# Tag Info

16

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 ...

13

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 ...

12

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 ...

10

Since the only normed division algebras are 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 this give nonzero products in only ...

9

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 ...

7

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 ...

7

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 ...

7

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

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 ...

6

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.

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.

5

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).

4

I'm not exactly sure what you are asking, however the following may be useful as a less verbose way of obtaining the same result. The key fact is that the cross product of $A,B$ is the unique element $A\times B$ such that $\langle x, A\times B \rangle = \det \begin{bmatrix} A & B & x\end{bmatrix}$, $\forall x$. Let $Q$ be a rotation (ie, $Q^TQ = ... 4 All quantities below are vectors. I will use the following properties of cross-products and dot-products: $$(x \times y) \times z = (x \cdot z) y - (y \cdot z)x \\ x \cdot ( y \times z) = y \cdot (z \times x) = z \cdot (x \times y) \\ x \cdot (x \times y) = 0$$ We start with the righthand side. For convenience, denote$a \times c = v$. Then ... 4 Since$u_r,u_{\phi},u_{\theta}$forms a right handed orthonormal frame of unit vectors the rules for computing vectors at a point$p$expressed in the frame at$p$is precisely the same as that for the globally constant Cartesian frame. For example, $$\vec{V}_1 \cdot \vec{V}_2 = 2(3)+\frac{\pi}{3}\frac{\pi}{6}+\frac{\pi}{4}\frac{\pi}{2}$$ More generally, ... 4 It's how the cross product is defined. Operations are generally things we define to have some nice properties. Even the notion of addition becomes non-intuitive and somewhat unnatural when you start involving irrational numbers. As far as anti-symmetry, this is why we use the "right hand rule." Align your right hand along the first vector, curl your fingers ... 4 The Cartesian product of sets is what is usually referred to as the product of two sets, and it is denoted with$\times$. It's certainly associative. The intuition is that when you form$(X\times Y)\times Z$and$X\times (Y\times Z)$, they are both basically the same as the set of ordered triples$X\times Y\times Z$. While the first two are not equal as ... 3 Your reasoning is correct. In fact, the essential part of a proper concept in geometry is the ability to define it without choosing particular coordinates. The "3D-only" character of the cross product comes from the fact that in higher dimensions two vectors, and the plane which they share do not define exactly one perpendicular direction (like in the 3D ... 3 Perhaps understanding the following definition of the cross product would eliminate your confusion: For two vectors$a$and$b$in$\mathbb{R}^3$the function from$\mathbb{R}^3$to$\mathbb{R}$determined by the rule$c \mapsto \det[a, b, c]$is a linear form on$\mathbb{R}^3$, that is, it is a real-valued linear function on$\mathbb{R}^3$. As such, it can ... 3 Assign$u$to be the unit normal vector to some plane$S$in$\mathbb{R}^3$, then$u\times x_n \in T(S)$for any$n$, which means$u\times x_n $is a vector field on the plane$S$. After the first cross product, all$x_n$'s ($n > 1$) lie on the plane$S$, because$x_{n}\cdot u = (u\times x_{n-1})\cdot u = 0$for all$n> 1$. Now$x_{n+2} = ...

3

This is basically a fleshing out of the hint by @GerryMyerson. We assume that ${\bf a}\ne {\bf 0}$. Let ${\bf d} = {\bf b}-{\bf c}$. Then ${\bf a}\times{\bf d} = {\bf 0}$ and ${\bf a}\cdot{\bf d} = 0$, and so $\|{\bf a}\| \|{\bf d}\|\sin\theta = 0$ and $\|{\bf a}\| \|{\bf d}\|\cos\theta = 0$, where $\theta$ is the angle between ${\bf a}$ and ${\bf d}$. ...

3

Suppose we have two linear functions, $f$ and $g$, which agree on all the basis vectors of some space. Then they must agree for every vector on that space, because they are both linear, and a linear function is completely determined by its values on the basis. In gory detail, suppose that we know that $f(\vec{e_i}) = g(\vec{e_i})$ for each basis vector ...

3

Here is an explanation which is nearer to linear algebra: In ${\mathbb R}^3$ we have a volume form $$\epsilon:\quad \bigl({\mathbb R}^3\bigr)^3\to{\mathbb R},\qquad (a,b,c)\mapsto \epsilon(a,b,c)\ ,$$ which produces for any three given vectors $a$, $b$, $c$ the signed volume of the parallelotope spanned by them. It is linear in all three entries, and when ...

3

You're supposed to curl your hand like so: $\hskip 1.3in$ Your knuckles are supposed to form an angle while your fingers are otherwise straight, and your knuckles themselves point in the direction of $a$ while your fingertips point in the direction of $b$. Using this, part (b) presumably wants you to use RHR to figure out which of the 8 octants $\vec{n}$ ...

3

Create a 3 by 3 matrix $$\left[ \begin{array}{ccc} x_1 & y_1 & z_1 \\ x_2& y_2 &z_2 \\ x_3 &y_3 &z_3 \end{array} \right]$$. Then the determinant of $z$ is given by $$z_3 \Bigl| \begin{array}{cc} x_1 & y_1 \\ x_2 & y_2 \end{array}\Bigr| - z_2 \Bigl| \begin{array}{cc} x_1 & y_1 \\ x_3 & y_3 \end{array}\Bigr| + ... 3 Orientation (handedness) is not about a set of vectors, it is about an ordered list of vectors. That is, a certain ordering, (i,j,k) is agreed to as right handed. Then (j,i,k) is left handed. This may or may not agree with some notion you have from physics, hard to predict. A smooth manifold is orientable...never mind. 3 Well, it depends on what you mean by "the vector cross product." There is a generalization to n dimensions which takes n-1 vectors as input and returns what can be thought of as a vector orthogonal to all of them. It generalizes to an operation taking k vectors as input where k \le n, but then the output is not something like a vector but something ... 3 Those formulas don't actually always hold. For example if you take the vectors$$a = \begin{pmatrix}1\\0\\0\end{pmatrix} \quad b=\begin{pmatrix}0\\1\\0\end{pmatrix} \quad c=\begin{pmatrix}0\\0\\-1\end{pmatrix} Then they are orthogonal, but $a \times b = \begin{pmatrix}0\\0\\1\end{pmatrix} =-c$ also $c \times a = -b$ and $b \times c = - a$.

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