Part 1 · Vectors

1.3 Vector product and volume

Source: Favata’s notes, appendix A.1, p. 11 (vector product, Ricci symbol, (A1)–(A2)); handedness and the triple product are added here

Objectives

Picture $u \times v$ : orthogonal to both, as long as the parallelogram is large, oriented by the right hand.
Compute it with the Ricci symbol, and use the Ricci–Kronecker identity.
Understand the scalar triple product as a signed volume, and handedness as its sign.
Recipe: compute the components of $u \times v$ ; decide whether a basis is right-handed.
Pictures: the parallelogram on $u, v$ ; the parallelepiped on $u, v, w$ ; a basis and its mirror image.
Theorem: $ϵ_{ijk} ϵ_{i pq} = δ_{j p} δ_{k q} - δ_{j q} δ_{k p}$ .
Vocabulary: vector product, Ricci (permutation) symbol, right-handed, left-handed, scalar triple product.

The vector product exists only in dimension 3. It is the tool the course uses to talk about areas (a parallelogram’s area and its normal at once) and, through the triple product, volumes. Both come back in kinematics, in how a deformation changes areas and volumes (§3.4 Change in area and volume).

The picture

Definition 1.3.1Vector product

For $u, v \in V$ , $u \times v$ is the vector

orthogonal to both $u$ and $v$ ,
of length the area of the parallelogram on them, $∣ u \times v ∣ = ∣ u ∣∣ v ∣ sin ϑ$ ,
oriented by the right-hand rule: curl the fingers of the right hand from $u$ towards $v$ , through the smaller angle; the thumb points along $u \times v$ .

Loading the demo…

Figure 1.3.1 Drag the tips of

u

and

v

(they move in the plane facing you); drag the background to turn the view.

u \times v

stays orthogonal to both, and its length follows the area of the parallelogram. Drag

v

onto the line of

u

: it snaps, the parallelogram goes flat and the product vanishes.

Consequences, all read off the picture:

$v \times u = - u \times v$ (the right hand curls the other way), so $u \times u = 0$ .
For nonzero vectors, $u \times v = 0$ iff they are parallel ( $sin ϑ = 0$ ). So $u \times v \neq = 0$ means “ $u, v$ linearly independent”, which is how Theorem 2.5.2 uses it.

Components: the Ricci symbol

For a right-handed orthonormal basis, $e_{1} \times e_{2} = e_{3}$ , $e_{2} \times e_{3} = e_{1}$ , $e_{3} \times e_{1} = e_{2}$ . All nine products are summarised by

$e_{i} \times e_{j} = ϵ_{ijk} e_{k},$

with the Ricci symbol (permutation symbol, Levi-Civita symbol):

Definition 1.3.2Ricci symbol (A1)

$ϵ_{ijk} = ⎩ ⎨ ⎧ + 1 - 1 0 (i, j, k) an even permutation of (1, 2, 3) : 123, 231, 312 an odd permutation : 132, 213, 321 two indices equal.$

Swapping any two indices flips its sign ( $ϵ_{ijk} = - ϵ_{jik}$ ); a cyclic shift does not ( $ϵ_{ijk} = ϵ_{jki} = ϵ_{kij}$ ). By bilinearity,

$u \times v = (u_{i} e_{i}) \times (v_{j} e_{j}) = ϵ_{ijk} u_{i} v_{j} e_{k}, (u \times v)_{k} = ϵ_{ijk} u_{i} v_{j} .$

Recipe 1.3.3Components of u × v

In $(u \times v)_{k} = ϵ_{ijk} u_{i} v_{j}$ , for each $k$ only two terms survive, the two orders of the other two indices: $u \times v = (u_{2} v_{3} - u_{3} v_{2}, u_{3} v_{1} - u_{1} v_{3}, u_{1} v_{2} - u_{2} v_{1}) .$ Mnemonic: component $k$ uses the two other indices in cyclic order, $(k + 1, k + 2)$ , minus the reverse.

The Ricci–Kronecker identity

Theorem 1.3.4Ricci–Kronecker identity (A2)

$ϵ_{ijk} ϵ_{i pq} = δ_{j p} δ_{k q} - δ_{j q} δ_{k p} .$

Proofbeyond the notes

Fix $j, k, p, q$ ; the left side sums over $i$ .

If $j = k$ , the left side is $0$ , and so is the right: $δ_{j p} δ_{j q} - δ_{j q} δ_{j p} = 0$ .
If $j \neq = k$ , only the $i$ different from both survives, so $ϵ_{i pq} \neq = 0$ needs ${p, q} = {j, k}$ .
- $(p, q) = (j, k)$ : left $= ϵ_{ijk}^{2} = 1$ , right $= 1 \cdot 1 - 0 = 1$ .
- $(p, q) = (k, j)$ : left $= ϵ_{ijk} ϵ_{ikj} = - 1$ , right $= 0 - 1 = - 1$ .
- otherwise both sides are $0$ .

Contracting it once more: set $p = j$ to get $ϵ_{ijk} ϵ_{ij q} = 3 δ_{k q} - δ_{k q} = 2 δ_{k q}$ ; contract again, $ϵ_{ijk} ϵ_{ijk} = 6$ .

The identity is what turns products of two cross products into scalar products. The picture of Definition 1.3.1 follows from the component formula with it:

The geometry of u × v from the componentsbeyond the notes

Orthogonality: $u \cdot (u \times v) = ϵ_{ijk} u_{i} v_{j} u_{k} = 0$ . $ϵ_{ijk}$ flips sign when $i \leftrightarrow k$ are swapped, $u_{i} u_{k}$ does not, so the terms cancel in pairs. The same holds for $v$ .
Length, using (A2) after a cyclic shift ( $ϵ_{ijk} ϵ_{pq k} = ϵ_{kij} ϵ_{k pq}$ ): $∣ u \times v ∣^{2} = ϵ_{kij} ϵ_{k pq} u_{i} v_{j} u_{p} v_{q} = (δ_{i p} δ_{j q} - δ_{i q} δ_{j p}) u_{i} v_{j} u_{p} v_{q} = ∣ u ∣^{2} ∣ v ∣^{2} - (u \cdot v)^{2} = ∣ u ∣^{2} ∣ v ∣^{2} sin^{2} ϑ .$

Right- and left-handed bases

Definition 1.3.5Right- and left-handedbeyond the notes

An orthonormal basis is right-handed if $e_{1} \times e_{2} = e_{3}$ : curling the right hand’s fingers from $e_{1}$ to $e_{2}$ , the thumb points along $e_{3}$ . It is left-handed if $e_{1} \times e_{2} = - e_{3}$ : only the left thumb would point along $e_{3}$ .

In both drawings

e_{1}

points out of the page and

e_{3}

up; the dashed arc turns

e_{1}

towards

e_{2}

. The two bases are mirror images of each other.

A reflection turns one handedness into the other; no rotation can, however you turn the basis (§2.6 Orthogonal tensors).
The vector $u \times v$ is defined by the right-hand rule, so it does not depend on the basis. Its component formula does: $(u \times v)_{k} = ϵ_{ijk} u_{i} v_{j}$ holds only in a right-handed basis. In a left-handed one the same sum gives the components of $- u \times v$ .
Favata, and continuum mechanics generally, always use right-handed bases.

Scalar triple product

Definition 1.3.6Scalar triple productbeyond the notes

$u \cdot (v \times w) = ϵ_{ijk} u_{i} v_{j} w_{k} = det u_{1} v_{1} w_{1} u_{2} v_{2} w_{2} u_{3} v_{3} w_{3} .$ In kinematics it is written $u \cdot v \times w$ , without brackets: the cross product binds first.

Key observation 1.3.7The triple product is a signed volumebeyond the notes

$∣ u \cdot (v \times w) ∣$ is the volume of the parallelepiped on $u, v, w$ : base area $∣ v \times w ∣$ times the height, the component of $u$ along the base’s normal. Its sign is the orientation:

positive iff ${u, v, w}$ is right-handed;
zero iff the three are coplanar, that is linearly dependent.

Loading the demo…

Figure 1.3.2 Drag the three tips. The readout is the triple product, a

3 \times 3

determinant with the vectors as rows. Push

w

through the plane of

u

and

v

: it snaps into the plane (volume

0

), and past it the box turns red, left-handed.

It is unchanged by a cyclic shift and changes sign when two vectors are swapped: $u \cdot (v \times w) = v \cdot (w \times u) = w \cdot (u \times v) = - v \cdot (u \times w) .$ It is linear in each slot (trilinear) and changes sign when two slots are swapped (alternating). These two properties are all the determinant needs.

Exercises

Exercise 1.3.1beyond the notespractice (A.6)

Use (A2) to prove $a \times (b \times c) = (a \cdot c) b - (a \cdot b) c$ .

Solution

$(b \times c)_{k} = ϵ_{pq k} b_{p} c_{q}$ . Then $[a \times (b \times c)]_{i} = ϵ_{jki} a_{j} ϵ_{pq k} b_{p} c_{q} = ϵ_{kij} ϵ_{k pq} a_{j} b_{p} c_{q} = (δ_{i p} δ_{j q} - δ_{i q} δ_{j p}) a_{j} b_{p} c_{q} = b_{i} (a_{j} c_{j}) - c_{i} (a_{j} b_{j}) .$

Exercise 1.3.2beyond the notes

Compute $(1, 2, 0) \times (0, 1, 3)$ and check that it is orthogonal to both factors.

Solution

$(2 \cdot 3 - 0 \cdot 1, 0 \cdot 0 - 1 \cdot 3, 1 \cdot 1 - 2 \cdot 0) = (6, - 3, 1)$ . Dot products: $6 - 6 + 0 = 0$ and $0 - 3 + 3 = 0$ .

Exercise 1.3.3beyond the notes

Show that $ϵ_{ijk} ϵ_{ijk} = 6$ by counting, without (A2).

Solution

The sum runs over the 27 triples; only the 6 permutations of $(1, 2, 3)$ are nonzero, each contributing $(\pm 1)^{2} = 1$ .