# Tag Info

32

$$\Large{\text{Bye_World's Second Treatise}} \\ \large{\text{On the Products of Vectors}}$$ Table of Contents$\bullet \$Preface$\bullet \$Vectors$\bullet \$The Dot Product$\bullet \$The Cross Product$\bullet \$The Real Skinny on the Cross Product$\bullet \$The Wedge Product$\bullet \$The Relationship between the Cross Product and Wedge ...

24

Yes, you are correct. You can generalize the cross product to $n$ dimensions by saying it is an operation which takes in $n-1$ vectors and produces a vector that is perpendicular to each one. This can be easily defined using the exterior algebra and Hodge star operator http://en.wikipedia.org/wiki/Hodge_dual: the cross product of $v_1,\ldots,v_{n-1}$ is ...

23

A little bit more of the 'how and why': the dot product comes about as a natural answer to the question: 'what functions do we have that take two vectors and produce a number?' Keep in mind that we have a natural function (addition) that takes two vectors and produces another vector, and another natural function (scalar multiplication) that takes a vector ...

18

Since the only normed division algebras are the real numbers, the complex numbers, the quaternions and the octonions, the cross product is formed from the product of the normed division algebra by restricting it to the $0, 1, 3, 7$ imaginary dimensions of the algebra. This gives nonzero products in only three and seven dimensions. Now why, you may ask, does ...

16

As Fabian wrote, $b$ is not uniquely determined by $a$ and $c$. Moreover, there is no solution unless $a$ and $c$ are orthogonal. If $a$ and $c$ are orthogonal, then the solutions are $(c \times a)/(a\, . a) + t a$ for arbitrary scalars $t$.

16

Yep, there is an easier way. Hint: $$|a \times b| = |a||b|\sin(\theta)$$ and $$a \cdot b = |a||b|\cos(\theta)$$ where $\theta$ is the angle between the vectors $a$ and $b$.

15

No, it's not an accident. The cross product is orthogonal to each factor, so the vector has to be orthogonal to $b\times c$, hence in the plane spanned by $b$ and $c$. But it also has to be orthogonal to $a$. So, writing $$a\times(b\times c) = xb + yc$$ and dotting with $a$, you get $x(b\cdot a) + y(c\cdot a)=0$. So the answer must be some scalar multiple of ...

14

I would have left this as a comment, but apparently I can't comment yet. To answer your final question, why introduce the wedge, the point is that the wedge product (as explained in the Wikipedia article) is a notion that generalizes to R^n and indeed any vector space---in general the output is what is called a "bivector". Now, it just so happens that in ...

13

Hint: the area of a parallelogram (see left-most image) is equal to the determinant of the $2\times 2$ matrix formed by the column vectors representing component vectors determined by the given points. $$A = \text{det}\,\left(\vec u \;\; \vec v\right)$$ The area of a parallelogram is also equal to the magnitutude of the cross product of the component ...

12

I am pretty sure you don't want this answer, but I couldn't resist. The term on the left is an alternating $3$-linear form in $(a,b,c)$. So it is equal to a constant times the determinant on the right.. Compute the constant with the canonical basis. It's $1$. So the lhs is equal to the rhs. Note: see here if you want to read about this characterization ...

11

Here's a slightly different perspective. We use the mnemonic device... $$\langle a_1,a_2,a_3 \rangle \times \langle b_1,b_2,b_3 \rangle = \begin{vmatrix} {\bf i} & {\bf j} & {\bf k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$$ ...to remember the formula for the cross product. If you think about the dot product in the ...

11

A longuish comment The cross product satisfies the Jacobi identity $$(a\times b)\times c+(b\times c)\times a+(c\times a)\times b=0.$$ Using this and the fact that it is antisymmetric, you can easily see that $$(a\times b)\times c=a\times(b\times c)\iff(c\times a)\times b=0.$$ This immediately explains your example where the equality holds and the one where ...

10

The determinant formula isn't so mysterious. Consider the cross product $\mathbf{v} = \langle a,b,c \rangle \times \langle d,e,f \rangle$ as the formal determinant $$\det \left(\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ a & b & c \\ d & e & f \end{array} \right)$$ where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are the ...

8

No difference at all. I've been trying to write a little proof, but the software on this page seems to have forgotten how to write maths. :-( Anyway: I assume that by "regular cross/vector product" you mean the definition with coordinates as in Wikipedia. Try to compute both sides of your equation $(u\wedge v ) \cdot w = \det (u, v, w)$ with your ...

8

The easiest demonstration is just to cmpute both sides of the equation: $$b\times c=(b_2c_3-b_3c_2,b_3c_1-b_1c_3,b_1c_2-b_2c_1)$$ $$a\cdot(b\times c)=a_1b_2c_3-a_1b_3c_2+a_2b_3c_1-a_2b_1c_3+a_3b_1c_2-a_3b_2c_1$$ That's the determinant. Note that each term of the sum is one possible permutation multiplied by its parity, which happens to be the definition of ...

8

This is most easily proved without coordinates. The cross product is the unique vector that is orthogonal to both factors, has length given by the area of the parallelogram they form and forms a right-handed triple with them. These properties are all invariant under rotations, and thus so is the cross product.

7

Jeff's answer is good, but I feel that it hides what is really going on a little bit. Given $n-1$ vectors $v_1,v_2,...,v_{n-1}$ in $\mathbb{R}^n$, you can form a linear map $$L: \mathbb{R}^n \to \mathbb{R}$$ by the rule $$L(w) = \det(v_1,v_2, \ldots ,v_{n-1},w)$$ In other words, $L$ is the linear map you get by partially applying the determinant. It ...

7

The answer to this problem is sadly not very well-known. Suppose an r-ary operation on certain d-dimensional space V. Then, a r-fold d-dimensional "cross product" multilinear operation exists: $$(C_1\times C_2\times \ldots\times C_r): V^{dr}=\underbrace{V^d\times \cdots \times V^d}_{r}\longrightarrow V^d$$ such as $$\forall i=1,2,...,r$$ we have that $$... 7 Here are a few references about the history of linear algebra: "A Brief History of Linear Algebra and Matrix Theory" "History of Linear Algebra" Cross Product - Wikipedia: History The Wikipedia article seems to address your question most directly. 7 No, \def\bfa{{\bf a}}\def\bfb{{\bf b}}\def\bfc{{\bf c}}\bfa \times \bfb \times \bfc simply doesn't make sense for vectors in \Bbb R^3, precisely because \times is nonassociative. The cross product does, however, satisfy the so-called Jacobi identity,$$\bfa \times (\bfb \times \bfc) + \bfb \times (\bfc \times \bfa) + \bfc \times (\bfa \times \bfb) = ...

6

Here are two ways to derive the formula for the dot product. I assume that $v_1$ and $v_2$ are vectors with spherical coordinates $(r_1, \varphi_1, \theta_1)$ and $(r_2, \varphi_2, \theta_2)$. First way: Let us convert these spherical coordinates to Cartesian ones. For the first point we get Cartesian coordinates $(x_1, y_1, z_1)$ like this: $$... 6 Yes, this definition is chosen for a reason, as the unique solution to a pedagogical problem. Do Carmo's definition is awkward and redundant in 3 dimensions, but it is the only one among the usual definitions for the cross product that when generalized to n dimensions (there is a cross-product of n-1 vectors in R^n) is rigorous, visibly basis ... 6 I would cite the beautiful method from W.H.Beyer to find the center and radius of the sphere (x-a)^2+(y-b^2)+(y-c)^2=R^2 6 I really truly believe it's hard to beat the 3\times 3 determinant mnemonic. However, there is a trick for 3 \times 3 determinants which make computing them a snap. Remember that you compute a 2 \times 2 determinant by multiplying diagonals and off-diagonals (the upward diagonal) and taking the difference. You can do the same for 3 \times ... 6 I usually teach this (which requires no additional writing and avoids cyclic permutations, which are often confusing for students):$$ \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \times \left[\begin{array}{c} u \\ v \\ w \end{array}\right]  1) ignore $x$ and $u$ (that is: mentally block view of the first row), compute $2\times 2$ ...

6

The name "product" for the cross product is unfortunate. It really should not be thought of as a product in the ordinary sense; for example, it is not even associative. Thus one should not expect it to have properties analogous to the properties of ordinary multiplication. What the cross product really is is a Lie bracket.

6

The best introduction I know of to the exterior product is Sergei Winitzki's free book Linear Algebra via Exterior Products. Chapter $2$ in particular I think addresses all of your questions (it is unclear how much of Chapter $1$ you need to read in order to read Chapter $2$, I guess that depends on how much linear algebra you've had).

6

The truth is you've been lied to, or at least that the usual notation makes the innate connection way more difficult to see than necessary. To see how the two are the same, let me tell you about the wedge product. The wedge product of vectors is like the cross product, in that it is anticommutative--$a \wedge b = -b \wedge a$--but it does not produce a ...

6

Here's an explanation in terms of the Hodge dual and the exterior (wedge) product. Let ${e_1, e_2, e_3}$ be the standard orthonormal basis for $\mathbb{R}^3$. Consider the two vectors $a = a_1 e_1 + a_2 e_2 + a_3 e_3$ and $b = b_1 e_1 + b_2 e_2 + b_3 e_3$. From the matrix computation we obtain the familiar formula \$a\times b = (a_2 b_3 - a_3 b_2) e_1 + ...

Only top voted, non community-wiki answers of a minimum length are eligible