Math 344: Calculus III

12.4 Cross Product

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12.4.1 Determinant

$$ \require{color} \definecolor{brightblue}{rgb}{.267, .298, .812} \definecolor{darkblue}{rgb}{.08, .18, .28} \definecolor{palepink}{rgb}{1, .73, .8} \definecolor{softmagenta}{rgb}{.99,.34,.86} \def\ihat{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}} \def\jhat{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}} \def\khat{\hat{\mathrm{k}}} \def\tombstone{\unicode{x220E}} \mathbf{P} = (x_1, x_2, \dots, x_n ) $$

Understanding of vector cross product and vector calculus will require us to understand determinants of $n\times n$ matrices. We will use the concept of determinant for cross products here in Chapter 12, and return to determinants in Chapter 15. In Chapter 15 the we will call our determinant a Jacobian. In this chapter we will use the determinant to compute the vector product or cross product of two vectors in three dimensional space. First let us study determinants.

Determinant

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12.4.2 Computing Two Dimensional Determinants

Determinant of a 2 x 2 Matrix

For a given $2\times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, its determinant is given by

$$ \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \text{det}(A) = |A| = \text{det}\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = ad - bc $$

Example 12.4.1

Compute the determinant of $A = \begin{bmatrix} -1\ &\ \ 8\ \\ \ \ 3\ & -2\ \end{bmatrix}$

$$ \text{det}(A) = |A| = \begin{vmatrix} -1\ &\ \ 8\ \\ \ \ 3\ & -2\ \end{vmatrix} = (-1)(-2) - (3)(8) = 2 - 24 = -22 $$
Recall that matrix $A$ maps $\ihat$ to the vector $\begin{bmatrix} -1\ \\ \ \ 3\ \end{bmatrix}$, and $\jhat$ to vector $\begin{bmatrix}\ \ 8\ \\ -2\ \end{bmatrix}$.

We determine from the graphs of $A\ihat$ and $A\jhat$, that the orientation is clockwise from $A\ihat$ to $A\jhat$ instead of counter-clockwise so the orientation is flipped. This means that the determinant is a negative number. Also, the square with sides $\ihat$ and $\jhat$, and area one, is stretched to a parallelogram with sides $A\ihat$ and $A\jhat$, and area 22. Thus the determinant of matrix $A$ is negative twenty two.

Exercise 12.4.1

Compute the determinant of matrix $A = \begin{bmatrix} 6 & 7 \\ 1 & 8 \end{bmatrix}$.

Check Your Work

$$ |A| = \begin{vmatrix} 6 & 7 \\ 1 & 8 \end{vmatrix} = (6)(8) - (7)(1) = 48 - 7 = 41 $$

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12.4.3 Computing Three Dimensional Determinants

Computing the determinant of a $3\times 3$ matrix requires breaking down the computation into $3$, $2\times 2$ determinants.

We are going to study the Laplace Expansion of the determinant because it is easier to remember and easier to compute. There are three steps:

1. Decide which row or column you are use for your expansion.
2. Determine the signs of the elements of the row or column you chose.
3. Determine the minor of the elements of the row or column you chose.
4. Multiply the sign * element * determinant of the minor for each element and add them all up.

Example 12.4.2

In order to compute the determinant of $A = \begin{bmatrix}\ \ 6\ &\ \ 7\ &\ \ 4\ \\ -3\ & \ \ 6\ & -4\ \\ \ \ 1\ &\ \ 8\ &\ \ 7\ \end{bmatrix}$

1. Choose to expand the matrix along the first column. $|A| = \begin{vmatrix}\ \ \color{#307fe2}{6}\ &\ \ 7\ &\ \ 4\ \\ \color{#307fe2}{-3}\ & \ \ 6\ & -4\ \\ \ \ \color{#307fe2}{1}\ &\ \ 8\ &\ \ 7\ \end{vmatrix}$
   
   
   
2. The signs of the elements of the determinant follow a pattern like a chess board $\begin{vmatrix} + & - & + \\ - & + & - \\ + & - & + \end{vmatrix}$
   In our example this means that we have
   
   $$ +\,6,\quad -\,(-3) = 3,\quad\text{and}\ + 1 $$
   
   
3. There are three minor matrices to determine. The minor of an element of a $3\times 3$ matrix is the $2\times 2$ matrix that is left when you remove the row and column the element resides.
   
   
   

4. Multiply and add:
$$ \begin{align*} |A| &= +\,6\begin{vmatrix}\ \ 6\ \ & -4\ \ \\ \ \ 8\ \ &\ \ 7\ \ \end{vmatrix} -\,(-3)\begin{vmatrix}\ \ 7\ \ &\ \ 4\ \ \\ \ \ 8\ \ &\ \ 7\ \ \end{vmatrix} +\,(1)\begin{vmatrix}\ \ 7\ \ &\ \ 4\ \ \\ \ \ 6\ \ & -4\ \ \end{vmatrix} \\ \\ &= 6\left[ (6)(7) - (-4)(8) \right] + 3\left[ (7)(7) - (8)(4) \right] + \left[ (7)(-4) - (6)(4) \right] \\ \\ &= 6\left[ 42 + 32 \right] + 3\left[ 49 - 32 \right] + \left[ -28 - 24 \right] \\ \\ &= 6(74) + 3(17) - 52 = 444 + 51 - 52 = 443 \\ \\ \end{align*} $$

So matrix $A$ preserves orientation and the unit cube is stretched into a parallelepiped with volume 443. The determinant of matrix $A$ is 443.

Exercise 12.4.2

Compute the determinant of matrix $A = \begin{bmatrix}\ \ 8\ \ &\ \ 5\ \ & -2\ \ \\ \ \ 4\ \ &\ \ 1\ \ &\ \ 5\ \ \\ \ \ 5\ \ &\ \ 4\ \ & -4\ \ \end{bmatrix}$.

Check Your Work

$$ \begin{align*} |A| &= \begin{bmatrix}\ \ 8\ \ &\ \ 5\ \ & -2\ \ \\ \ \ 4\ \ &\ \ 1\ \ &\ \ 5\ \ \\ \ \ 5\ \ &\ \ 4\ \ & -4\ \ \end{bmatrix} \\ \\ &= 8\begin{vmatrix}\ \ 1\ \ &\ \ 5\ \ \\ \ \ 4\ \ & -4\ \ \end{vmatrix} - 4\begin{vmatrix}\ \ 5\ \ & -2\ \ \\ \ \ 4\ \ & -4\ \ \end{vmatrix} + 5\begin{vmatrix}\ \ 5\ \ & -2\ \ \\ \ \ 1\ \ &\ \ 5\ \ \end{vmatrix} \\ \\ &= 8\left[ (1)(-4) - (5)(4) \right] - 4\left[ (5)(-4) - (-2)(4) \right] + 5\left[ (5)(5) - (-2)(1) \right] \\ \\ &= 8\left[ -4 - 20 \right] - 4\left[ -20 + 8 \right] + 5\left[ 25 + 2 \right] \\ \\ &= 8(-24) - 4(-12) + 5(27) = -192 + 48 + 135 = -192 + 183 = -9 \end{align*} $$

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12.4.4 Cross Product

Cross Product

The cross product, vector product, or outer product of two $n\times 1$ vectors is another $n\times 1$ vector.

We use the times symbol $\times$ to indicate the vector product of two vectors.

Example 12.4.3

Compute the outer product of vectors $\mathbf{x} = -3\ihat - \jhat + 3\khat$ and $\mathbf{y} = 8\ihat + 8\jhat - 2\khat$.

$$ \begin{align*} \mathbf{x}\times\mathbf{y} &= \begin{vmatrix}\ \ \ihat & -3 &\ \ 8\ \ \\ \ \ \jhat & -1 &\ \ 8\ \ \\ \ \ \khat &\ \ 3 & -2\ \ \end{vmatrix} \\ \\ &= \begin{vmatrix} -1 &\ \ 8\ \ \\ \ \ 3 & -2\ \ \end{vmatrix}\ihat - \begin{vmatrix} -3 &\ \ 8\ \ \\ \ \ 3 & -2\ \ \end{vmatrix}\jhat + \begin{vmatrix} -3 &\ \ 8\ \ \\ -1 &\ \ 8\ \ \end{vmatrix}\khat \\ \\ &= \left((-1)(-2) - (3)(8)\right)\ihat - \left((-3)(-2) - (3)(8)\right)\jhat + \left((-3)(8) - (-1)(8)\right)\khat \\ \\ &= (2 - 24)\ihat - (6 - 24)\jhat + (-24 + 8)\khat \\ \\ &= -22\ihat + 18\jhat - 16\khat = \langle -22, 18, -16\rangle \end{align*} $$

Exercise 12.4.3

Compute the cross product of vectors $\mathbf{u} = \langle 6, -3, 1 \rangle$ and $\mathbf{v} = \langle 7, 6, 8 \rangle$.

Check Your Work

$$ \begin{align*} \mathbf{u}\times\mathbf{v} &= \begin{vmatrix}\ \ \ihat &\ \ 6 & 7\ \ \\ \ \ \jhat & -3 & 6\ \ \\ \ \ \khat &\ \ 1 & 8\ \ \end{vmatrix} \\ \\ &= \begin{vmatrix} -3 & 6\ \ \\ \ \ 1 & 8\ \ \end{vmatrix}\ihat - \begin{vmatrix}\ \ 6 & 7\ \ \\ \ \ 1 & 8\ \ \end{vmatrix}\jhat + \begin{vmatrix}\ \ 6 & 7\ \ \\ -3 & 6\ \ \end{vmatrix}\khat \\ \\ &= ((-3)(8) - (1)(6))\ihat - ((6)(8) - (1)(7))\jhat + ((6)(6) - (-3)(7))\khat \\ \\ &= (-24 - 6)\ihat - (48 - 7)\jhat + (36 + 21)\khat \\ \\ &= -30\ihat - 41\jhat + 57\khat = \langle -30, -41, 57 \rangle \end{align*} $$

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12.4.5 Area and Length of Cross Product

The length of the cross product of two vectors is equal to the area of the parallelogram inscribed by the two vectors. Two vectors in three dimensional space lie in the same plane and the area of the parallelogram is given by

$$ \text{Area} = \left|\mathbf{b}\right|\left|\mathbf{a}\right|\sin(\theta) = \left|\mathbf{a}\right|\left|\mathbf{b}\right|\sin(\theta) $$

Length of Cross Product of two Vectors

The length of the outer product, cross product, or vector product of two vectors in three dimensional space is given by

$$ \left|\mathbf{a}\times\mathbf{b}\right| = \left|\mathbf{a}\right|\left|\mathbf{b}\right|\sin(\theta) $$

As two nonzero vectors are parallel if and only if the angle between them is zero, we see that the vector product of two parallel vectors $\mathbf{a}$ and $\mathbf{b}$ must have a length of zero.

$$ \left|\mathbf{a}\times\mathbf{b}\right| = \left|\mathbf{a}\right|\left|\mathbf{b}\right|\sin(0) = 0 $$
Recall that the only vector with a length of zero is the zero vector.

Determine when Two Vectors are Parallel

Two nonzero vectors $\mathbf{a}$ and $\mathbf{b}$ are parallel if and only if

$$ \mathbf{a}\times\mathbf{b} = \mathbf{0} $$

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12.4.6 Direction of Cross Product

Cross Product of a Vector with Itself is the Zero Vector

Every vector is parallel to itself, so for every vector $\,\mathbf{a}$

$$ \mathbf{a}\times\mathbf{a} = \mathbf{0} $$

If we say that vector $\mathbf{a} = \langle a_1, a_2, a_3\rangle$, and $\mathbf{b} = \langle b_1, b_2, b_3 \rangle$ we can compute the cross product using the special determinant

$$ \mathbf{a}\times\mathbf{b} = \begin{vmatrix} \ihat & a_1 & b_1 \\ \jhat & a_2 & b_2 \\ \khat & a_3 & b_3 \end{vmatrix} = \begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}\ihat - \begin{vmatrix} a_1 & b_1 \\ a_3 & b_3 \end{vmatrix}\jhat + \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}\khat $$
If we compute the dot product of $\mathbf{a}\times\mathbf{b}$ and $\mathbf{a}$ one obtains

$$ \begin{align*} (\mathbf{a}\times\mathbf{b})\cdot\mathbf{a} &= \left\langle \begin{vmatrix}\ a_2 & b_2 \\ a_3 & b_3 \end{vmatrix},\ -\begin{vmatrix} a_1 & b_1 \\ a_3 & b_3 \end{vmatrix},\ \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}\right\rangle\cdot\left\langle a_1, a_2, a_3 \right\rangle \\ \\ &= \begin{vmatrix}\ a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}a_1 - \begin{vmatrix} a_1 & b_1 \\ a_3 & b_3 \end{vmatrix}a_2 + \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}a_3 \\ \\ &= \left(a_2b_3 - a_3b_2\right)a_1 - \left(a_1b_3 - a_3b_1\right)a_2 + \left(a_1b_2 - a_2b_1\right)a_3 \\ \\ &= a_1a_2b_3 - a_1a_3b_2 - a_1a_2b_3 + a_2a_3b_1 + a_1a_3b_2 - a_2a_3b_1 \\ \\ &= 0 \end{align*} $$
We verified that the cross product vector $\mathbf{a}\times\mathbf{b}$ is perpendicular to $\mathbf{a}$. Similarly $\mathbf{a}\times\mathbf{b}$ is perpendicular to $\mathbf{b}$.

Vector Product of Two Vectors

For two nonzero vectors $\,\mathbf{a}$ and $\,\mathbf{b}$, the vector $\,\mathbf{a}\times\mathbf{b}$ is orthogonal to both $\,\mathbf{a}$ and $\,\mathbf{b}$.

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12.4.7 Problems Involving Cross Products

Example 12.4.4

Find the area of the parallelogram inscribed in the points $P(8,4,5)$, $Q(5,1,4)$, $R(-2,5,-4)$, and $S(2,-5,-5)$.

The length of the cross product of vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ is the area of the parallelogram inscribed by these two vectors.

$$ \begin{align*} \overrightarrow{PQ} &= Q-P = \langle 5-8, 1-4, 4-5 \rangle = \langle -3, -3, -1 \rangle \\ \\ \overrightarrow{PR} &= R-P = \langle -2-8, 5-4, -4-5) \rangle = \langle -10, 1, -9 \rangle \\ \\ \overrightarrow{PQ}\times\overrightarrow{PR} &= \begin{vmatrix} \ihat & -3 & -10 \\ \jhat & -3 & 1 \\ \khat & -1 & -9 \end{vmatrix} \\ \\ &= \begin{vmatrix} -3 & 1 \\ -1 & -9 \end{vmatrix}\ihat - \begin{vmatrix} -3 & -10 \\ -1 & -9 \end{vmatrix}\jhat + \begin{vmatrix} -3 & -10 \\ -3 & 1 \end{vmatrix}\khat \\ \\ &= \left((-3)(-9) - (1)(-1)\right)\ihat - \left((-3)(-9) - (-10)(-1)\right)\jhat + \left((-3)(1) - (-10)(-3)\right)\khat \\ \\ &= (27 + 1)\ihat - (27 - 10)\jhat + (-3 - 30)\khat \\ \\ &= 28\ihat - 17\jhat - 33\khat = \langle 28, -17, -33 \rangle \\ \\ \left|\,\overrightarrow{PQ}\times\overrightarrow{PR}\,\right| &= \sqrt{28^2 + (-17)^2 + (-33)^2} = \sqrt{784 + 289 + 1089} = \sqrt{2162} \end{align*} $$
Hence the area of the parallelogram is $\sqrt{2162}\approx 46.5 $.

Example 12.4.5

Find a vector perpendicular to the plane passing through the points $P(1,4,3)$, $Q(6,5,0)$, and $R(8,-4,1)$.

$$ \begin{align*} \overrightarrow{PQ} &= Q-P = \langle 6-1, 5-4, 0-3 \rangle = \langle 5, 1, -3 \rangle \\ \\ \overrightarrow{PR} &= R-P = \langle 8-1, -4-4, 1-3 \rangle = \langle 7, -8, -2 \rangle \\ \\ \overrightarrow{PQ}\times\overrightarrow{PR} &= \begin{vmatrix} \ihat & 5 & 7 \\ \jhat & 1 & -8 \\ \khat & -3 & -2 \end{vmatrix} \\ \\ &= \begin{vmatrix} 1 & -8 \\ -3 & -2 \end{vmatrix}\ihat - \begin{vmatrix}\ \ 5 &\ \ 7 \\ -3 & -2 \end{vmatrix}\jhat + \begin{vmatrix} 5 & 7 \\ 1 & -8 \end{vmatrix}\khat \\ \\ &= \left((1)(-2) - (-8)(-3)\right)\ihat - \left((5)(-2) - (7)(-3)\right)\jhat + \left((5)(-8) - (7)(1)\right)\khat \\ \\ &= (-2-24)\ihat - (-10+21)\jhat + (-40-7)\khat \\ \\ &= -26\ihat - 11\jhat - 47\khat \end{align*} $$

12.4.8 Properties of Cross Product

Cross Product is a Linear Transformation

The DUAL of a vector represented by a column of coordinates is the transpose of the $n\times 1$ matrix. This $1\times n$ transpose is a linear transformation from $\mathbb{R}^n \rightarrow \mathbb{R}$.

Since the cross product is anti-commutative and has the right-hand rule, we have the following identities for the cross products of the standard basis vectors:

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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12.4.9 Triple Product

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

$$\text{Area} = \left|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})\right|$$
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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