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# 5.2.3: Cross Products


## Cross Products

Cross products are related to dot products in a number of ways. Both are vector calculations, both are related not only to the magnitude of each vector, but also to the relative directions of both vectors.

Dot products in a sense calculate the joined force of two vectors in a given direction, and so are greatest when the vectors are parallel. Cross products, however, are greatest when the vectors are perpendicular... what then do they calculate?

## Cross Products

Whereas a dot product of two vectors produces a scalar value; the cross product of the same two vectors produces a vector quantity having a direction perpendicular to the original two vectors.

The cross product of two vector quantities is another vector whose magnitude varies as the angle between the two original vectors changes. The cross product is sometimes referred to as the vector product of two vectors. The magnitude of the cross product represents the area of the parallelogram whose sides are defined by the two vectors, as shown in the figure below. Therefore, the maximum value for the cross product occurs when the two vectors are perpendicular to one another, but when the two vectors are parallel to one another the magnitude of the cross product is equal to zero.

The algebraic form of the cross product equation is more complicated than that for the dot product. For two 3D vectors $$\ \vec{A}$$ and $$\ \vec{B}$$,

$$\ \vec{A} \times \vec{B}=\left\langle\left(A_{2} B_{3}-A_{3} B_{2}\right),\left(A_{3} B_{1}-A_{1} B_{3}\right),\left(A_{1} B_{2}-A_{2} B_{1}\right)\right\rangle$$

Another way to describe the process is to say that the cross product is the multiplication of one vector by the component of the other vector which is perpendicular to the first vector. In the diagram below are two vectors, A and B. A perpendicular line has been drawn radially outward from B towards A to create a right triangle with A as the hypotenuse.

The component of $$\ \vec{A}$$ which is perpendicular to $$\ \vec{B}$$ is given by A sin θ so the magnitude of the cross product can be written as $$\ |\vec{A} \times \vec{B}|=\vec{A}(\vec{B} \sin \theta)=|\vec{A}||\vec{B}| \sin \theta$$

The direction of the cross product is perpendicular to the plane defined by the two crossed vectors. For example, the cross product of two vectors in the x-y plane will be parallel to the z-axis. This still leaves two possible directions for the cross product, though: either +ẑ or −ẑ .

We use a right-hand-rule to indicate the direction of the cross product. Position the thumb and index finger of your right hand with the first vector along your thumb and the second vector along your index finger. Your middle finger, when extended perpendicular to your palm, will indicate the direction of the cross product of the two vectors.

As you can see in the diagram above, $$\ \vec{A} \times \vec{B}$$ is along +ẑ (coming up out of the page) while $$\ \vec{B} \times \vec{A}$$ is along −ẑ (going down into the page) and $$\ \vec{A} \times \vec{B}=-\vec{B} \times \vec{A}$$

## The Normal Vector

We can use the cross product and the definition of the unit vector to determine the direction which is perpendicular to a plane.

In general, we can define a normal vector, n̂ , which has a unity magnitude (i.e. magnitude equal to one) and which is perpendicular to a plane occupied by a pair of vectors, U and V.

$$\ \hat{n}=\frac{\vec{U} \times \vec{V}}{|\vec{U} \times \vec{V}|}$$

## Examples

###### Example 1

Calculate the cross product of the two vectors shown below.

Solution

Use the components of the two vectors to determine the cross product.

$$\ \vec{A} \times \vec{B}=\left\langle\left(A_{y} B_{z}-A_{z} B_{y}\right),\left(A_{z} B_{x}-A_{x} B_{z}\right),\left(A_{x} B_{y}-A_{y} B_{x}\right)\right\rangle$$

Since these two vectors are both in the x-y plane, their own z-components are both equal to 0 and the vector product will be parallel to the z axis.

$$\ \begin{array}{l} \vec{A} \times \vec{B}=\langle[(3 \cdot 0)-(0 \cdot 2)],[(0 \cdot-4)-(2.5 \cdot 0)],[(2.5 \cdot 2)-(3 \cdot-4)]\rangle \\ \vec{A} \times \vec{B}=\langle[(0)-(0)],[(0)-(0)],[(5)-(-12)]\rangle=\langle 0,0,(5+12)\rangle=\langle 0,0,17\rangle \end{array}$$

We can check our answer using the sine version of the cross product, but first we need to know the angle between the two vectors. We can use the dot product to find θ. First use the components to find the dot product.

$$\ \vec{A} \times \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}=(2.5 *-4)+(3 * 2)+(0 * 0)=-10+6+0=-4$$

Then find the magnitudes of the two vectors:

$$\ \begin{array}{l} |\vec{A}|=\sqrt{A_{x}^{2}+A_{y}^{2}+A_{z}^{2}}=\sqrt{2.5^{2}+3^{2}+0^{2}}=\sqrt{6.25+9+0}=\sqrt{15.25} \\ |\vec{B}|=\sqrt{B_{x}^{2}+B_{y}^{2}+B_{z}^{2}}=\sqrt{(-4)^{2}+2^{2}+0^{2}}=\sqrt{16+4+0}=\sqrt{20} \end{array}$$

Then use these magnitudes with the cosine version of the dot product to find θ.

$$\ \begin{array}{l} \vec{A} \times \vec{B}=|A||B| \cos \theta \\ -4=\sqrt{15.25} \sqrt{20} \cos \theta \\ \cos \theta=\frac{-4}{\sqrt{305}} \approx \frac{-4}{17.5}=-0.229 \\ \theta=103^{\circ} \end{array}$$

Now use the sine of this angle and the two magnitudes to determine the cross product:

$$\ \begin{array}{l} |\vec{A} \times \vec{B}|=|\vec{A}||\vec{B}| \sin \theta \\ |\vec{A} \times \vec{B}|=\sqrt{15.25} \sqrt{20} \sin 103^{\circ}=\sqrt{305} \sin 103^{\circ}=17 \end{array}$$

This is the same answer that we obtained from the component notation, which is good. We use the right-hand rule to determine the direction of the vector product. If you place your thumb along vector A and your forefinger along vector B, your middle finger will point along +ẑ and $$\ |\vec{A} \times \vec{B}|=\langle 0,0,17\rangle$$

###### Example 2

The diagram shows two vectors A and B which define a plane passing through the origin. Use these two vectors to determine the normal vector to this plane. $$\ \vec{A}=\langle 3,0,4\rangle$$ and $$\ \vec{B}=\langle 5,10,0\rangle$$

Solution

The normal vector is defined by

$$\ \hat{n}=\frac{\vec{U} \times \vec{V}}{|\vec{U} \times \vec{V}|}$$

In this case, we obtain

$$\ \hat{n}=\frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}$$

Use the component version of the cross-product equation to find the components of $$\ \vec{A} \times \vec{B}$$

$$\ \begin{array}{l} \vec{A} \times \vec{B}=\left\langle\left(A_{y} B_{z}-A_{z} B_{y}\right),\left(A_{z} B_{x}-A_{x} B_{z}\right),\left(A_{x} B_{y}-A_{y} B_{x}\right)\right\rangle \\ \vec{A} \times \vec{B}=\langle[(0 \cdot 0)-(4 \cdot 10)],[(4 \cdot 5)-(3 \cdot 0)],[(3 \cdot 10)-(0 \cdot 5)]\rangle \\ \vec{A} \times \vec{B}=\langle(0-40),(20-0),(30-0)\rangle=\langle-40,20,30\rangle \end{array}$$

Next, calculate the magnitude of the cross product, $$\ |\vec{A} \times \vec{B}|$$

$$\ \begin{array}{l} |\vec{A} \times \vec{B}|=\sqrt{(-40)^{2}+20^{2}+30^{2}}=\sqrt{1600+400+900}=\sqrt{2900}=53.8516 \\ \hat{n}=\frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}=\frac{\langle-40,20,30\rangle}{53.9}=\left\langle\frac{-40}{53.9}, \frac{20}{53.9}, \frac{30}{53.9}\right\rangle=\langle-0.743,0.371,0.557\rangle \end{array}$$

###### Example 3

Determine the cross product $$\ \vec{F} \times \vec{r}$$ for the two vectors $$\ \vec{F}=\langle 2,3,4\rangle$$ and $$\ \vec{r}=\langle 7,6,5\rangle$$. Then use the cross product to determine the angle between the two vectors.

Solution

One of the two ways to determine the magnitude of the cross product of two vectors uses the components of the two vectors:

$$\ \begin{array}{l} \vec{F} \times \vec{r}=\left\langle\left(F_{y} r_{z}-F_{z} r_{y}\right),\left(F_{z} r_{x}-F_{x} r_{z}\right),\left(F_{x} r_{y}-F_{y} r_{x}\right)\right\rangle \\ \vec{F} \times \vec{r}=\langle(3 \cdot 5-4 \cdot 6),(4 \cdot 7-2 \cdot 5),(2 \cdot 6-3 \cdot 7)\rangle=\langle(15-24),(28-10),(12-21)\rangle \\ \vec{F} \times \vec{r}=\langle-9,18,-9\rangle \end{array}$$

Now we can use the cross product and the second definition of the cross product to determine the angle between the two vectors.

$$\ |\vec{F} \times \vec{r}|=|F||r| \sin \theta$$

We need to calculate the magnitudes of the vectors and of the cross product.

$$\ \begin{array}{l} |\vec{F}|=\sqrt{F_{x}^{2}+F_{y}^{2}+F_{z}^{2}}=\sqrt{2^{2}+3^{2}+4^{2}}=\sqrt{4+9+16}=\sqrt{29}=5.385 \\ |\vec{r}|=\sqrt{r_{x}^{2}+r_{y}^{2}+r_{z}^{2}}=\sqrt{7^{2}+6^{2}+5^{2}}=\sqrt{49+36+25}=\sqrt{110}=10.488 \\ |\vec{F} \times \vec{r}|=\sqrt{(-9)^{2}+18^{2}+(-9)^{2}}=\sqrt{81+324+81}=\sqrt{486}=22.0454 \\ \sin \theta=\frac{|\vec{F} \times \vec{r}|}{|F||r|}=\frac{22.0454}{(5.385)(10.488)}=0.390 \\ \theta=\sin ^{-1}(0.390)=22.98^{\circ} \end{array}$$

We can use the dot product of the two vectors to check our solution.

$$\ \begin{array}{l} \vec{F} \times \vec{r}=|\vec{F} \| \vec{r}| \cos \theta \\ \vec{F} \times \vec{r}=F_{x} r_{x}+F_{y} r_{y}+F_{z} r_{z}=2 * 7+3 * 6+4 * 5=14+18+20=52 \\ \cos \theta=\frac{\vec{F} \times \vec{r}}{|\vec{F}||\vec{r}|}=\frac{52}{(5.385)(10.488)}=0.920714 \\ \theta=\cos ^{-1}(0.920714)=22.97 \end{array}$$

This answer matches our value from the cross product to within rounding variations.

###### Example 4

Determine the magnitude of the cross product of the two vectors shown below.

Solution

First we need to identify the components of the two vectors by using the information given on the graph. In this case, $$\ \overrightarrow{M N}=\langle-2.25,0,0\rangle$$ and $$\ \overrightarrow{K L}=\langle 1.5,2,0\rangle$$

$$\ \begin{array}{l} \overrightarrow{M N} \times \overrightarrow{K L}=\left\langle\left(M N_{y} K L_{z}-M N_{z} K L_{y}\right),\left(M N_{z} K L_{x}-M N_{x} K L_{z}\right),\left(M N_{x} K L_{y}-M N_{y} K L_{x}\right)\right\rangle \\ \overrightarrow{M N} \times \overrightarrow{K L}=\langle(0 \cdot 0-0 \cdot 2),(0 \cdot 1.5-(-2.25) \cdot 0),((-2.25) \cdot 2-0 \cdot 1.5)\rangle \\ \overrightarrow{M N} \times \overrightarrow{K L}=\langle 0-0,0-0,-4.5-0\rangle=\langle 0,0,-4.5\rangle \end{array}$$

As we can see by the components, this vector has a magnitude of 4.5 units and lies in the –z direction. We can also use the right-hand-rule to see the direction of the cross product. As shown in the figure below, if we align the right thumb with vector MN and the right fore-finger with vector KL, the palm and extended middle-finger point in the –z direction.

###### Example 5

A plane passing through the origin is defined by the two vectors, $$\ \vec{W}=\langle 4,5,2\rangle$$ and $$\ \vec{L}=\langle 8,1,9\rangle$$. Determine the equation of a unit vector representing a direction perpendicular to this plane.

Solution

To solve this problem we need to use the definition of the normal vector $$\ \hat{n}=\frac{\vec{W} \times \vec{L}}{|\vec{W} \times \vec{L}|}$$, the component form of the definition of the cross product,

\ \begin{aligned} &\vec{W} \times \vec{L}=\left\langle\left(W_{y} L_{z}-W_{z} L_{y}\right),\left(W_{z} L_{x}-W_{x} L_{z}\right),\left(W_{x} L_{y}-W_{y} L_{x}\right)\right\rangle . \text { In this case, we obtain }\\ &\vec{W} \times \vec{L}=\langle(5 \cdot 9-2 \cdot 1),(2 \cdot 8-4 \cdot 9),(4 \cdot 1-5 \cdot 8)\rangle\\ &\vec{W} \times \vec{L}=\langle(45-2),(16-36),(4-40)\rangle=\langle 43,-20,-36\rangle \end{aligned}

We also need to know the magnitude of this cross product

$$\ \begin{array}{l} |\vec{W} \times \vec{L}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{(43)^{2}+(-20)^{2}+(-36)^{2}}=\sqrt{1849+400+1296}= \\ \sqrt{3545}=59.54 \end{array}$$

Now we can determine the normal vector

$$\ \hat{n}=\frac{\vec{W} \times \vec{L}}{|\vec{W} \times \vec{L}|}=\frac{\langle 43,-20,-36\rangle}{59.54}=\left\langle\frac{43}{59.54}, \frac{-20}{59.54}, \frac{-36}{59.54}\right\rangle=\langle 0.7222,-0.3359,-0.6046\rangle$$

###### Example 6

Determine the area of a parallelogram whose sides are defined by the vectors $$\ \vec{w}=\langle 85,89,91\rangle$$ and $$\ \vec{h}=\langle 67,70,88\rangle$$, lengths measured in centimeters.

Solution

The area of the parallelogram whose sides are defined by a pair of vectors is equal to the magnitude of the cross product of the two vectors, $$\ |\vec{w} \times \vec{h}|$$. First we need to find the cross product of the two vectors:

$$\ \begin{array}{l} \vec{w} \times \vec{h}=\left\langle\left(w_{y} h_{z}-w_{z} h_{y}\right),\left(w_{z} h_{x}-w_{x} h_{z}\right),\left(w_{x} h_{y}-w_{y} h_{x}\right)\right\rangle \\ \vec{w} \times \vec{h}=\langle(89 \cdot 88-91 \cdot 70),(91 \cdot 67-85 \cdot 88),(85 \cdot 70-89 \cdot 67)\rangle \\ \vec{w} \times \vec{h}=\langle(7832-6370),(6097-7480),(5950-5963)\rangle=\langle 1462,-1383,-13\rangle \\ |\vec{w} \times \vec{h}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{1462^{2}+(-1383)^{2}+(13)^{2}}=\sqrt{4050302} \approx 2012.5 \end{array}$$

Since the lengths of the two vectors were measured in centimeters, the area of the parallelogram is 2013 cm2 measured to the nearest square centimeter.

###### Example 7

Determine the cross product of the two vectors $$\ \vec{f}=\langle 3,13,11\rangle$$ and $$\ \vec{g}=\langle 9,6,15\rangle$$

Solution

$$\ \begin{array}{l} \vec{f} \times \vec{g}=\left\langle\left(f_{y} g_{z}-f_{z} g_{y}\right),\left(f_{z} g_{x}-f_{x} g_{z}\right),\left(f_{x} g_{y}-f_{y} g_{x}\right)\right\rangle \\ \vec{f} \times \vec{g}=\langle(13 \cdot 15-11 \cdot 6),(11 \cdot 9-3 \cdot 15),(3 \cdot 6-13 \cdot 9)\rangle \\ \vec{f} \times \vec{g}=\langle(195-66),(99-45),(18-117)\rangle=\langle 129,54,-99\rangle \end{array}$$

###### Example 8

Determine the magnitude of the cross-product of these two vectors.

Solution

Since we know the magnitudes of the two vectors and the angle between them, we can use the angle-version of the cross-product equation to determine the magnitude of the cross-product:

$$\ |\vec{A} \times \vec{B}|=|\vec{A}||\vec{B}| \sin \theta=(61)(45) \sin 58=2328$$

Since these two vectors lie in the x-y plane, the direction of the cross-product will be parallel to the z-axis.

## Review

Calculate the cross products:

1. Vectors $$\ c=-6 i+2 j+3 k$$ and $$\ a=-6 i+2 j+13 k$$
2. Vectors $$\ v=\langle-1,4,7\rangle$$ and $$\ u=\langle-5,10,3\rangle$$
3. Vectors $$\ f=-6 i+8 j-6 k$$ and $$\ s=-3 i+15 j+19 k$$
4. Vectors $$\ \mathrm{j}=-3 i+15 j-4 k$$ and $$\ \mathrm{t}=7 i+10 j+6 k$$
5. Vectors $$\ r=\langle 3,13,-1\rangle$$ and $$\ v=\langle 7,6,1\rangle$$
6. Vectors $$\ e=\langle-1,8,-3\rangle$$ and $$\ a=\langle-2,1,19\rangle$$
7. Vectors $$\ \mathrm{j}=-3 i+17 j+6 k$$ and $$\ h=8 i+9 j+7 k$$
8. Vectors $$\ a=9 i+10 j+9 k$$ and $$\ g=-5 i+19 j+15 k$$
9. Vectors $$\ \mathrm{j}=4 i+18 j-8 k$$ and $$\ m=2 i+j+19 k$$
10. What is the cross product of $$\ \langle-2,1,-2\rangle$$ and $$\ \langle 5,6,9\rangle$$?
11. Find a vector orthogonal to both $$\ \langle 1,20,2\rangle$$ and $$\ \langle 4,2,3\rangle$$
12. Vectors $$\ y=5 i+6 j+6 k$$ and $$\ f=-4 i+9 j+3 k$$. What is the area of the parallelogram formed by having y and f as adjacent sides?
13. What is the area of the parallelogram formed by having $$\ \langle-2,1,7\rangle$$ and $$\ \langle 5,7,16\rangle$$ as adjacent sides?
14. What is the cross product between $$\ \langle 8,6,8\rangle$$ and $$\ \langle 8,6,8\rangle$$?
15. Vectors $$\ g=-6 i+9 j-7 k$$ and $$\ y=-24 i+36 j-28 k$$ What is the cross product between g and y?
16. A boat is sailing on a bearing of 89° east of north at 564 feet per min. A tail wind is adding to the plane's velocity and blowing 78° west of north at 25 feet per min. Determine the actual speed of the plane in mph.
17. A plane is flying on a bearing of 77° east of south at 606 mph. A tail wind is adding to the plane's velocity and blowing 33° west of south at 80 mph. Determine the direction of the plane.

## Review (Answers)

To see the Review answers, open this PDF file and look for section 5.6.

## Vocabulary

Term Definition
Cross product The cross product of two vectors is a third vector that is perpendicular to both of the original vectors.
dot product The dot product is also known as inner product or scalar product. The two forms of the dot product are $$\ \vec{a} \cdot \vec{b}=\|\vec{a}\|\|\vec{b}\| \cos \theta$$ and $$\ \vec{a} \cdot \vec{b}=x_{a} x_{b}+y_{a} y_{b}$$
normal vector A normal vector is a vector that is perpendicular to a given surface or plane. A unit normal vector is a normal vector with a magnitude of one.
Right-hand-rule The right-hand-rule is used to indicate the direction of a cross product. Position the thumb and index finger of your right hand with the first vector along your thumb and the second vector along your index finger. Your middle finger, when extended perpendicular to your palm, will indicate the direction of the cross product of the two vectors.
unit vector A unit vector is a vector with a magnitude of one.

5.2.3: Cross Products is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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