Practice: Deriving the 3D cross product
Self-check
Section titled “Self-check”Six short questions. Answer each one in your head (or on paper) before opening the collapsible. Trying to retrieve the answer is where the learning sticks; rereading feels productive but does much less.
1. What are the three geometric properties of the 3D cross product v × w?
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It is perpendicular to both v and w; its length equals the area of the parallelogram they span (|v|·|w|·sin θ); and its direction follows the right-hand rule (curl the right hand’s fingers from v toward w, the thumb points along v × w).
2. What single function is at the heart of the derivation?
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f(p) = signed volume of the parallelepiped spanned by p, v, and w, which equals det([p | v | w]). With v and w fixed, this is a linear function from a 3D input p to a single number.
3. What does duality say, lifted to 3D?
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Every linear function from 3D to a single number is a dot product with one unique 3D vector. So the volume function f(p) must equal u · p for some specific u, and that u is defined to be v × w.
4. Why is v × w perpendicular to both v and w?
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Because u · v is the signed volume of the box [v | v | w], which has a repeated edge and is therefore flat (zero volume). So u · v = 0, meaning u is perpendicular to v; the same argument with w gives u · w = 0. Perpendicularity is forced by the definition, not an extra rule.
5. Why does |v × w| equal the area of the parallelogram?
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Take p to be a unit vector along u itself. Then u · p = |u|, and that same dot product is the volume of a box with a unit-length edge perpendicular to the v-w plane sitting on the parallelogram base, which is a height-1 prism. Its volume equals the base area, so |u| equals the parallelogram’s area.
6. Is the 3D cross product commutative?
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No, it is anti-commutative: v × w = -(w × v). Swapping the inputs reverses the right-hand-rule direction, so the result points the opposite way (same as the 2D sign flip, now a flipped vector).
Try it yourself, part 1: compute and verify perpendicularity
Section titled “Try it yourself, part 1: compute and verify perpendicularity”For each pair, compute v × w using [v_y·w_z - v_z·w_y, v_z·w_x - v_x·w_z, v_x·w_y - v_y·w_x], then dot the result with both inputs to confirm it is perpendicular to each. About 8 minutes, pen and paper.
- a)
[1, 0, 2] × [0, 1, 1] - b)
[2, 1, 0] × [1, 3, 0]
Check your work
- a) First:
(0)(1) - (2)(1) = -2. Second:(2)(0) - (1)(1) = -1. Third:(1)(1) - (0)(0) = 1. Sov × w = [-2, -1, 1]. Perpendicularity:[-2,-1,1] · [1,0,2] = -2 + 0 + 2 = 0and[-2,-1,1] · [0,1,1] = 0 - 1 + 1 = 0. Both zero. - b) First:
(1)(0) - (0)(3) = 0. Second:(0)(1) - (2)(0) = 0. Third:(2)(3) - (1)(1) = 5. Sov × w = [0, 0, 5]. Both inputs lie in the xy-plane, so the cross product points straight up the z-axis, and its third component, 5, is exactly the 2D cross product[2,1] × [1,3]from the previous lesson. Perpendicularity: dotting[0,0,5]with either input (both have zero z-component) gives 0.
Try it yourself, part 2: basis cross products and anti-commutativity
Section titled “Try it yourself, part 2: basis cross products and anti-commutativity”The standard basis vectors have a clean cyclic pattern. About 6 minutes.
Step 1. Given that i × j = k (computed in the lesson), compute j × k and k × i with the formula. What pattern do you see?
Step 2. Compute k × j and confirm it equals -(j × k).
Check your work
Step 1. j × k = [0,1,0] × [0,0,1]: first (1)(1) - (0)(0) = 1, second (0)(0) - (0)(1) = 0, third (0)(0) - (1)(0) = 0, so j × k = [1, 0, 0] = i. k × i = [0,0,1] × [1,0,0]: first (0)(0) - (1)(0) = 0, second (1)(1) - (0)(0) = 1, third (0)(0) - (0)(1) = 0, so k × i = [0, 1, 0] = j. The pattern is cyclic: i × j = k, j × k = i, k × i = j (each pair of basis vectors crosses to the third, going around the cycle).
Step 2. k × j = [0,0,1] × [0,1,0]: first (0)(0) - (1)(1) = -1, second (1)(0) - (0)(0) = 0, third (0)(1) - (0)(0) = 0, so k × j = [-1, 0, 0] = -i. Since j × k = i, this is -(j × k): anti-commutativity confirmed.
Flashcards
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Q. What are the three geometric properties of the 3D cross product?
Perpendicular to both inputs; length equal to the area of the parallelogram they span (|v|·|w|·sin θ); direction by the right-hand rule (fingers curl v to w, thumb points along v × w).
Q. What function is at the heart of the cross-product derivation?
f(p) = signed volume of the box spanned by p, v, w = det([p | v | w]). With v, w fixed, it is a linear function from a 3D vector p to a single number.
Q. What does the duality argument say in 3D?
Every linear function from 3D to a single number is a dot product with one unique 3D vector. The volume function f(p) therefore equals u · p for some u, and that u is v × w.
Q. How is the 3D cross product defined via duality?
As the unique vector u such that u · p equals the signed volume of the parallelepiped spanned by p, v, and w, for every p. The formula is then read off, not memorized.
Q. Why is v × w perpendicular to both v and w?
u · v is the volume of the box [v | v | w], which has a repeated edge and so is flat (zero volume). Hence u · v = 0, and likewise u · w = 0. Perpendicularity is forced by the definition.
Q. Why does the length of v × w equal the parallelogram's area?
With p a unit vector along u, u · p = |u| is the volume of a height-1 prism on the parallelogram base (since p is perpendicular to the v-w plane). A height-1 prism’s volume equals its base area, so |u| is that area.
Q. Is the 3D cross product a vector or a number?
A vector. Unlike the 2D cross product (a single number), the 3D cross product has a direction (perpendicular to both inputs) as well as a magnitude (the spanned area).
Q. Is the 3D cross product anti-commutative?
Yes: v × w = -(w × v). Swapping the inputs reverses the right-hand-rule direction, so the result vector points the opposite way.
Q. What is the cyclic pattern of basis-vector cross products?
i × j = k, j × k = i, k × i = j: each consecutive pair crosses to the third, cycling around. Reversing any pair flips the sign (e.g. j × i = -k).
Q. Why don't you need to memorize the criss-cross formula?
Because the components are the cofactor expansion of det([p | v | w]). Remembering “the cross product is the vector whose dot with p gives that volume” lets you reconstruct the components any time.