2.4 The Cross Product

← Chapter 2 §2.4 The Cross Product 叉积 · Chapter 2 · Vectors in Space

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	Theorem 2.6: 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	Takeaways; say them first, then click
2.4.1 Calculate the cross product of two given vectors.	(2.9) slot by slot; right hand; $\mathbf v\times\mathbf u=-(\mathbf u\times\mathbf v)$ ; expand keeping the order
2.4.2 Use determinants to calculate a cross product.	$\mathbf i,\mathbf j,\mathbf k$ in row 1, expand $+\,-\,+$ ; dot-check
2.4.3 Find a vector orthogonal to two given vectors.	$\mathbf u\times\mathbf v$ ; unit: divide by its length
2.4.4 Determine areas and volumes by using the cross product.	area $\\|\mathbf u\times\mathbf v\\|$ , triangle ½; volume $\|\mathbf u\cdot(\mathbf v\times\mathbf w)\|$
2.4.5 Calculate the torque of a given force and position vector.	$\boldsymbol\tau=\mathbf r\times\mathbf F$ , $\\|\boldsymbol\tau\\|=\\|\mathbf r\\|\\|\mathbf F\\|\sin\theta$

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	$\langle a,b,0\rangle\times\langle c,d,0\rangle=(ad-bc)\mathbf k$

#	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

#	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

The Cross Product

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

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⟩

Two perpendicularity equations force the cross-product formula

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

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

Each property comes down to swapping or expanding components

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

u×v\mathbf u\times\mathbf vu×v is square to both and as long as their area

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

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⟩

∥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

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

The dot product and the cross product share one fixed budget

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

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

Lean the box and the volume stays; only the height counts

Base times height is ∣u⋅(v×w)∣|\mathbf u\cdot(\mathbf v\times\mathbf w)|∣u⋅(v×w)∣; the sign records the hand

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

Torque grows with the wrench, the force and the sine of the angle

The cross product is an area with a direction

Common mistakes

Practice and answers

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: $(2\mathbf i\times3\mathbf j)\times\mathbf j=-6\mathbf i$

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

$\mathbf u\times\mathbf v$ is square to both and as long as their area

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

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

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

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

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: $|-40|=40$

Base times height is $|\mathbf u\cdot(\mathbf v\times\mathbf w)|$ ; the sign records the hand

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