Skip to main content
11 votes
Accepted

How to prove that the cross product doesn't satisfy any kind of generalized associativity?

Let $p(v_1, \dots v_n)$ and $q(v_1, \dots v_n)$ be two parenthesizations of the cross product of $n$ vectors. First note that if $p$ and $q$ have any "bottom-level" cross products $v_i \...
Qiaochu Yuan's user avatar
6 votes

What is the operation between a bivector and a vector that outputs another vector?

$ \newcommand\proj[1]{\langle#1\rangle} \newcommand\lcontr{\mathbin\rfloor} \newcommand\rcontr{\mathbin\lfloor} $peek-a-boo's answer is good, but because geometric algebra (GA) is mentioned in the ...
Nicholas Todoroff's user avatar
6 votes
Accepted

What is the operation between a bivector and a vector that outputs another vector?

It is 'almost' the interior product (not to be confused with the inner product, which will also play a role, hence the ‘almost’). In general, suppose $V$ is a real vector space, and you have integers $...
peek-a-boo's user avatar
4 votes
Accepted

How do you show that $u+w+v = 0$ given $u \times v = v \times w = w \times u$

It isn't true. Try $u = v = w \ne 0$. EDIT: If you add the extra condition that the vectors are pairwise non-collinear, it is true. Note that $a \times b = 0$ if and only if $a$ and $b$ are collinear....
Robert Israel's user avatar
3 votes

Evaluation of surface integral $\iint F ds$; $F(x,y,z)=(x,y,z)$

Yes, your calculation gives the correct answer. A minor point on notation: usually we use $\iint_S \vec{F}\cdot d\vec{s}$ to mean integrating a vector field on a surface. Writing $\iint_S Fds$ can be ...
Mysterium's user avatar
  • 185
3 votes

Do the BAC-CAB identity for triple vector product have some intepretation?

Well, we can prove the BAC-CAB identity using only geometric arguments as follows. First consider a coordinate system where the $x$- and $y$-axes are skewed (non-orthogonal). We project the line $\...
Engelmark's user avatar
  • 123
3 votes

Derivative of cross product w.r.t. a vector

$ \def\LR#1{\left(#1\right)} \def\qiq{\quad\implies\quad} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\gradLR#1#2{\LR{\grad{#1}{#2}}} $...
greg's user avatar
  • 36.8k
3 votes

Derivative of cross product w.r.t. a vector

In tensorial form, we can write $$ (a\times b)_i=\epsilon_{ijk}a^jb^k, $$ where $\epsilon_{ijk}$ is the Levi-Civita's symbol. Therefore, what we are looking for, can be expressed as $$ \frac{\partial (...
Mostafa Ayaz's user avatar
  • 32.5k
3 votes
Accepted

$\hat i\times\hat j$

My rubric for computation of $\mathbf{x} \times \mathbf{y}$ is $$ \left|\begin{array}z & \mathbf{i}\quad\;\; \mathbf{j}\quad\;\;\mathbf{k} \\ &x_1 \quad x_2 \quad x_3 \\ &y_1 \quad y_2 \...
GEdgar's user avatar
  • 113k
3 votes

Why is the magnitude of the cross product equal to the parallelogram spanned by the two vectors?

Q1. What is the purpose of wanting the cross product to have magnitude equal to the parallelogram spanned by $a$ and $b$? I'm not sure what sort of answer you expect to get to this question. The ...
Blitzer's user avatar
  • 2,163
2 votes
Accepted

Show that the sum of these four vectors is $0$.

In these types of problems, you must carefully account for the direction of the vectors. Let's choose $ABC$ in such a way that they form a triangle going clockwise, and $O$ a point above the $ABC$ ...
Andrei's user avatar
  • 37.7k
2 votes

How to prove that the cross product doesn't satisfy any kind of generalized associativity?

Here is my hand-wavy Proof. When it is associativity for some $N$ terms , & it so occurs that both sides contain some same sub-term $U \times V$ (somewhere at start or at end or in middle) , then ...
Prem's user avatar
  • 11.6k
2 votes
Accepted

On vector multiplication

Point 2 is an axiom, used to define vector multiplication. I wouldn't say that this is used to define the dot product. Instead, one must already have a definition of the dot product, specifically, ...
Peeter Joot's user avatar
  • 2,786
2 votes
Accepted

Is it true that $\vec{\nabla}\times[(\vec{a}\cdot\vec{\nabla})\vec{a}]=\vec{\nabla}\times[\vec{a}\times(\vec{\nabla}\times\vec{a})]$?

You are off by a sign. Recall $$ V\times(\nabla\times V)=\underbrace{V\cdot (\nabla V)^T}_{=\nabla(\tfrac12 V\cdot V)} - (V\cdot\nabla)V $$ (This is the usual "BAC-CAB" identity for triple ...
user10354138's user avatar
  • 33.3k
2 votes

Evaluation of surface integral $\iint F ds$; $F(x,y,z)=(x,y,z)$

The end results $54\pi$ is correct but the chain of equations by which it was derived is very confusing. This answer shows various methods to calculate a surface integral $$ \iint_S\mathbf{F\cdot n}\,...
Kurt G.'s user avatar
  • 15.3k
2 votes

Why is the magnitude of the cross product equal to the parallelogram spanned by the two vectors?

Regarding the second question: The formula for $|a\times b|$ follows from four axioms: $|a\times b| = |a||b|$ if $a\perp b$. $a\times a = 0$. Bilinearity of $\times$. $a\cdot b = |a||b|\cos\theta$. ...
Nicholas Todoroff's user avatar
1 vote
Accepted

Generalize an identity of cross products to $n$ dimensions

This isn't a complete answer but it's too long for a comment. I'm going to use the wedge product instead of the cross product and its higher-dimensional analogue (sometimes called the "external ...
blargoner's user avatar
  • 3,190
1 vote
Accepted

Evaluation of surface integral $\iint F ds$; $F(x,y,z)=(x,y,z)$

Just to give something different… If we close off the surface by adding the disk of radius 3 in the xy plane, we can apply the divergence theorem. $\iint F dS + \iint F dA = \iiint \nabla \cdot F dV$ ...
user317176's user avatar
  • 11.5k
1 vote

Loop integral of $\mathbf{r}\times d\mathbf{r}$ is equal to twice the enclosed area?

By using the very definition of the cross-product, one finds: $$ \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x& y & z \\ dx& dy & dz \\ \end{array} \right|...
Egor Larionov's user avatar
1 vote
Accepted

Loop integral of $\mathbf{r}\times d\mathbf{r}$ is equal to twice the enclosed area?

We may compute $\mathbf r\times \mathrm d\mathbf r$ first: $$\mathbf r\times \mathrm d\mathbf r=\begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ x & y & 0\\ \mathrm dx & \mathrm ...
Conreu's user avatar
  • 2,302
1 vote

Finding vector equation of a line

Let the required line be $\vec{r} = \vec{b} + \beta \vec{l}$. Now as per the diagram below, we can say that vector $\vec{p} = (\vec{a}-\vec{b}) \times \vec{c}$ is perpendicular to the plane containing ...
redshellspy's user avatar
1 vote

Finding vector equation of a line

In terms of unknown, it may be abstract. Let's recall in our first multivariable class, with numerical example. Like find a line passing through $b=(\pi,1,2)$ such that it is perpendicular to the line ...
Angae MT's user avatar
  • 1,230
1 vote
Accepted

Interior, cross and outer products between two multivectors?

$ \newcommand\form[1]{\langle#1\rangle} \newcommand\G{\mathscr G} \newcommand\lcontr{\mathbin\rfloor} \newcommand\rcontr{\mathbin\lfloor} $Any decent book on geometric algebra should have these ...
Nicholas Todoroff's user avatar
1 vote
Accepted

Computing $U \times V$

The answer given is correct subject to a typo. That is, the second vectori is $V$ not $U$.
Bob's user avatar
  • 4,004
1 vote
Accepted

Intuition for why a 90 degree rotation of a vector about an arbitrary axis can be expressed as 3 90 degree rotations of the vector's projections.

Note that $ \mathbf{\hat{u}} $ and $ \mathbf{v} $ necessarily lie in a plane, so it's possible to decompose $ \mathbf{v} $ into components that lie strictly in that plane. Suppose $ \{ \mathbf{\hat{f}...
Peeter Joot's user avatar
  • 2,786
1 vote
Accepted

Prove that the trajectory of $P$ is a circle in $3D$ and find its properties

I have I think a solution to your question, but it might not be the most straightforward approach. Start with the given equation $$ \tfrac{\rm d}{{\rm d}t} \boldsymbol{p} = \boldsymbol{a} \times \...
John Alexiou's user avatar
  • 14.2k
1 vote

What's the opposite of a cross product?

@Robert Israel already gave the answer, but didn't give a derivation. @Oscillon hinted at using the triple product, however, I agree with @Aidan Jalili, a written out derivation is very helpful, ...
Caesar.tcl's user avatar
1 vote

Is it true that $\vec{\nabla}\times[(\vec{a}\cdot\vec{\nabla})\vec{a}]=\vec{\nabla}\times[\vec{a}\times(\vec{\nabla}\times\vec{a})]$?

start from the right hand side's inner cross product: $a\times (\nabla \times a)$ It is best to use tensor algebra here: $\epsilon_{ijk}\epsilon_{klm}a_j\frac{\partial a_m}{\partial x_l}=(\delta_{il}\...
ryaron's user avatar
  • 1,105
1 vote

Is it true that $\vec{\nabla}\times[(\vec{a}\cdot\vec{\nabla})\vec{a}]=\vec{\nabla}\times[\vec{a}\times(\vec{\nabla}\times\vec{a})]$?

In component Einstein notation $$\varepsilon_{i\ j\ k} \partial_i ( a_l \partial_l a_l) = \varepsilon_{i\ j\ k} \partial_i \varepsilon_{j\ l \ m} a_l \varepsilon_{m \ s\ t} \partial_s a_t\ \text{?}$...
Roland F's user avatar
  • 3,021

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