0.8 The Cross Product

The dot product works in every $\mathbb{R}^n$ and returns a number. The cross product is its opposite in both respects: it is defined only in $\mathbb{R}^3$ , and it returns a vector. For $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$ ,

$\mathbf{u} \times \mathbf{v} = (u_2 v_3 - u_3 v_2, \; u_3 v_1 - u_1 v_3, \; u_1 v_2 - u_2 v_1)$

Each component is a $2 \times 2$ determinant — the number $ad - bc$ that Chapter 4 studies at length — built from the other two components of $\mathbf{u}$ and $\mathbf{v}$ : the first component uses the second and third, the second uses the third and first, the third uses the first and second. The middle sign is the one people get wrong; written as $u_3 v_1 - u_1 v_3$ it follows the same cyclic pattern as the others.

Rather than memorize the formula, watch what it produces. The figure fixes $\mathbf{u} = (2, 1, 0)$ in the floor, puts $\mathbf{v} = (1, 2, c)$ on a slider that lifts its tip out of the floor, and draws $\mathbf{u} \times \mathbf{v}$ together with the parallelogram the two vectors span.

A translucent parallelogram with one corner at the origin, spanned by a thick shaded arrow u ending at (2, 1, 0) and a second thick arrow v ending at (1, 2, c), where c is set by a slider that starts at 0. A third, thicker arrow rises from the origin to (c, -2c, 3): the cross product u times v. At c = 0 the parallelogram lies flat on the floor and that arrow stands straight up, 3 units tall, and 3 is also the area of the parallelogram. Dragging c to 2 lifts the arrow v, tilting the parallelogram out of the floor; the cross-product arrow leans away to (2, -4, 3), staying perpendicular to both u and v, and grows to length the square root of 29, which is the new area. Dragging c to -2 leans it the other way, to (-2, 4, 3), with the same length. A hidden fourth arrow can be revealed: v times u, pointing to (-c, 2c, -3), below the floor and exactly opposite to the cross-product arrow at every slider value. — The two arrows spanning the translucent parallelogram are $\mathbf{u} = (2, 1, 0)$ and $\mathbf{v} = (1, 2, c)$ ; the third and thickest is the cross product $\mathbf{u} \times \mathbf{v} = (c, -2c, 3)$ . A hidden fourth arrow is $\mathbf{v} \times \mathbf{u}$ .

Explore the figure

At the starting value $c = 0$ both vectors lie in the floor, and the cross-product arrow stands straight up: $(0, 0, 3)$ . Orbit to confirm it is perpendicular to the floor, and so to both vectors at once.
Its length is $3$ . Now compute the area of the flat parallelogram: base $|\mathbf{u}| = \sqrt{5}$ times the height of $\mathbf{v}$ above the line of $\mathbf{u}$ , which by Section 0.7 is the perpendicular part of $\mathbf{v}$ , of length $3/\sqrt{5}$ . The area is $3$ — the same number.
Drag $c$ to $2$ . The arrow $\mathbf{v}$ lifts off the floor, the parallelogram tilts, and the cross-product arrow leans away to $(2, -4, 3)$ to stay perpendicular to the tilted sheet. Orbit until you look along it: the parallelogram is seen face-on, which is what "perpendicular to the sheet" looks like. Its length is now $\sqrt{4 + 16 + 9} = \sqrt{29}$ , and so is the area.
Reveal the hidden object labeled $\mathbf{v} \times \mathbf{u}$ . It is the cross-product arrow exactly reversed, $(-c, 2c, -3)$ , at every slider value: swapping the factors flips the sign.
Drag $c$ to $-2$ . The cross-product arrow leans the other way, to $(-2, 4, 3)$ , but its third component is still $3$ : that component comes from the floor shadow of the parallelogram, which the slider never changes.

What the formula guarantees

Three properties hold for every pair of vectors, and each is a short computation with the components.

Orthogonality. Dot $\mathbf{u}$ with $\mathbf{u} \times \mathbf{v}$ : $u_1 (u_2 v_3 - u_3 v_2) + u_2 (u_3 v_1 - u_1 v_3) + u_3 (u_1 v_2 - u_2 v_1)$ . Every product appears twice with opposite signs — $u_1 u_2 v_3$ from the first term cancels $-u_2 u_1 v_3$ from the second, and so on — so the sum is $0$ . The same happens with $\mathbf{v}$ . So $\mathbf{u} \times \mathbf{v}$ is orthogonal to both $\mathbf{u}$ and $\mathbf{v}$ , which is why it stood perpendicular to the sheet in every position of the slider.

Anticommutativity. Swapping $\mathbf{u}$ and $\mathbf{v}$ swaps the two products in every component, so $\mathbf{v} \times \mathbf{u} = -(\mathbf{u} \times \mathbf{v})$ . In particular $\mathbf{u} \times \mathbf{u} = -(\mathbf{u} \times \mathbf{u})$ , which forces $\mathbf{u} \times \mathbf{u} = \mathbf{0}$ : a vector crossed with itself, or with any multiple of itself, gives nothing, because two parallel vectors span no parallelogram. Which of the two perpendicular directions $\mathbf{u} \times \mathbf{v}$ takes is fixed by the formula and remembered by the right-hand rule: curl the fingers of your right hand from $\mathbf{u}$ toward $\mathbf{v}$ , and the thumb points along $\mathbf{u} \times \mathbf{v}$ . The standard unit vectors are the check: $\mathbf{e}_1 \times \mathbf{e}_2 = (0 - 0, \; 0 - 0, \; 1 - 0) = \mathbf{e}_3$ .

Length. Multiply out $|\mathbf{u} \times \mathbf{v}|^2$ — three squared components, nine terms after expansion — and compare with $|\mathbf{u}|^2 |\mathbf{v}|^2 - (\mathbf{u} \cdot \mathbf{v})^2$ , also expanded. They agree term by term, an identity worth verifying once with patience:

$|\mathbf{u} \times \mathbf{v}|^2 = |\mathbf{u}|^2 |\mathbf{v}|^2 - (\mathbf{u} \cdot \mathbf{v})^2$

Now substitute the angle formula $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| \, |\mathbf{v}| \cos \theta$ : the right side becomes $|\mathbf{u}|^2 |\mathbf{v}|^2 (1 - \cos^2 \theta) = |\mathbf{u}|^2 |\mathbf{v}|^2 \sin^2 \theta$ , and since $\sin \theta \ge 0$ for angles between $0$ and $\pi$ ,

$|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}| \, |\mathbf{v}| \sin \theta$

That is base times height: $|\mathbf{u}|$ is the base of the parallelogram spanned by the two vectors, and $|\mathbf{v}| \sin \theta$ is its height. The length of the cross product is the area of the parallelogram — the fact step 2 checked by hand and step 3 checked again after the tilt.

Example 1 A cross product, checked two ways

Compute $\mathbf{u} \times \mathbf{v}$ for $\mathbf{u} = (1, -2, 3)$ and $\mathbf{v} = (2, 0, -1)$ , verify that it is orthogonal to both, and find the area of the parallelogram they span.

By the formula, the first component is $(-2)(-1) - (3)(0) = 2$ , the second is $(3)(2) - (1)(-1) = 7$ , and the third is $(1)(0) - (-2)(2) = 4$ : $\mathbf{u} \times \mathbf{v} = (2, 7, 4)$ . Orthogonality: $\mathbf{u} \cdot (2, 7, 4) = 2 - 14 + 12 = 0$ and $\mathbf{v} \cdot (2, 7, 4) = 4 + 0 - 4 = 0$ . The area is $|(2, 7, 4)| = \sqrt{4 + 49 + 16} = \sqrt{69} \approx 8.31$ .

Example 2 The area of a triangle in space

Find the area of the triangle with vertices $P = (1, 0, 0)$ , $Q = (0, 2, 0)$ and $R = (0, 0, 3)$ .

Two edges from $P$ are $\mathbf{a} = Q - P = (-1, 2, 0)$ and $\mathbf{b} = R - P = (-1, 0, 3)$ . Their cross product is $(2 \cdot 3 - 0 \cdot 0, \; 0 \cdot (-1) - (-1) \cdot 3, \; (-1) \cdot 0 - 2 \cdot (-1)) = (6, 3, 2)$ , of length $\sqrt{36 + 9 + 4} = 7$ . That is the area of the parallelogram on the two edges; the triangle is half of it, with area $7/2$ . Keep the vector $(6, 3, 2)$ in mind — it is perpendicular to the triangle's plane, and Section 0.10 will use it to write that plane's equation.

c

The first component of u x v on the slider ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The $2 \times 2$ determinant of the second and third components of $\mathbf{u} = (2, 1, 0)$ and $\mathbf{v} = (1, 2, c)$ , computed by a computer algebra system: $1 \cdot c - 0 \cdot 2 = c$ , the first component of the cross-product arrow.

2 \cdot c

The second component, before its sign ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The determinant of the first and third components, $2 \cdot c - 0 \cdot 1 = 2c$ . The second component of the cross product is its negative, $-2c$ — the middle sign of the formula.

3

The third component ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The determinant of the first and second components, $2 \cdot 2 - 1 \cdot 1 = 3$ : the third component, which the slider never touches, and the height of the cross-product arrow at $c = 0$ .

\left[ c,\; -2 \cdot c,\; 3 \right]

The whole figure, in one card ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The figure's two vectors with the slider left as the symbol $c$ , crossed by a computer algebra system: $(c, -2c, 3)$ . Every position of the slider at once, including the ones you do not drag to — and the third component really is free of $c$ , which is why the arrow's height never changes.

\sqrt{9 + 5 \cdot c^{2}}

Its length, which is the area ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The length of that cross product, still symbolic: $\sqrt{5c^2 + 9}$ . It gives $3$ at $c = 0$ and $\sqrt{29}$ at $c = 2$ — the two areas the exploration measured — and it says what happens between and beyond them.

0

Perpendicular to u, at every slider value ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The orthogonality argument, checked rather than trusted: $\mathbf{u}$ dotted with $\mathbf{u} \times \mathbf{v}$ is $2c - 2c + 0 = 0$ for every $c$ .

0

And to v ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The same for $\mathbf{v}$ : $c - 4c + 3c = 0$ , again for every $c$ . Two symbolic zeros are the whole content of "the cross product is perpendicular to both".

\left[ 2,\; 7,\; 4 \right]

The worked example's cross product ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

$(1, -2, 3) \times (2, 0, -1) = (2, 7, 4)$ , computed rather than assembled by hand from the three $2 \times 2$ determinants.

\sqrt{69}

And the area it measures ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

Its length, $\sqrt{69}$ : the area of the parallelogram those two vectors span.

\left[ 6,\; 3,\; 2 \right]

The triangle's normal ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The two edges of the triangle example, crossed: $(6, 3, 2)$ . Section 0.10 uses this vector again, as the normal of the plane through the three points.

7

Twice the triangle's area ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

Its length is exactly $7$ , the area of the parallelogram on the two edges — so the triangle, which is half of it, has area $7/2$ .

An original work of XYZ Homework, built around interactive XYZ 3D figures. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.