Calculus Volume 3 · Chapter 2 · Section 2.4

The Cross Product

Two vectors in, one vector out: perpendicular to both, as long as their parallelogram's area, pointing the way the right hand says.

Idea 1 · perpendicular to bothA product that returns a direction.The dot product gives a number; this gives a vector. Idea 2 · length is area‖u × v‖ is the parallelogram's area.Parallel vectors give the zero vector. Idea 3 · order is directionv × u = −(u × v).Fingers along the first, curl to the second: the thumb.

Colour key u v w u × v, τ key idea · 3D figures turn when you drag them · page links open the textbook

Level 1 · see itWhat the cross product isbefore any formula

Which way is perpendicular?

A wrench turns a bolt, which moves along its own axis — at right angles to the handle and to the push (p149). Up or down, and how hard?

Without the wrench: $\mathbf u=\langle 0,2,1\rangle$ , $\mathbf v=\langle 3,-1,0\rangle$ ; find $\mathbf w$ with $\mathbf u\cdot\mathbf w=0$ , $\mathbf v\cdot\mathbf w=0$ .

2w_2+w_3=0,\qquad 3w_1-w_2=0 \quad\Longrightarrow\quad \mathbf w=t\,\langle 1,3,-6\rangle

Two conditions in three dimensions leave a whole line of answers. The cross product picks $t=1$ : $\mathbf u\times\mathbf v=\langle 1,3,-6\rangle$ (Example 2.32).

A dot product is a number, and a number has no direction: answering "which way" needs a product that returns a vector.

Wrench r, push F. Both candidates lie on the bolt's axis. Drag to turn.

Three facts, one figure

Perpendicular to both $\mathbf u$ and $\mathbf v$ , so to the plane they span.
Length = area of the parallelogram on $\mathbf u$ , $\mathbf v$ . Parallel vectors span no area: $\mathbf u\times\mathbf v=\mathbf 0$ .
Right hand: fingers along $\mathbf u$ , curl toward $\mathbf v$ ; the thumb is $\mathbf u\times\mathbf v$ (p150). Seen from its tip, $\mathbf u$ turns to $\mathbf v$ counter-clockwise.

Example 2.32: $\mathbf u\times\mathbf v=\langle 1,3,-6\rangle$ points below the floor; length $\sqrt{46}\approx 6.78$ = the shaded area.

parallelogram right angles also v × u

Example 2.32 in space. Press from the tip of u × v to look straight down the red arrow.

The right-hand rule

Keep u on the $x$ -axis and turn v in the floor.

turn of v, φ60°

also show v × u

Red: $z$ -component of $\mathbf u\times\mathbf v$ , $5\sin\varphi$ ; dashed: $\mathbf u\cdot\mathbf v=5\cos\varphi$ . Past 180°, v is clockwise from u seen from above, so the thumb points down.

Curl your right fingers along the grey arrow: the thumb matches the red one.

With the arrows: positive. Against: negative.

Unit vectors (p152):

\begin{aligned} \mathbf i\times\mathbf j&=\mathbf k, & \mathbf j\times\mathbf k&=\mathbf i, & \mathbf k\times\mathbf i&=\mathbf j,\\ \mathbf j\times\mathbf i&=-\mathbf k, & \mathbf k\times\mathbf j&=-\mathbf i, & \mathbf i\times\mathbf k&=-\mathbf j,\\ \mathbf i\times\mathbf i&=\mathbf 0, & \mathbf j\times\mathbf j&=\mathbf 0, & \mathbf k\times\mathbf k&=\mathbf 0 . \end{aligned}

$\mathbf i\times\mathbf j=\mathbf k$ is what "right-handed axes" means. A vector crossed with itself is the zero vector.

One picture

\mathbf u=\langle 3,0,0\rangle

\mathbf v=\langle 1,2,0\rangle

: area 6, and

\mathbf u\times\mathbf v=\langle 0,0,6\rangle

is 6 long. Reverse the order and only the direction changes.

The cross product is an area with a direction.

Level 2 · compute itThe determinant, the rules, what it measuresand four labs

The determinant

Definition · p149 · as a determinant · p156

\mathbf u\times\mathbf v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\ u_1&u_2&u_3\\ v_1&v_2&v_3\end{vmatrix} =\begin{vmatrix}u_2&u_3\\ v_2&v_3\end{vmatrix}\mathbf i-\begin{vmatrix}u_1&u_3\\ v_1&v_3\end{vmatrix}\mathbf j+\begin{vmatrix}u_1&u_2\\ v_1&v_2\end{vmatrix}\mathbf k

Cover a column, take the $2\times2$ left over: $\begin{vmatrix}a&b\\ c&d\end{vmatrix}=ad-bc$ . Signs $+\ -\ +$ .

Check every answer by dotting it with both vectors: a wrong sign almost never stays perpendicular.

Rules and length

Theorem 2.6 (p153).

rule	name	means
$\mathbf u\times\mathbf v=-(\mathbf v\times\mathbf u)$	anticommutative	order is direction
$\mathbf u\times(\mathbf v+\mathbf w)=\mathbf u\times\mathbf v+\mathbf u\times\mathbf w$	distributive	expand, keeping the order
$c(\mathbf u\times\mathbf v)=(c\mathbf u)\times\mathbf v=\mathbf u\times(c\mathbf v)$	scalars factor out	stretch a side, scale the area
$\mathbf u\times\mathbf 0=\mathbf 0,\quad \mathbf v\times\mathbf v=\mathbf 0$	zero	a flat parallelogram
$\mathbf u\cdot(\mathbf v\times\mathbf w)=(\mathbf u\times\mathbf v)\cdot\mathbf w$	triple scalar product	dot and cross trade places

Length · Theorem 2.7 · p154

\|\mathbf u\times\mathbf v\|=\|\mathbf u\|\,\|\mathbf v\|\sin\theta

Beside $\mathbf u\cdot\mathbf v=\|\mathbf u\|\|\mathbf v\|\cos\theta$ .

Not true

$\mathbf u\times\mathbf v=\mathbf v\times\mathbf u$ and $(\mathbf u\times\mathbf v)\times\mathbf w=\mathbf u\times(\mathbf v\times\mathbf w)$ (Level 3).

The dot product measures how much two vectors agree; the cross product, how far they open.

Area, volume, torque

Area · Theorem 2.8 · p157

Parallelogram on $\mathbf u,\mathbf v$ : $A=\|\mathbf u\times\mathbf v\|$ . A triangle is half.

Volume · Theorems 2.9–2.10 · p158

V=|\mathbf u\cdot(\mathbf v\times\mathbf w)|=\left|\begin{vmatrix}u_1&u_2&u_3\\ v_1&v_2&v_3\\ w_1&w_2&w_3\end{vmatrix}\right|

Zero exactly when the three vectors are coplanar.

Torque · p162

Example 2.39:

\overrightarrow{PQ}\times\overrightarrow{PR}=\langle 1,1,1\rangle

, length

\sqrt3

— twice the triangle. The normal is drawn at half length.

Examples 2.31–2.44

Example	given	result
2.31 · p150	$\mathbf p=\langle -1,2,5\rangle$ , $\mathbf q=\langle 4,0,-3\rangle$	$\mathbf p\times\mathbf q=\langle -6,17,-8\rangle$
2.32 · p151	$\mathbf u=\langle 0,2,1\rangle$ , $\mathbf v=\langle 3,-1,0\rangle$	$\mathbf u\times\mathbf v=\langle 1,3,-6\rangle$ , $\mathbf v\times\mathbf u=\langle -1,-3,6\rangle$
2.33 · p152	$\mathbf i\times(\mathbf j\times\mathbf k)$	$\mathbf i\times\mathbf i=\mathbf 0$
2.34 · p154	$(2\mathbf i\times3\mathbf j)\times\mathbf j$	$6(\mathbf k\times\mathbf j)=-6\mathbf i$
2.35 · p154	$\\|\langle 0,4,0\rangle\times\langle 0,0,-3\rangle\\|$	$4\cdot3\cdot\sin\tfrac\pi2=12$
2.36 · p155	$3\times3$ determinant, rows $2,5,-1;\ -1,1,3;\ -2,3,4$	$-19$
2.37 · p156	Example 2.31 by determinant	$\langle -6,17,-8\rangle$ ; dots with $\mathbf p$ , $\mathbf q$ : $0,\ 0$
2.38 · p156	unit vector ⟂ $\langle 5,2,-1\rangle,\ \langle 0,-1,4\rangle$	$\tfrac{1}{\sqrt{474}}\langle 7,-20,-5\rangle$
2.39 · p157	triangle $P(1,0,0),Q(0,1,0),R(0,0,1)$	area $\tfrac12\sqrt3$
2.40 · p159	$\langle 1,3,5\rangle\cdot(\langle 2,-1,0\rangle\times\langle -3,0,-1\rangle)$	$-8$
2.41 · p161	box on $\langle -1,-2,1\rangle,\langle 4,3,2\rangle,\langle 0,-5,-2\rangle$	triple product $-40$ , $V=40$
2.42 · p161	$\langle 2,0,5\rangle,\langle 2,2,4\rangle,\langle 1,-1,3\rangle$	triple product 0: coplanar
2.43 · p162	⟂ the plane of $P(9,-3,-2),Q(1,3,0),R(-2,5,0)$	$\overrightarrow{PQ}\times\overrightarrow{QR}=\langle -4,-6,2\rangle$
2.44 · p163	0.15 m wrench, 6 N at 40°	$\approx 0.58\ \text{N}\cdot\text{m}$

Lab 1 · compute any u × v

u₁−1 u₂2 u₃5 v₁4 v₂0 v₃−3

presets

component guides

Step through the determinant

Try: parallel — the area collapses, $\mathbf u\times\mathbf v=\mathbf 0$ . swap u ↔ v — every component flips sign. i, j — you get $\mathbf k$ . A long cross product is drawn shorter; its label says by how much.

Lab 2 · sine against cosine

‖u‖2.5 ‖v‖2 turn of v, φ60°

Try: φ = 90° — dot product 0, area as large as it gets. Both lengths 3 at 30° — area 4.5, half of 9. Agreement and spread share one fixed budget, $\|\mathbf u\|^2\|\mathbf v\|^2$ .

Lab 3 · the box and the triple product

Fixed base: $\mathbf v=\langle 3,0,0\rangle$ , $\mathbf w=\langle 1,2.5,0\rangle$ , area 7.5. Move the third edge u.

lean in x, u₁1 lean in y, u₂1 height, u₃2

Try: play the lean — the volume stays $7.5\times\text{height}$ . Height 0 — coplanar, volume 0. Negative height — the triple product turns negative, the volume does not. The absolute value is the volume; the sign says which side of the base u is on.

Lab 4 · torque on a bolt

Bolt on the $z$ -axis, wrench along $x$ . Opens on Example 2.44.

wrench ‖r‖ (m)0.15 force ‖F‖ (N)6 angle θ in the floor40° lift out of the floor φ0°

Not to one scale: r in m, F in N, τ in N·m.

Try: play the angle — largest at 90°, zero along the handle. Double the wrench: double the torque. Lift the push with φ: only $\tau_z$ , dashed, turns the bolt. A longer wrench and a square push: $\|\mathbf r\|\|\mathbf F\|\sin\theta$ in the hands.

Common mistakes

Mistake	Result	Fix
Dropping the minus on the $\mathbf j$ term	$\langle -6,-17,-8\rangle$ in Example 2.37: not perpendicular	signs $+\ -\ +$ ; dot the answer with $\mathbf u$
$\mathbf v\times\mathbf u$ for $\mathbf u\times\mathbf v$	right size, opposite direction	first vector in row 2
$\mathbf i\times\mathbf i=0$ , a number	a scalar added to a vector later	$\mathbf i\times\mathbf i=\mathbf 0$ , the zero vector
Angle to the horizontal in $\\|\mathbf r\\|\\|\mathbf F\\|\sin\theta$	Exercise 235: $\sin30^\circ$ gives 5.00, not 8.66 ft·lb	angle between $\mathbf r$ and $\mathbf F$ , tail to tail: 120°
Triangle area $=\\|\overrightarrow{PQ}\times\overrightarrow{PR}\\|$	twice the area	halve it
$\overrightarrow{OP}\times\overrightarrow{OQ}$ for triangle $PQR$	a different triangle, cornered at the origin	edge vectors from one vertex
$-40$ reported as a volume	a negative volume in Example 2.41	$V=\|\mathbf u\cdot(\mathbf v\times\mathbf w)\|$
$(\mathbf a\times\mathbf b)\times\mathbf c=\mathbf a\times(\mathbf b\times\mathbf c)$	$(\mathbf i\times\mathbf i)\times\mathbf j=\mathbf 0$ , but $\mathbf i\times(\mathbf i\times\mathbf j)=-\mathbf j$	brackets matter
$\langle a,b\rangle\times\langle c,d\rangle$	undefined in the plane	$\langle a,b,0\rangle\times\langle c,d,0\rangle=(ad-bc)\mathbf k$

Practice and answers

Exercises from p163; the book's key: p835, p840–842.

Checkpoints 2.30–2.42

#	task	answer
2.30	$\langle 5,1,2\rangle\times\langle -2,0,1\rangle$	$\mathbf i-9\mathbf j+2\mathbf k$
2.31	$\mathbf u,\mathbf v$ in the $xy$ -plane, $u_1,u_2,v_2>0$ , $v_1<0$ : direction of $\mathbf u\times\mathbf v$ ?	up: $u_1v_2-u_2v_1>0$
2.32	$(\mathbf i\times\mathbf j)\times(\mathbf k\times\mathbf i)$	$\mathbf k\times\mathbf j=-\mathbf i$
2.33	$(\mathbf i\times\mathbf k)\times(\mathbf k\times\mathbf j)$	$(-\mathbf j)\times(-\mathbf i)=-\mathbf k$
2.34	$\\|\langle -8,0,0\rangle\times\langle 0,2,0\rangle\\|$	$8\cdot2\cdot\sin90^\circ=16$
2.35	$\det$ with rows $1,-2,-1;\ 3,2,-3;\ 1,5,4$	$23+30-13=40$
2.36	$\langle 8,2,3\rangle\times\langle -1,0,4\rangle$	$\langle 8,-35,2\rangle$
2.37	unit vector ⟂ $\langle 4,0,3\rangle,\langle 1,1,4\rangle$	$\tfrac{1}{\sqrt{194}}\langle -3,-13,4\rangle$
2.38	area of $PQRS$ , $P(1,1,0),Q(7,1,0),R(9,4,2),S(3,4,2)$	$\\|\langle 6,0,0\rangle\times\langle 2,3,2\rangle\\|=6\sqrt{13}\approx21.63$
2.39	$\mathbf a\cdot(\mathbf b\times\mathbf c)$ , $\langle 2,-4,1\rangle,\langle 0,3,-1\rangle,\langle 5,-3,3\rangle$	17
2.40	volume, $3\mathbf i+4\mathbf j-\mathbf k,\ 2\mathbf i-\mathbf j-\mathbf k,\ 3\mathbf j+\mathbf k$	$\|-8\|=8$
2.41	coplanar? $\mathbf i+\mathbf j-\mathbf k,\ \mathbf i-\mathbf j+\mathbf k,\ \mathbf i+\mathbf j+\mathbf k$	no: triple product $-4$
2.42	force for 15 N·m at 30° on a 150-cm rod	$15/(1.5\cdot\tfrac12)=20$ N

Exercises

#	task	answer	key step
183	$\langle 2,0,0\rangle\times\langle 2,2,0\rangle$	$\langle 0,0,4\rangle$	both in the floor → along $z$
185	$(2\mathbf i+3\mathbf j)\times(\mathbf j+2\mathbf k)$	$\langle 6,-4,2\rangle$	missing components are 0
187	$(\mathbf i\times\mathbf i-2\mathbf i\times\mathbf j-4\mathbf i\times\mathbf k+3\mathbf j\times\mathbf k)\times\mathbf i$	$-2\mathbf j-4\mathbf k$	$(3\mathbf i+4\mathbf j-2\mathbf k)\times\mathbf i$
189	unit $\mathbf w$ along $\langle 3,-1,2\rangle\times\langle -2,0,1\rangle$ (p164)	$\tfrac{1}{3\sqrt6}\langle -1,-7,-2\rangle$	$\sqrt{54}=3\sqrt6$
197	determinant, rows $\mathbf i,\mathbf j,\mathbf k;\ 1,-1,7;\ 2,0,3$	$-3\mathbf i+11\mathbf j+2\mathbf k$	$\langle 1,-1,7\rangle\times\langle 2,0,3\rangle$
209	area, sides $\langle 3,2,0\rangle,\langle 0,2,1\rangle$ (p165)	7	$\\|\langle 2,-3,6\rangle\\|=\sqrt{49}$
211	$A(3,-1,2),B(2,1,5),C(1,-2,-2)$ : parallelogram, triangle, distance from A to BC	$5\sqrt6$ ; $\tfrac{5\sqrt6}{2}$ ; $\tfrac{5\sqrt6}{\sqrt{59}}\approx1.59$	height = area ÷ base
213	volume on $\mathbf i+\mathbf j,\ \mathbf j+\mathbf k,\ \mathbf i+\mathbf k$	2	3 × 3 determinant
219	$A(2,1,0),B(1,2,0),C(0,1,\alpha)$ : α for volume 3; height (p166)	$\alpha=1$ ; $h=1$	determinant $=3\alpha$
235	12-in wrench at 30°, 10 lb down (p168)	8.66 ft·lb	angle between r and F: 120°
237	20-cm wrench on $+y$ , force along $\langle 0,1,-2\rangle$ , 100 N·m	559 N	$0.4k=100$ , $\\|\mathbf F\\|=250\sqrt5$
239	$\mathbf F=1.6\times10^{-19}(\mathbf v\times\mathbf B)$ , $\mathbf v=10^5\mathbf i+10^5\mathbf j$ , $\mathbf B=0.3\mathbf j$	$4.8\times10^{-15}\,\mathbf k$ N	magnetic force is a cross product

Level 3 · why it worksWhere the formulas come fromshort arguments

Where the formula comes from

Solve for a perpendicular vector (p149):

u_1w_1+u_2w_2+u_3w_3=0,\qquad v_1w_1+v_2w_2+v_3w_3=0

Multiply by $v_3$ and $u_3$ , subtract; $w_3$ drops out:

(u_1v_3-u_3v_1)\,w_1+(u_2v_3-u_3v_2)\,w_2=0

Take $w_1=u_2v_3-u_3v_2$ , $w_2=-(u_1v_3-u_3v_1)$ ; then $w_3=u_1v_2-u_2v_1$ . That is (2.9). A determinant with a repeated row is 0, so $\mathbf u\cdot(\mathbf u\times\mathbf v)=0$ . Nothing about area went in; the length is a bonus.

Why the length is ‖u‖‖v‖ sin θ

Square and add the components of (2.9) (p154):

\|\mathbf u\times\mathbf v\|^2=\|\mathbf u\|^2\|\mathbf v\|^2-(\mathbf u\cdot\mathbf v)^2=\|\mathbf u\|^2\|\mathbf v\|^2\sin^2\theta

Check · Example 2.32

$5\cdot10-(-2)^2=46=1^2+3^2+(-6)^2$ .

Read it as Pythagoras

The sign is an orientation

Swapping two rows of a determinant flips its sign (p159); a cyclic shift is two swaps:

\mathbf u\cdot(\mathbf v\times\mathbf w)=\mathbf v\cdot(\mathbf w\times\mathbf u)=\mathbf w\cdot(\mathbf u\times\mathbf v)=-\,\mathbf u\cdot(\mathbf w\times\mathbf v)

V=\underbrace{\|\mathbf v\times\mathbf w\|}_{\text{base}}\cdot\underbrace{\frac{|\mathbf u\cdot(\mathbf v\times\mathbf w)|}{\|\mathbf v\times\mathbf w\|}}_{\text{height}}=|\mathbf u\cdot(\mathbf v\times\mathbf w)|

Positive: the edges in order are right-handed, like $\mathbf i,\mathbf j,\mathbf k$ . The determinant is a volume that remembers which hand built it.

Example 2.41: $\mathbf v\times\mathbf w=\langle 4,8,-20\rangle$ , $\mathbf u\cdot\langle 4,8,-20\rangle=-4-16-20=-40$ ; in the order $\mathbf u,\mathbf w,\mathbf v$ it is $+40$ .

Not associative

\mathbf i\times(\mathbf i\times\mathbf j)=-\mathbf j,\qquad (\mathbf i\times\mathbf i)\times\mathbf j=\mathbf 0

What holds instead:

\mathbf a\times(\mathbf b\times\mathbf c)=\mathbf b\,(\mathbf a\cdot\mathbf c)-\mathbf c\,(\mathbf a\cdot\mathbf b)

$\mathbf a\times(\mathbf b\times\mathbf c)$ is perpendicular to $\mathbf b\times\mathbf c$ , so it lies back in the plane of $\mathbf b$ and $\mathbf c$ .

Only in three dimensions

The cross product is defined only in three dimensions.

In the plane

\langle a,b,0\rangle\times\langle c,d,0\rangle=\langle 0,0,ad-bc\rangle

In two dimensions it survives as a signed area.

Count the directions

In $\mathbb R^n$ , two independent vectors leave $n-2$ perpendicular directions. Exactly one only when $n=3$ .

Where it is used next

Where	The cross product gives
§2.5 Planes	the normal $\mathbf n=\overrightarrow{PQ}\times\overrightarrow{PR}$ (Example 2.43)
§2.5 Distances	point to line = parallelogram area ÷ base (Exercise 211)
§3.3 Curves	the binormal $\mathbf B=\mathbf T\times\mathbf N$
Physics	torque $\mathbf r\times\mathbf F$ , angular momentum $\mathbf r\times\mathbf p$ , magnetic force $q\,\mathbf v\times\mathbf B$
§6.6 Surface integrals	$\\|\mathbf r_u\times\mathbf r_v\\|$ , the area of a small patch

Each use is one of the three facts: perpendicular, length = area, right-handed.

	Property	Rule	As arrows
i	anticommutative	$\mathbf u\times\mathbf v=-(\mathbf v\times\mathbf u)$	the thumb flips
ii	distributive	$\mathbf u\times(\mathbf v+\mathbf w)=\mathbf u\times\mathbf v+\mathbf u\times\mathbf w$	areas with direction add
iii	multiplication by a constant	$c(\mathbf u\times\mathbf v)=(c\mathbf u)\times\mathbf v=\mathbf u\times(c\mathbf v)$	stretch a side: area × c
iv	zero vector	$\mathbf u\times\mathbf 0=\mathbf 0\times\mathbf u=\mathbf 0$	no parallelogram
v	a vector with itself	$\mathbf v\times\mathbf v=\mathbf 0$	flat parallelogram
vi	scalar triple product	$\mathbf u\cdot(\mathbf v\times\mathbf w)=(\mathbf u\times\mathbf v)\cdot\mathbf w$	same box volume

Objective	You can now
2.4.1 cross product	(2.9) slot by slot; right-hand rule; $\mathbf v\times\mathbf u=-(\mathbf u\times\mathbf v)$
2.4.2 determinants	$\mathbf i,\mathbf j,\mathbf k$ in row 1, expand; dot-check
2.4.3 orthogonal vector	$\mathbf u\times\mathbf v$ ; unit: divide by its length
2.4.4 areas, volumes	$\\|\mathbf u\times\mathbf v\\|$ , triangle ½; $V=\|\mathbf u\cdot(\mathbf v\times\mathbf w)\|$
2.4.5 torque	$\\|\boldsymbol\tau\\|=\\|\mathbf r\\|\\|\mathbf F\\|\sin\theta$

Which way is perpendicular?

Three facts, one figure

The right-hand rule

One picture

The determinant

Rules and length

Area, volume, torque

Examples 2.31–2.44

Lab 1 · compute any u × v

Step through the determinant

Lab 2 · sine against cosine

Lab 3 · the box and the triple product

Lab 4 · torque on a bolt

Common mistakes

Practice and answers

Where the formula comes from

Why the length is ‖u‖‖v‖ sin θ

The sign is an orientation

Not associative

Only in three dimensions

Where it is used next

The Cross Product

Turning a bolt needs a direction perpendicular to both the wrench and the push

Two perpendicularity equations force the cross-product formula

The cross product is the vector of cross-differences of the components

Slot by slot, ⟨−1,2,5⟩×⟨4,0,−3⟩=⟨−6,17,−8⟩\langle -1,2,5\rangle\times\langle 4,0,-3\rangle=\langle -6,17,-8\rangle⟨−1,2,5⟩×⟨4,0,−3⟩=⟨−6,17,−8⟩

Right-hand rule: fingers along u\mathbf uu, curl toward v\mathbf vv, the thumb is u×v\mathbf u\times\mathbf vu×v

u×v=⟨1,3,−6⟩\mathbf u\times\mathbf v=\langle 1,3,-6\rangleu×v=⟨1,3,−6⟩ and v×u=⟨−1,−3,6⟩\mathbf v\times\mathbf u=\langle -1,-3,6\ranglev×u=⟨−1,−3,6⟩: one line, opposite ways

i×j=k, j×k=i, k×i=j\mathbf i\times\mathbf j=\mathbf k,\ \mathbf j\times\mathbf k=\mathbf i,\ \mathbf k\times\mathbf i=\mathbf ji×j=k, j×k=i, k×i=j; against the cycle, a minus sign

Work from the inside out: i×(j×k)=i×i=0\mathbf i\times(\mathbf j\times\mathbf k)=\mathbf i\times\mathbf i=\mathbf 0i×(j×k)=i×i=0

The cross product distributes and scales like a product — but order matters

Each property comes down to swapping or expanding components

Pull the numbers out first: (2i×3j)×j=−6i(2\mathbf i\times3\mathbf j)\times\mathbf j=-6\mathbf i(2i×3j)×j=−6i

∥u×v∥=∥u∥∥v∥sin⁡θ\|\mathbf u\times\mathbf v\|=\|\mathbf u\|\|\mathbf v\|\sin\theta∥u×v∥=∥u∥∥v∥sinθ: largest when perpendicular, zero when parallel

Lagrange's identity turns ∥u×v∥2\|\mathbf u\times\mathbf v\|^2∥u×v∥2 into ∥u∥2∥v∥2sin⁡2θ\|\mathbf u\|^2\|\mathbf v\|^2\sin^2\theta∥u∥2∥v∥2sin2θ

Only in three dimensions is there exactly one direction square to two vectors

For perpendicular vectors, ∥u×v∥\|\mathbf u\times\mathbf v\|∥u×v∥ is simply ∥u∥∥v∥\|\mathbf u\|\|\mathbf v\|∥u∥∥v∥

A 3 × 3 determinant expands along row 1 into three 2 × 2 determinants, signs + − +

Expanding along row 1, the determinant of Example 2.36 is −19

Put i,j,k\mathbf i,\mathbf j,\mathbf ki,j,k in the top row: the determinant expansion is the cross product

By determinant, p×q=−6i+17j−8k\mathbf p\times\mathbf q=-6\mathbf i+17\mathbf j-8\mathbf kp×q=−6i+17j−8k — the same as Example 2.31

Dot-check every cross product: both dot products must be 0

Cross two vectors, then divide by the length, for a unit vector orthogonal to both

The length of u×v\mathbf u\times\mathbf vu×v is the area of the parallelogram on u\mathbf uu and v\mathbf vv

Triangle PQRPQRPQR has area 12∥PQ→×PR→∥=32\tfrac12\|\overrightarrow{PQ}\times\overrightarrow{PR}\|=\tfrac{\sqrt3}{2}21​∥PQ​×PR∥=23​​

u⋅(v×w)\mathbf u\cdot(\mathbf v\times\mathbf w)u⋅(v×w), the triple scalar product, is the determinant with rows u,v,w\mathbf u,\mathbf v,\mathbf wu,v,w

Example 2.40: u⋅(v×w)=1+6−15=−8\mathbf u\cdot(\mathbf v\times\mathbf w)=1+6-15=-8u⋅(v×w)=1+6−15=−8

Swapping two rows flips the sign; a cyclic shift keeps it

The box on u,v,w\mathbf u,\mathbf v,\mathbf wu,v,w has volume ∣u⋅(v×w)∣|\mathbf u\cdot(\mathbf v\times\mathbf w)|∣u⋅(v×w)∣

A negative triple product still means a positive volume: ∣−40∣=40|-40|=40∣−40∣=40

A triple product of 0 means the three vectors lie in one plane

A normal to the plane through P,Q,RP,Q,RP,Q,R is PQ→×QR→=⟨−4,−6,2⟩\overrightarrow{PQ}\times\overrightarrow{QR}=\langle -4,-6,2\ranglePQ​×QR​=⟨−4,−6,2⟩

Torque τ=r×F\boldsymbol\tau=\mathbf r\times\mathbf Fτ=r×F: only the part of the push square to the wrench turns the bolt

Solve the torque formula for the force: 15 N·m at 30° on a 1.5 m rod needs 20 N

The cross product is an area with a direction

Slot by slot, $\langle -1,2,5\rangle\times\langle 4,0,-3\rangle=\langle -6,17,-8\rangle$

Right-hand rule: fingers along $\mathbf u$ , curl toward $\mathbf v$ , the thumb is $\mathbf u\times\mathbf v$

$\mathbf u\times\mathbf v=\langle 1,3,-6\rangle$ and $\mathbf v\times\mathbf u=\langle -1,-3,6\rangle$ : one line, opposite ways

$\mathbf i\times\mathbf j=\mathbf k,\ \mathbf j\times\mathbf k=\mathbf i,\ \mathbf k\times\mathbf i=\mathbf j$ ; against the cycle, a minus sign

Work from the inside out: $\mathbf i\times(\mathbf j\times\mathbf k)=\mathbf i\times\mathbf i=\mathbf 0$

Pull the numbers out first: $(2\mathbf i\times3\mathbf j)\times\mathbf j=-6\mathbf i$

$\|\mathbf u\times\mathbf v\|=\|\mathbf u\|\|\mathbf v\|\sin\theta$ : largest when perpendicular, zero when parallel

Lagrange's identity turns $\|\mathbf u\times\mathbf v\|^2$ into $\|\mathbf u\|^2\|\mathbf v\|^2\sin^2\theta$

For perpendicular vectors, $\|\mathbf u\times\mathbf v\|$ is simply $\|\mathbf u\|\|\mathbf v\|$

Put $\mathbf i,\mathbf j,\mathbf k$ in the top row: the determinant expansion is the cross product

By determinant, $\mathbf p\times\mathbf q=-6\mathbf i+17\mathbf j-8\mathbf k$ — the same as Example 2.31

The length of $\mathbf u\times\mathbf v$ is the area of the parallelogram on $\mathbf u$ and $\mathbf v$

Triangle $PQR$ has area $\tfrac12\|\overrightarrow{PQ}\times\overrightarrow{PR}\|=\tfrac{\sqrt3}{2}$

$\mathbf u\cdot(\mathbf v\times\mathbf w)$ , the triple scalar product, is the determinant with rows $\mathbf u,\mathbf v,\mathbf w$

Example 2.40: $\mathbf u\cdot(\mathbf v\times\mathbf w)=1+6-15=-8$

The box on $\mathbf u,\mathbf v,\mathbf w$ has volume $|\mathbf u\cdot(\mathbf v\times\mathbf w)|$

A negative triple product still means a positive volume: $|-40|=40$

A normal to the plane through $P,Q,R$ is $\overrightarrow{PQ}\times\overrightarrow{QR}=\langle -4,-6,2\rangle$

Torque $\boldsymbol\tau=\mathbf r\times\mathbf F$ : only the part of the push square to the wrench turns the bolt