vmatrix} - \jhat\begin{vmatrix}\ 0\ &\ 3\ \\ \ \ 1\ &\ 0\ \end{vmatrix} + \khat\begin{vmatrix}\ \ 0\ &\ \ 3\ \\ \ \ 2\ & -1\ \end{vmatrix} \\ \\ &= \ihat(1) - \jhat(-3) + \khat(-6) \\ \\ &= \ihat + 3\jhat - 6\khat = \langle 1, 3, -6 \rangle \\ \\ \mathbf{v} \times \mathbf{u} &= \begin{vmatrix}\ \ihat\ &\ \ 3\ &\ \ 0\ \\ \ \jhat\ & -1\ &\ \ 2\ \\ \ \khat\ &\ \ 0\ &\ \ 1\ \end{vmatrix} \\ \\ &= \ihat\begin{vmatrix} -1\ &\ \ 2\ \\ \ \ 0\ &\ \ 1\ \end{vmatrix} - \jhat\begin{vmatrix}\ 3\ &\ 0\ \\ \ 0\ &\ 1\ \end{vmatrix} + \khat\begin{vmatrix}\ \ 3\ &\ \ 0\ \\ -1\ &\ \ 2\ \end{vmatrix} \\ \\ &= \ihat(-1) - \jhat(3) + \khat(6) \\ \\ &= -\ihat - 3\jhat + 6\khat = \langle -1, -3, 6 \rangle = -\mathbf{u}\times\mathbf{v} \end{align*} $$

Theorem 12.4.1

Vectors $\mathbf{u}$ and $\mathbf{v}$ are parallel in $\mathbb{R}^3$ if and only if the angle between them is $0$ and magnitude of their cross product becomes

$$ |\mathbf{u}\times\mathbf{v}| = |\mathbf{u}|\,|\mathbf{v}|\,|\sin(0)| = 0 $$

Since the only vector in $\mathbb{R}^3$ with a magnitude of zero is the zero vector, two vectors in $\mathbb{R}^3$ are parallel if and only if

$$ \mathbf{u}\times\mathbf{v} = \mathbf{0} $$

Using this theorem, if $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^3$ are parallel, then

$$ \begin{align*} \mathbf{0} = \mathbf{u}\times\mathbf{v} &= \begin{vmatrix}\ \ihat\ &\ u_1\ &\ v_1\ \\ \ \jhat\ &\ u_2\ &\ v_2\ \\ \ \khat\ &\ u_3\ &\ v_3\ \end{vmatrix} \\ \\ &= \ihat\begin{vmatrix}\ u_2\ &\ v_2\ \\ \ u_3\ &\ v_3\ \end{vmatrix} - \jhat\begin{vmatrix}\ u_1\ &\ v_1\ \\ \ u_3\ &\ v_3\ \end{vmatrix} + \khat\begin{vmatrix}\ u_1\ &\ v_1\ \\ \ u_2\ &\ v_2\ \end{vmatrix} \\ \\ &= \ihat(u_2v_3 - u_3v_2) - \jhat(u_1v_3 - u_3v_1) + \khat(u_1v_2 - u_2v_1) \end{align*} $$

This means that each of the coordinates of $\mathbf{u}\times\mathbf{v}$ is zero, hence we have a system of equations

$$ \begin{align*} u_2v_3 - u_3v_2 &= 0 \\ u_1v_3 - u_3v_1 &= 0 \\ u_1v_2 - u_2v_1 &= 0 \end{align*} $$

This system can be rewritten

$$ \begin{align*} u_2v_3 &= u_3v_2 \\ u_1v_3 &= u_3v_1 \\ u_1v_2 &= u_2v_1 \end{align*} $$

Finally, ignoring the special cases when some of the coordinates are zeros,

$$ \begin{align*} \frac{u_3}{v_3} &= \frac{u_2}{v_2} = \frac{u_1}{v_1} \\ u_1 &= \frac{u_3}{v_3}v_1 \\ u_2 &= \frac{u_3}{v_3}v_2 \\ u_3 &= \frac{u_1}{v_1}v_3 = \frac{u_3}{v_3}v_3 \end{align*} $$

Hence vector $\mathbf{u}$ is the a scalar multiple of $\mathbf{v}$.

Corollary 12.4.2

Vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^3$ are parallel if and only if they are scalar multiples of each other.

Exercise 2

Determine a vector perpendicular to the plane that passes through the points

$$ P_1(1,-1,1),\ P_2(1,2,-4),\ P_3(0,-3,5) $$

View Solution

In order to find a vector perpendicular to a __plane__ of vectors, one should think of cross product since the cross product of two vectors is a vector perpendicular to both. __How does one find two vectors in the plane?__ Given three points, pick one to be the tail of two possible vectors (there are three choices that will all yield a correct result). This results in two vectors
$$ \begin{align*} \overrightarrow{P_1P_2} &= \langle 1 - 1, 2 - (-1), -4 - 1 \rangle = \langle 0, 3, -5 \rangle,\ \text{and} \\ \overrightarrow{P_1P_3} &= \langle 0 - 1, -3 - (-1), 5 - 1 \rangle = \langle -1, -2, 4 \rangle \end{align*} $$
A vector perpendicular to these two vectors on the plane will be perpendicular to all vectors on the plane.
$$ \begin{align*} \overrightarrow{P_1P_2}\times\overrightarrow{P_1P_3} &= \begin{vmatrix}\ \ihat\ &\ \ 0\ & -1\ \\ \ \jhat\ &\ \ 3\ & -2\ \\ \ \khat\ & -5\ &\ \ 4\ \end{vmatrix} \\ \\ &= \ihat\begin{vmatrix}\ \ 3\ & -2\ \\ -5\ &\ \ 4\ \end{vmatrix} - \jhat\begin{vmatrix}\ \ 0\ & -1\ \\ -5\ &\ \ 4\ \end{vmatrix} + \khat\begin{vmatrix}\ \ 0\ & -1\ \\ \ \ 3\ & -2\ \end{vmatrix} \\ \\ &= \ihat(2) - \jhat(-5) + \khat(3) = \langle 2, 5, 3 \rangle \end{align*} $$

Exercise 3

Determine the area of the triangle with vertices

$$ P_1(1,-1,1),\ P_2(1,2,-4),\ P_3(0,-3,5) $$

given that the units of the plane is in centimeters.

View Solution

The area of the triangle is half of the area of the parallelogram spanned by vectors $\overrightarrow{P_1P_2}$ and $\overrightarrow{P_1P_3}$. Thus the area of the triangle is half of the magnitude of the cross product of the two vectors
$$ \frac{1}{2}\left|\overrightarrow{P_1P_2}\times\overrightarrow{P_1P_3}\right| = \frac{1}{2}\sqrt{2^2 + 5^2 + 3^3} = \frac{1}{2}\sqrt{4 + 25 + 9} = \frac{\sqrt{38}}{2}\,\text{cm}^2 $$

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12.4.4 Properties of Cross Product

Cross Product is a Linear Transformation

Cross Products of Standard Basis Vectors

$$ \begin{align*} \ihat\times\jhat &= \khat &\qquad \jhat\times\khat &= \ihat &\qquad \khat\times\ihat &= \jhat \\ \\ \jhat\times\ihat &= -\khat &\qquad \khat\times\jhat &= -\ihat &\qquad \ihat\times\khat &= -\jhat \end{align*} $$

Useful Identities

If $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ are vectors in three dimensional space and $\lambda$ is a scalar then,

$$ \begin{align*} 1. &\ \mathbf{a}\times\mathbf{b} = -\mathbf{b}\times\mathbf{a} \\ \\ 2. &\ (\lambda\mathbf{a})\times\mathbf{b} = \lambda(\mathbf{a}\times\mathbf{b}) = \mathbf{a}\times(\lambda\mathbf{b}) \\ \\ 3. &\ \mathbf{a}\times(\mathbf{b} + \mathbf{c}) = \mathbf{a}\times\mathbf{b} + \mathbf{a}\times\mathbf{c} \\ \\ 4. &\ (\mathbf{a} + \mathbf{b})\times\mathbf{c} = \mathbf{a}\times\mathbf{c} + \mathbf{b}\times\mathbf{c} \\ \\ 5. &\ \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = (\mathbf{a}\times\mathbf{b})\cdot\mathbf{c} \\ \\ 6. &\ \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = (\mathbf{a}\cdot\mathbf{c})\mathbf{b} - (\mathbf{a}\cdot\mathbf{b})\mathbf{c} \end{align*} $$

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12.4.5 Triple Product

Figure 2

The area of the base of the parallelepiped inscribed by vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ has area $\left|\mathbf{a}\times\mathbf{b}\right|$, the length of the cross product. The volume of the parallelepiped is the area of the base times its height. The height of the parallelepiped is the orthogonal height. The best way to compute this orthogonal height is the component of $\mathbf{c}$ onto a vector perpendicular to the base, $\mathbf{a}\times\mathbf{b}$. Thus the height is given by

$$ \text{Comp}_{\mathbf{a}\times\mathbf{b}}\mathbf{c} = \dfrac{\mathbf{c}\cdot(\mathbf{a}\times\mathbf{b})}{|\mathbf{a}\times\mathbf{b}|} $$
The base of the parallelepiped has area

$$ \text{Area} = |\mathbf{a}\times\mathbf{b}| $$
Hence the volume of the parallelepiped is given by

$$ \text{Volume} = \text{area of base} \times \text{height} = \left(|\mathbf{a}\times\mathbf{b}|\right)\left(\dfrac{\mathbf{c}\cdot(\mathbf{a}\times\mathbf{b})}{|\mathbf{a}\times\mathbf{b}|}\right) = (\mathbf{a}\times\mathbf{b})\cdot\mathbf{c} = \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) $$
This is not quite correct because the inner product can be a negative number. Since area is always positive we have

$$\text{Volume} = \left|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})\right|$$
This scalar result

$$ \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) $$
is called the triple product of vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$.

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