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Topos Uranos
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The cross product and the scalar triple product

Summary

This lesson defines the cross product of two vectors in space by their coordinates, as a formal determinant with the vectors i⃗\vec{i}, j⃗\vec{j} and k⃗\vec{k}, and proves with the properties of determinants that it is anticommutative and bilinear, but not associative. It proves that the cross product is perpendicular to its two factors, Lagrange’s identity and, from it, that its norm is the area of the parallelogram determined by the factors and that it vanishes exactly when they are parallel. It establishes the vector triple product formula and the Jacobi identity, defines the scalar triple product as a determinant of order three, proves its cyclic property and interprets it as the volume of a parallelepiped and of a tetrahedron, with the coplanarity criterion and the expression of the volume through the dot products. It states precisely which aspect of the right-hand rule is proved and which is a convention, shows that the cross product is the vector part of the product of two pure quaternions, and applies it to the moment of a force, to the magnetic force on a moving charge, to the area of a triangle in space and to the normals of computer graphics.

Learning objectives

  1. Compute cross products and scalar triple products from coordinates and by means of determinants, and operate with them using anticommutativity, bilinearity, the cyclic property and the vector triple product formula.
  2. Prove the perpendicularity of the cross product, Lagrange’s identity, the criterion for parallelism, the vector triple product formula and the coplanarity criterion.
  3. Compute areas of parallelograms and of triangles in space and volumes of parallelepipeds and of tetrahedra, and decide whether four points are coplanar.
  4. Distinguish, in the right-hand rule, what is proved from what is agreed upon, and model with the cross product the moment of a force, the magnetic force and the orientation of surfaces in computer graphics.

The cross product

In the lesson on vectors in space and the dot product the vectors of R3\mathbb{R}^3 were written as triples u⃗=(u1,u2,u3)\vec{u} = \left( u_1, u_2, u_3 \right), and the dot product u⃗⋅v⃗=u1v1+u2v2+u3v3\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 made it possible to measure lengths and angles, to decide perpendicularity and to project one vector onto another. The dot product of two vectors is a number. Now, many problems in space call for a vector: the direction perpendicular to a flat surface, the axis about which a force makes a body turn, the direction in which a magnetic field deflects a charge. In the plane, each direction has a single perpendicular; in space, the perpendiculars to a direction fill an entire plane, and two nonparallel directions are needed to fix a common perpendicular. If this is so, the natural question is this: given two vectors u⃗\vec{u} and v⃗\vec{v}, which vector x⃗=(x1,x2,x3)\vec{x} = \left( x_1, x_2, x_3 \right) satisfies u⃗⋅x⃗=0\vec{u} \cdot \vec{x} = 0 and v⃗⋅x⃗=0\vec{v} \cdot \vec{x} = 0?

These are two linear equations in three unknowns. If the first, u1x1+u2x2+u3x3=0u_1 x_1 + u_2 x_2 + u_3 x_3 = 0, is multiplied by v1v_1, the second by u1u_1, and they are subtracted, x1x_1 disappears and what remains is (u2v1−u1v2)x2+(u3v1−u1v3)x3=0\left( u_2 v_1 - u_1 v_2 \right) x_2 + \left( u_3 v_1 - u_1 v_3 \right) x_3 = 0, which holds with x2=u3v1−u1v3x_2 = u_3 v_1 - u_1 v_3 and x3=u1v2−u2v1x_3 = u_1 v_2 - u_2 v_1; reasoning in the same way with the other unknowns yields x1=u2v3−u3v2x_1 = u_2 v_3 - u_3 v_2. The three expressions are determinants of order two formed with the components of u⃗\vec{u} and of v⃗\vec{v}, like those of the lesson on determinants. This search suggests the definition; the section “The geometry of the cross product” proves that the vector so formed is indeed perpendicular to both factors.

Let u⃗=(u1,u2,u3)\vec{u} = \left( u_1, u_2, u_3 \right) and v⃗=(v1,v2,v3)\vec{v} = \left( v_1, v_2, v_3 \right) be vectors of R3\mathbb{R}^3. Their cross product, or vector product, is the vector u⃗×v⃗=(u2v3−u3v2,u3v1−u1v3,u1v2−u2v1)\vec{u} \times \vec{v} = \left( u_2 v_3 - u_3 v_2, u_3 v_1 - u_1 v_3, u_1 v_2 - u_2 v_1 \right).

The definition has a regularity worth remembering. Each component is obtained from the one before it by shifting the indices circularly, 1→2→3→11 \rightarrow 2 \rightarrow 3 \rightarrow 1: the first is u2v3−u3v2u_2 v_3 - u_3 v_2; adding 11 to each index, counting 3+13 + 1 as 11, gives u3v1−u1v3u_3 v_1 - u_1 v_3, which is the second, and repeating this gives the third. To write it from memory one uses the vectors of the standard basis, i⃗=(1,0,0)\vec{i} = \left( 1, 0, 0 \right), j⃗=(0,1,0)\vec{j} = \left( 0, 1, 0 \right) and k⃗=(0,0,1)\vec{k} = \left( 0, 0, 1 \right), with which every vector is written u⃗=u1i⃗+u2j⃗+u3k⃗\vec{u} = u_1 \vec{i} + u_2 \vec{j} + u_3 \vec{k}, and a formal determinant: an array of order three whose first row contains the vectors i⃗\vec{i}, j⃗\vec{j} and k⃗\vec{k}, and which is expanded along that row as in the lesson on determinants.

u⃗×v⃗=∣i⃗j⃗k⃗u1u2u3v1v2v3∣=∣u2u3v2v3∣i⃗−∣u1u3v1v3∣j⃗+∣u1u2v1v2∣k⃗\vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{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} \vec{i} - \begin{vmatrix} u_1 & u_3 \\ v_1 & v_3 \end{vmatrix} \vec{j} + \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix} \vec{k}

The determinant is formal because its entries are not all numbers, and no property of determinants is applied to it: it only summarizes the formula of the expansion along the first row. The minus sign of the second term reproduces the second component, because −(u1v3−u3v1)=u3v1−u1v3-\left( u_1 v_3 - u_3 v_1 \right) = u_3 v_1 - u_1 v_3. On the other hand, each component is a genuine determinant of order two, formed by two columns of the matrix with rows u⃗\vec{u} and v⃗\vec{v}; the properties in the section “The algebraic properties” rest on this observation.

ExampleA cross product and the products of the standard basis

Compute u⃗×v⃗\vec{u} \times \vec{v} for u⃗=(2,1,0)\vec{u} = \left( 2, 1, 0 \right) and v⃗=(0,2,1)\vec{v} = \left( 0, 2, 1 \right), and the products of the vectors of the standard basis with one another.

Proof

  1. ∣u2u3v2v3∣=∣1021∣=1⋅1−0⋅2=1\begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} = \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} = \resaltar{1 \cdot 1 - 0 \cdot 2} = \dato{c1}{1}

    First component: we delete the column of the first components and compute the determinant of order two that remains.

  2. −∣u1u3v1v3∣=−∣2001∣=−(2−0)=−2-\begin{vmatrix} u_1 & u_3 \\ v_1 & v_3 \end{vmatrix} = -\begin{vmatrix} 2 & 0 \\ 0 & 1 \end{vmatrix} = \resaltar{-\left( 2 - 0 \right)} = \dato{c2}{-2}

    Second component: we delete the middle column and change the sign of the determinant.

  3. ∣u1u2v1v2∣=∣2102∣=4−0=4\begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix} = \begin{vmatrix} 2 & 1 \\ 0 & 2 \end{vmatrix} = \resaltar{4 - 0} = \dato{c3}{4}

    Third component: we delete the last column.

  4. u⃗×v⃗=(1,−2,4)\vec{u} \times \vec{v} = \resaltar{\left( 1, -2, 4 \right)}

    We gather the three components. The figure draws the three vectors with a common origin; u⃗\vec{u} lies in the horizontal plane, and u⃗×v⃗\vec{u} \times \vec{v} points upward.

  5. i⃗×j⃗=k⃗,j⃗×k⃗=i⃗,k⃗×i⃗=j⃗,j⃗×i⃗=−k⃗\vec{i} \times \vec{j} = \resaltar{\vec{k}}, \quad \vec{j} \times \vec{k} = \resaltar{\vec{i}}, \quad \vec{k} \times \vec{i} = \resaltar{\vec{j}}, \qquad \vec{j} \times \vec{i} = -\vec{k}

    By the definition, i⃗×j⃗=(0⋅0−0⋅1,0⋅0−1⋅0,1⋅1−0⋅0)=(0,0,1)\vec{i} \times \vec{j} = \left( 0 \cdot 0 - 0 \cdot 1, 0 \cdot 0 - 1 \cdot 0, 1 \cdot 1 - 0 \cdot 0 \right) = \left( 0, 0, 1 \right), and likewise with the other pairs. Each vector of the basis, multiplied by the one that follows it in the circular order i⃗\vec{i}, j⃗\vec{j}, k⃗\vec{k}, gives the third; in the opposite order, its opposite.

The algebraic properties

The properties of the cross product can be read off its components. Let us call D1D_1, D2D_2 and D3D_3 the three determinants of order two that form them, with the columns in circular order: D1D_1 uses the second and third columns of the matrix with rows u⃗\vec{u} and v⃗\vec{v}; D2D_2, the third and the first, in that order; and D3D_3, the first and the second.

u⃗×v⃗=(∣u2u3v2v3∣,∣u3u1v3v1∣,∣u1u2v1v2∣)\vec{u} \times \vec{v} = \left( \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix}, \begin{vmatrix} u_3 & u_1 \\ v_3 & v_1 \end{vmatrix}, \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix} \right)
TheoremAlgebraic properties of the cross product

For all vectors u⃗\vec{u}, v⃗\vec{v} and w⃗\vec{w} of R3\mathbb{R}^3 and all real numbers ss and tt: the cross product is anticommutative, v⃗×u⃗=−u⃗×v⃗\vec{v} \times \vec{u} = -\vec{u} \times \vec{v}; it is bilinear, that is, linear in each factor, (su⃗+tw⃗)×v⃗=s u⃗×v⃗+t w⃗×v⃗\left( s \vec{u} + t \vec{w} \right) \times \vec{v} = s \, \vec{u} \times \vec{v} + t \, \vec{w} \times \vec{v} and v⃗×(su⃗+tw⃗)=s v⃗×u⃗+t v⃗×w⃗\vec{v} \times \left( s \vec{u} + t \vec{w} \right) = s \, \vec{v} \times \vec{u} + t \, \vec{v} \times \vec{w}; and u⃗×u⃗=0⃗\vec{u} \times \vec{u} = \vec{0}.

The three properties reduce to those of determinants of order two, applied to each of the three components: the lemma on the interchange of two rows and the lemma on linearity in each row, proved in the lesson on determinants for every order. The figure follows the example “A cross product and the products of the standard basis”.

Proof

  1. (u⃗×v⃗)1=∣u2u3v2v3∣,(u⃗×v⃗)2=∣u3u1v3v1∣,(u⃗×v⃗)3=∣u1u2v1v2∣\left( \vec{u} \times \vec{v} \right)_1 = \begin{vmatrix} \resaltar{u_2} & \resaltar{u_3} \\ v_2 & v_3 \end{vmatrix}, \quad \left( \vec{u} \times \vec{v} \right)_2 = \begin{vmatrix} \resaltar{u_3} & \resaltar{u_1} \\ v_3 & v_1 \end{vmatrix}, \quad \left( \vec{u} \times \vec{v} \right)_3 = \begin{vmatrix} \resaltar{u_1} & \resaltar{u_2} \\ v_1 & v_2 \end{vmatrix}

    Each component DmD_m is the determinant of a matrix of order two whose first row comes from u⃗\vec{u} and whose second row comes from v⃗\vec{v}, with the same two columns.

  2. (v⃗×u⃗)1=∣v2v3u2u3∣=−∣u2u3v2v3∣→v⃗×u⃗=−u⃗×v⃗\left( \vec{v} \times \vec{u} \right)_1 = \begin{vmatrix} \resaltar{v_2} & \resaltar{v_3} \\ \resaltar{u_2} & \resaltar{u_3} \end{vmatrix} = -\begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} \rightarrow \dato{a}{\vec{v} \times \vec{u} = -\vec{u} \times \vec{v}}

    If u⃗\vec{u} and v⃗\vec{v} are interchanged, the two rows of each determinant are interchanged, and by the interchange lemma each component changes sign. In the figure, v⃗×u⃗=(−1,2,−4)\vec{v} \times \vec{u} = \left( -1, 2, -4 \right) points downward.

  3. ∣su2+tw2su3+tw3v2v3∣=s∣u2u3v2v3∣+t∣w2w3v2v3∣→(su⃗+tw⃗)×v⃗=s u⃗×v⃗+t w⃗×v⃗\begin{vmatrix} \resaltar{s u_2 + t w_2} & \resaltar{s u_3 + t w_3} \\ v_2 & v_3 \end{vmatrix} = s \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} + t \begin{vmatrix} w_2 & w_3 \\ v_2 & v_3 \end{vmatrix} \rightarrow \dato{l}{\left( s \vec{u} + t \vec{w} \right) \times \vec{v} = s \, \vec{u} \times \vec{v} + t \, \vec{w} \times \vec{v}}

    If the first factor is su⃗+tw⃗s \vec{u} + t \vec{w}, the first row of each determinant is ss times the row taken from u⃗\vec{u} plus tt times the one taken from w⃗\vec{w}. By the linearity lemma, each component splits in the same way.

  4. v⃗×(su⃗+tw⃗)=−(su⃗+tw⃗)×v⃗=−s u⃗×v⃗−t w⃗×v⃗=s v⃗×u⃗+t v⃗×w⃗\vec{v} \times \left( s \vec{u} + t \vec{w} \right) = \resaltar{-\left( s \vec{u} + t \vec{w} \right) \times \vec{v}} = -s \, \vec{u} \times \vec{v} - t \, \vec{w} \times \vec{v} = \resaltar{s \, \vec{v} \times \vec{u} + t \, \vec{v} \times \vec{w}}

    Linearity in the second factor follows from the two properties just proved: we change the order, apply linearity in the first factor and change the order back.

  5. u⃗×u⃗=(u2u3−u3u2,u3u1−u1u3,u1u2−u2u1)=0⃗\vec{u} \times \vec{u} = \left( \resaltar{u_2 u_3 - u_3 u_2}, \resaltar{u_3 u_1 - u_1 u_3}, \resaltar{u_1 u_2 - u_2 u_1} \right) = \vec{0}

    If the two factors coincide, each determinant has two equal rows and is zero. Thus, i⃗×i⃗=j⃗×j⃗=k⃗×k⃗=0⃗\vec{i} \times \vec{i} = \vec{j} \times \vec{j} = \vec{k} \times \vec{k} = \vec{0}.

With bilinearity, the cross product can also be computed by distributing, like a product of sums, and using the products of the standard basis. For example, (i⃗+2j⃗)×(3k⃗)=3 i⃗×k⃗+6 j⃗×k⃗=−3j⃗+6i⃗\left( \vec{i} + 2 \vec{j} \right) \times \left( 3 \vec{k} \right) = 3 \, \vec{i} \times \vec{k} + 6 \, \vec{j} \times \vec{k} = -3 \vec{j} + 6 \vec{i}. Anticommutativity says that the order of the factors matters, as in the product of matrices, although in a very peculiar way: reversing it only changes the sign.

ExampleThe cross product is not associative

Compare (i⃗×i⃗)×j⃗\left( \vec{i} \times \vec{i} \right) \times \vec{j} with i⃗×(i⃗×j⃗)\vec{i} \times \left( \vec{i} \times \vec{j} \right).

Proof

  1. (i⃗×i⃗)×j⃗=0⃗×j⃗=0⃗\left( \resaltar{\vec{i} \times \vec{i}} \right) \times \vec{j} = \resaltar{\vec{0}} \times \vec{j} = \dato{p}{\vec{0}}

    By the theorem, the product of a vector with itself is zero, and by bilinearity the product of the zero vector with any other is zero as well.

  2. i⃗×(i⃗×j⃗)=i⃗×k⃗\vec{i} \times \left( \resaltar{\vec{i} \times \vec{j}} \right) = \vec{i} \times \resaltar{\vec{k}}

    In the other grouping, the product inside the parentheses is k⃗\vec{k}, by the example of the standard basis.

  3. i⃗×k⃗=−k⃗×i⃗=−j⃗\vec{i} \times \vec{k} = \resaltar{-\vec{k} \times \vec{i}} = \dato{q}{-\vec{j}}

    By anticommutativity, i⃗×k⃗\vec{i} \times \vec{k} is the opposite of k⃗×i⃗=j⃗\vec{k} \times \vec{i} = \vec{j}.

  4. (i⃗×i⃗)×j⃗=0⃗≠−j⃗=i⃗×(i⃗×j⃗)\left( \vec{i} \times \vec{i} \right) \times \vec{j} = \resaltar{\vec{0}} \neq \resaltar{-\vec{j}} = \vec{i} \times \left( \vec{i} \times \vec{j} \right)

    The two results are different: the cross product is not associative, and an expression such as u⃗×v⃗×w⃗\vec{u} \times \vec{v} \times \vec{w}, without parentheses, makes no sense.

Nor is there an identity element: no vector e⃗\vec{e} satisfies e⃗×u⃗=u⃗\vec{e} \times \vec{u} = \vec{u} for every u⃗\vec{u}, because e⃗×e⃗=0⃗\vec{e} \times \vec{e} = \vec{0} would force e⃗=0⃗\vec{e} = \vec{0}, and then e⃗×u⃗=0⃗\vec{e} \times \vec{u} = \vec{0} for every u⃗\vec{u}. The cross product therefore has none of the properties of products of numbers except distributivity. The section “The vector triple product” will give the exact rule that replaces associativity.