New answers tagged

1

It surprises me that nobody has brought up the rule of Sarrus yet: For $\mathbf{u} \times \mathbf{v}$, make the table $\begin{array}{|ccc|cc} \mathbf{i}& \mathbf{j}& \mathbf{k}& \mathbf{i}& \mathbf{j} \\ u_1 & u_2 & u_3 & u_1 & u_2 \\ v_1 & v_2 & v_3 & v_1 & v_2 \end{array}$ , then sum up products of diagonals ...


1

In my opinion $2 \times 2$ matrices, particularly $2 \times 2$ real matrices are fairly easy to understand as mathematical objects, even to folks who haven't been exposed to linear algebra. $2 \times 2$ real matrices can be motivated by linear functions in $\mathbb{R}^2$, (equivalently) as scaled rotations of the plane that fix the origin, as a convenient ...


1

The question of whether formal relationships "make sense" is an old one in mathematics. People asked the question about whether negative numbers "made sense" (how can you start with three pebbles and take away six of them?) or complex numbers (what number when squared gives minus 1?). But we found that putting them in algebraic ...


6

this makes no sense But it does. A determinant exists for any $n\times n$ matrix for which the Leibniz formula$$\det A:=\sum_{\sigma\in S_n}\epsilon_\sigma\prod_{i=1}^nA_{i\sigma_i}$$is fully antisymmetric. This is why, for example, quaternionic matrices lack determinants: their elements don't commute. But if we apply it to the case at hand, we don't have ...


2

Just go with the Cartesian formula. The first time I saw a cross product was in computer code and I always felt like the typical way it's coded is the easiest way to compute $\mathbf{c} = \textbf{a} \times \mathbf{b}$: $c_x = a_y b_z - a_z b_y$ $c_y = a_z b_x - a_x b_z$ $c_z = a_x b_y - a_y b_x$ The pattern is obvious. Each component of $\mathbf{c}$ is the ...


7

I had a lecturer who thought the same as you, so suggested we just learnt the definition of the cross product (which he derived along the same lines as James S Cook's answer). Nobody liked his lectures because he pitched them way to high for our first-year brains. The point is, it's a very useful mnemonic device, and when you teach it make that clear, but ...


14

I think you're correct, strictly speaking, but overreacting. I just checked two calculus books, Rogawski and Strang. Rogawski defines the cross product as the "symbolic" determinant (his words) in question but then defines exactly what he means by this determinant in the same line, so I don't see a problem there. Strang mentions the determinant ...


16

My answer is probably not very useful when teaching in high school. I'll just mention here a few reasons why this definition is in fact a good one, and why it's a good idea to teach this formula at a university level mathematics course. $\newcommand{\u}{\mathbf{u}}$ $\newcommand{\v}{\mathbf{v}}$ $\newcommand{\w}{\mathbf{w}}$ $\newcommand{\R}{\mathbb{R}}$ Let ...


3

As a preliminary, it's important to realize the incredibly low level of most students who are learning this stuff. Low means a low intellectual level and previous exposure to low-intellectual-level instruction, with low expectations for intellectual understanding. Low means that students who have previously been exposed to $\hat{\textbf{i}}$, $\hat{\textbf{j}...


4

I agree with the OP: the determinant is only a mnemonic device, but (a) students may not know how to evaluate a determinant, so it's useless as a mnemonic, and (b) it is not a "legal" determinant and so mathematicallly misleading. I would rather start with $|a \times b| = |a| |b| \sin \theta$ as a contrast to $a \cdot b = |a| |b| \cos \theta$, and ...


13

Sure, give up on the mixed-object determinant formula. Instead use tensor arithmetic. Let $\epsilon_{ijk}$ be the completely antisymmetric symbol. Also, while we're at it, let's give up on quaternionic notation for unit-vectors and instead use $\hat{x},\hat{y},\hat{z}$ or $\hat{x_i}$ for $i=1,2,3$ then you have an elegant formula: $$ \vec{A} \times \vec{B} = ...


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