Unit 1, lesson 3

The Cross Product

How two vectors in space produce a third vector perpendicular to both, how its right-handed direction and its length give normals, areas and torque, and how the triple product measures volume.

  • Theory part 5
  • Practice part 6
  • Coordinates, Vectors, and Products

Turn a wrench and the bolt moves along its axis, at a right angle to both the handle and your push. Space has a product that captures exactly this. It takes two vectors and returns a vector perpendicular to both, whose length measures the area they span. It is defined for vectors in three-dimensional space, and its direction depends on the right-handed axes from the first lesson.

Key terms

Cross product

Definition For vectors a=⟨a1,a2,a3⟩\mathbf{a} = \langle a_1, a_2, a_3 \rangle and b=⟨b1,b2,b3⟩\mathbf{b} = \langle b_1, b_2, b_3 \rangle in three-dimensional space, the cross product is the vector whose components are listed below as a column, first component on top,

a×b=[a2b3−a3b2a3b1−a1b3a1b2−a2b1].\mathbf{a} \times \mathbf{b} = \begin{bmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{bmatrix}.

The column is only a narrower way to write the same vector. In angle brackets it is ⟨a2b3−a3b2, a3b1−a1b3, a1b2−a2b1⟩\langle a_2 b_3 - a_3 b_2, \ a_3 b_1 - a_1 b_3, \ a_1 b_2 - a_2 b_1 \rangle. A convenient way to remember it is the symbolic determinant

a×b=∣ijka1a2a3b1b2b3∣,\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix},

expanded along the first row with the middle term subtracted,

(a2b3−a3b2) i−(a1b3−a3b1) j+(a1b2−a2b1) k.\begin{aligned} &(a_2 b_3 - a_3 b_2)\,\mathbf{i} \\[0.4em] &\quad - (a_1 b_3 - a_3 b_1)\,\mathbf{j} \\[0.4em] &\quad + (a_1 b_2 - a_2 b_1)\,\mathbf{k}. \end{aligned}

It is only a memory aid, since a real determinant has numbers in every entry.

In plain words Each component comes from the other two components of the vectors, multiplied crosswise. The minus sign on the middle term is the most common place to slip.

Example For a=⟨1,2,3⟩\mathbf{a} = \langle 1, 2, 3 \rangle and b=⟨4,5,6⟩\mathbf{b} = \langle 4, 5, 6 \rangle,

a×b=⟨12−15,12−6,5−8⟩=⟨−3,6,−3⟩.\begin{aligned} \mathbf{a} \times \mathbf{b} &= \langle 12 - 15, 12 - 6, 5 - 8 \rangle \\[0.4em] &= \langle -3, 6, -3 \rangle. \end{aligned}

Perpendicularity and algebraic rules

Definition For all vectors a\mathbf{a}, b\mathbf{b}, c\mathbf{c} in space and every real number tt,

(a×b)⋅a=0,(a×b)⋅b=0,b×a=−(a×b),a×a=0,a×(b+c)=a×b+a×c,(t a)×b=t (a×b).\begin{gathered} (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} = 0, \quad (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = 0, \\[0.4em] \mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b}), \qquad \mathbf{a} \times \mathbf{a} = \mathbf{0}, \\[0.4em] \mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}, \\[0.4em] (t\,\mathbf{a}) \times \mathbf{b} = t\,(\mathbf{a} \times \mathbf{b}). \end{gathered}

The cross product is not commutative, and it is not associative. In general (a×b)×c≠a×(b×c)(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} \ne \mathbf{a} \times (\mathbf{b} \times \mathbf{c}).

In plain words The result is always perpendicular to both factors. Swapping the order flips it to point the opposite way.

Example With a×b=⟨−3,6,−3⟩\mathbf{a} \times \mathbf{b} = \langle -3, 6, -3 \rangle from above, check ⟨−3,6,−3⟩⋅⟨1,2,3⟩=−3+12−9=0\langle -3, 6, -3 \rangle \cdot \langle 1, 2, 3 \rangle = -3 + 12 - 9 = 0 and ⟨−3,6,−3⟩⋅⟨4,5,6⟩=−12+30−18=0\langle -3, 6, -3 \rangle \cdot \langle 4, 5, 6 \rangle = -12 + 30 - 18 = 0. This check takes seconds and catches most arithmetic slips. In the other order, b×a=⟨3,−6,3⟩\mathbf{b} \times \mathbf{a} = \langle 3, -6, 3 \rangle. Non-associativity shows up already with basis vectors, since (i×i)×j=0(\mathbf{i} \times \mathbf{i}) \times \mathbf{j} = \mathbf{0} but i×(i×j)=i×k=−j\mathbf{i} \times (\mathbf{i} \times \mathbf{j}) = \mathbf{i} \times \mathbf{k} = -\mathbf{j}.

Right-hand rule and length

Definition Let a\mathbf{a} and b\mathbf{b} be nonzero, nonparallel vectors with angle θ\theta, where 0<θ<π0 < \theta < \pi. In right-handed coordinates, a×b\mathbf{a} \times \mathbf{b} is the unique vector that

  1. is perpendicular to both a\mathbf{a} and b\mathbf{b},
  2. has length ∣a∣ ∣b∣sin⁡θ\lvert \mathbf{a} \rvert \, \lvert \mathbf{b} \rvert \sin \theta,
  3. points the way your right thumb points when the fingers curl from a\mathbf{a} toward b\mathbf{b} through the angle θ\theta.

The length formula also holds for nonzero parallel vectors, where θ=0\theta = 0 or θ=π\theta = \pi and both sides are 00. Since sin⁡θ>0\sin \theta > 0 for 0<θ<π0 < \theta < \pi, two nonzero vectors are parallel exactly when a×b=0\mathbf{a} \times \mathbf{b} = \mathbf{0}. If a\mathbf{a} or b\mathbf{b} is 0\mathbf{0}, the cross product is 0\mathbf{0} and no angle is needed. For the basis vectors,

i×j=k,j×k=i,k×i=j.\mathbf{i} \times \mathbf{j} = \mathbf{k}, \quad \mathbf{j} \times \mathbf{k} = \mathbf{i}, \quad \mathbf{k} \times \mathbf{i} = \mathbf{j}.

In plain words Of the two directions perpendicular to both vectors, the right hand picks one. The order of the factors decides which.

Example The vectors a=⟨2,0,0⟩\mathbf{a} = \langle 2, 0, 0 \rangle and b=⟨0,3,0⟩\mathbf{b} = \langle 0, 3, 0 \rangle lie along the positive xx and yy axes. Curling from a\mathbf{a} to b\mathbf{b} points the thumb up, and the formula agrees, since a×b=⟨0,0,6⟩\mathbf{a} \times \mathbf{b} = \langle 0, 0, 6 \rangle, with length 2⋅3⋅sin⁡π2=62 \cdot 3 \cdot \sin \frac{\pi}{2} = 6. Reversing the order gives ⟨0,0,−6⟩\langle 0, 0, -6 \rangle, pointing down. The vectors ⟨1,−2,3⟩\langle 1, -2, 3 \rangle and ⟨−2,4,−6⟩\langle -2, 4, -6 \rangle have cross product 0\mathbf{0}, and indeed the second is −2-2 times the first.

Parallelogram area

Definition The parallelogram with adjacent sides a\mathbf{a} and b\mathbf{b} has area

A=∣a×b∣=∣a∣ ∣b∣sin⁡θ.A = \lvert \mathbf{a} \times \mathbf{b} \rvert = \lvert \mathbf{a} \rvert \, \lvert \mathbf{b} \rvert \sin \theta.

The triangle with sides a\mathbf{a} and b\mathbf{b} from a common vertex has half this area. If a\mathbf{a} and b\mathbf{b} are not parallel, a×b\mathbf{a} \times \mathbf{b} is a nonzero normal vector to every plane that contains both directions.

In plain words Base times height is ∣a∣\lvert \mathbf{a} \rvert times ∣b∣sin⁡θ\lvert \mathbf{b} \rvert \sin \theta, and the cross product packages that as a length.

Example Take the triangle with vertices P(1,0,0)P(1, 0, 0), Q(0,2,0)Q(0, 2, 0) and R(0,0,3)R(0, 0, 3). With PQ→=⟨−1,2,0⟩\overrightarrow{PQ} = \langle -1, 2, 0 \rangle and PR→=⟨−1,0,3⟩\overrightarrow{PR} = \langle -1, 0, 3 \rangle,

PQ→×PR→=⟨6,3,2⟩.\overrightarrow{PQ} \times \overrightarrow{PR} = \langle 6, 3, 2 \rangle.

Its length is 36+9+4=7\sqrt{36 + 9 + 4} = 7, so the triangle has area 72\frac{7}{2}, and ⟨6,3,2⟩\langle 6, 3, 2 \rangle is normal to the plane through the three points.

Torque

Definition If a force F\mathbf{F} acts at a point whose position relative to a pivot OO is r\mathbf{r}, the torque about OO is

τ=r×F,∣τ∣=∣r∣ ∣F∣sin⁡θ,\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}, \qquad \lvert \boldsymbol{\tau} \rvert = \lvert \mathbf{r} \rvert \, \lvert \mathbf{F} \rvert \sin \theta,

where θ\theta is the angle between r\mathbf{r} and F\mathbf{F}. When the torque is nonzero it points along the axis about which the force tends to rotate the object, and by the right-hand rule the rotation looks counterclockwise from the tip of τ\boldsymbol{\tau}. With meters and newtons, torque is in newton meters.

In plain words A longer handle, a stronger push, and a push closer to a right angle all turn the bolt harder. Pushing straight along the handle does nothing.

Example A wrench handle runs from the bolt along r=⟨0.25,0,0⟩\mathbf{r} = \langle 0.25, 0, 0 \rangle meters, and a force F=⟨0,40,−30⟩\mathbf{F} = \langle 0, 40, -30 \rangle newtons acts at its end. Then

τ=r×F=⟨0,7.5,10⟩,\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} = \langle 0, 7.5, 10 \rangle,

with magnitude 12.512.5 newton meters. As a check, ∣F∣=50\lvert \mathbf{F} \rvert = 50 and F\mathbf{F} is perpendicular to r\mathbf{r}, so the magnitude is 0.25⋅50⋅1=12.50.25 \cdot 50 \cdot 1 = 12.5.

Scalar triple product

Definition The scalar triple product of a\mathbf{a}, b\mathbf{b}, c\mathbf{c} is the number a⋅(b×c)\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}). It equals the determinant

a⋅(b×c)=∣a1a2a3b1b2b3c1c2c3∣.\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}.

It is unchanged by a cyclic shift of the three vectors and changes sign when any two of them are swapped, so

a⋅(b×c)=b⋅(c×a)=− a⋅(c×b).\begin{aligned} \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) &= \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) \\[0.4em] &= -\,\mathbf{a} \cdot (\mathbf{c} \times \mathbf{b}). \end{aligned}

In plain words Put the three vectors in the rows of a matrix and take the determinant. The order of the rows matters only through the sign.

Example For a=⟨1,2,0⟩\mathbf{a} = \langle 1, 2, 0 \rangle, b=⟨0,1,3⟩\mathbf{b} = \langle 0, 1, 3 \rangle and c=⟨2,0,1⟩\mathbf{c} = \langle 2, 0, 1 \rangle, expanding along the first row gives

1(1−0)−2(0−6)+0=13.1(1 - 0) - 2(0 - 6) + 0 = 13.

Swapping b\mathbf{b} and c\mathbf{c} gives −13-13.

Signed and geometric volume

Definition The parallelepiped with edges a\mathbf{a}, b\mathbf{b}, c\mathbf{c} from one vertex has volume

V=∣a⋅(b×c)∣.V = \lvert \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \rvert.

The triple product itself is a signed volume. It is zero exactly when the three vectors are coplanar, that is, when they lie in one plane once they are drawn from a common initial point. Three vectors that are not coplanar form a right-handed ordered triple (a,b,c)(\mathbf{a}, \mathbf{b}, \mathbf{c}) when c\mathbf{c} lies on the side of the plane of a\mathbf{a} and b\mathbf{b} that your right thumb points to as your fingers curl from a\mathbf{a} toward b\mathbf{b}, which is the side a×b\mathbf{a} \times \mathbf{b} points to. Otherwise the triple is left-handed. In right-handed coordinates the identity a⋅(b×c)=(a×b)⋅c\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} shows that the triple product is positive for a right-handed triple and negative for a left-handed one. For example i⋅(j×k)=1\mathbf{i} \cdot (\mathbf{j} \times \mathbf{k}) = 1. The tetrahedron with the same three edges has volume 16V\frac{1}{6}V.

In plain words The size of the triple product is a volume, and its sign records the handedness of the three edges. Zero means the box is flat.

Example The vectors above give a box of volume 1313, and since the triple product is positive, (a,b,c)(\mathbf{a}, \mathbf{b}, \mathbf{c}) is right-handed. For ⟨1,2,3⟩\langle 1, 2, 3 \rangle, ⟨4,5,6⟩\langle 4, 5, 6 \rangle and ⟨7,8,9⟩\langle 7, 8, 9 \rangle the determinant is 00, so they are coplanar. Indeed ⟨7,8,9⟩=2⟨4,5,6⟩−⟨1,2,3⟩\langle 7, 8, 9 \rangle = 2\langle 4, 5, 6 \rangle - \langle 1, 2, 3 \rangle.

Check yourself

  1. Compute ⟨2,−1,1⟩×⟨1,3,−2⟩\langle 2, -1, 1 \rangle \times \langle 1, 3, -2 \rangle and check your answer.
Show the answer

The components are (−1)(−2)−(1)(3)=−1(-1)(-2) - (1)(3) = -1, then (1)(1)−(2)(−2)=5(1)(1) - (2)(-2) = 5, then (2)(3)−(−1)(1)=7(2)(3) - (-1)(1) = 7, so the cross product is ⟨−1,5,7⟩\langle -1, 5, 7 \rangle. Its dot products with the two factors are −2−5+7=0-2 - 5 + 7 = 0 and −1+15−14=0-1 + 15 - 14 = 0.

  1. Find the area of the triangle with vertices (0,0,0)(0, 0, 0), (1,2,2)(1, 2, 2) and (2,0,1)(2, 0, 1).
Show the answer

The cross product ⟨1,2,2⟩×⟨2,0,1⟩=⟨2,3,−4⟩\langle 1, 2, 2 \rangle \times \langle 2, 0, 1 \rangle = \langle 2, 3, -4 \rangle has length 29\sqrt{29}, so the area is 292≈2.69\frac{\sqrt{29}}{2} \approx 2.69.

  1. Are the points A(1,0,1)A(1, 0, 1), B(2,1,3)B(2, 1, 3), C(0,2,1)C(0, 2, 1) and D(1,3,3)D(1, 3, 3) coplanar?
Show the answer

Yes. With AB→=⟨1,1,2⟩\overrightarrow{AB} = \langle 1, 1, 2 \rangle, AC→=⟨−1,2,0⟩\overrightarrow{AC} = \langle -1, 2, 0 \rangle and AD→=⟨0,3,2⟩\overrightarrow{AD} = \langle 0, 3, 2 \rangle, the triple product is

1(4−0)−1(−2−0)+2(−3−0)=4+2−6=0,\begin{aligned} &1(4 - 0) - 1(-2 - 0) + 2(-3 - 0) \\[0.4em] &= 4 + 2 - 6 = 0, \end{aligned}

so the four points lie in one plane. In fact AD→=AB→+AC→\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{AC}.

How this lesson connects

It also uses Vectors in an orthonormal right-handed coordinate system, dot product, sine and cosine.