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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 ...


14

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 $t$.


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 ...


13

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 ...


12

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 ...


10

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

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 ...


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

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 ...


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

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 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

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 ...


5

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 = ...


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).


5

It means that you can verify the relation just using the standard basis $\{e_1, e_2, e_3 \}$ of three dimensional space. For example you should check $(e_1 \times e_2)\cdot (e_2 \times e_3) =0$ which is the same as the right hand side.


5

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 ...


5

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 ...


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

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: $$ ...


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

$$(\vec a\times \vec b)\times\vec c=-\vec c\times (\vec a\times \vec b)=\vec c\times(\vec b \times \vec a)=\vec b(\vec c\cdot\vec a)-\vec a(\vec c\cdot \vec b).$$


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 ...


4

I think the best way to answer your question is, it's a mnemonic. This mnemonic lets you get your hands on a collection of mathematical objects called "exterior forms". From this perspective, it's not a "hack" but to explain exactly what it is essentially requires discussing dual spaces and things not appropriate for standard multivariable calculus ...


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 problem amounts to finding a vector potential $\mathbf G$ such that $\mathbf F = \nabla \times \mathbf G$. The existence of a vector potential for $\mathbf F$ is equivalent to saying that $\mathbf F$ is divergence free: $$\nabla \cdot \mathbf F = \nabla \cdot (\nabla \times \mathbf G) = 0.$$ In our case we have $\nabla \cdot \mathbf F = 0$, so there ...



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