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Let $\mathbf{a},\mathbf{b} \in \mathbb{R^3}$. Let $a_1,a_2,a_3,b_1,b_2,b_3 \in \mathbb{R}$ such that $\mathbf{a} = (a_1,a_2,a_3)$ and $\mathbf{b} = (b_1,b_2,b_3)$. Let $\hat{i}$, $\hat{j}$ and $\hat{k}$ denote unit vectors along the positive $x$,$y$ and $z$ axis of a right handed coordinate system.

Definition: The cross product of $\mathbf{a}$ and $\mathbf{b} $, denoted by $\mathbf{a} \times \mathbf{b}$ is defined as:- $$\mathbf{a} \times \mathbf{b} := (a_2b_3 - a_3b_2)\hat{i} + (a_3b_1 - a_1b_3)\hat{j} + (a_1b_2 - a_2b_1)\hat{k}$$ Using this definition of $\mathbf{a} \times \mathbf{b}$, I wanted to show the following property: $$\mathbf{a} \times \mathbf{b} = |\mathbf{a} | |\mathbf{b} | \sin (\theta) \hat{n} $$ where $\theta$ is the non-negative non-reflex angle between $\mathbf{a}$ and $\mathbf{b}$ and $\hat{n}$ is a unit vector perpendicular to $\mathbf{a}$ and $\mathbf{b}$ whose direction is given by the right hand thumb rule.

Right-Hand Thumb Rule for $\mathbf{a} \times \mathbf{b} $: If the index finger of your right hand is along the direction of $\mathbf{a}$, and your middle finger is along the direction of $\mathbf{b}$, then the thumb gives the direction of $\mathbf{a} \times \mathbf{b} $:

I was able to show that $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a} | |\mathbf{b} | \sin (\theta) $ and $\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 0$ and $\mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) = 0$ using plain algebra. Hence I have shown that $\hat{n}$ is perpendicular to $\mathbf{a}$ and $\mathbf{b}$. However, I don't know how to show that the direction of $\hat{n} $ by the right hand thumb rule is the same as the direction of $\mathbf{a} \times \mathbf{b} $ as given by the definition of cross product. How can I show this?

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To prove that the Right-Hand Thumb Rule holds we can start by the definition for the unitary vectors along coordinates axes

  • $\vec i \times \vec j=\vec k, \qquad \vec j \times \vec k=\vec i, \qquad \vec k\times \vec i=\vec j$

  • $\vec j \times \vec i=-\vec k, \qquad \vec k \times \vec j=-\vec i, \qquad \vec i\times \vec k=-\vec j$

and assume wlog

  • $\vec a = a_x \vec i $
  • $\vec b = b_x \vec i +b_y \vec j $
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One can show, that it's possible to continuously transform pair of vectors from $\mathbf a$, $\mathbf b$ to $\hat i$, $\hat j$, so that $|\mathbf a\times\mathbf b|$ is never zero (if it wasn't zero in the first place). Since $\mathbf a\times\mathbf b$ is continuous and non-zero, it never jumps from one half-space to the other. Thus, $\mathbf a\times\mathbf b$ and $\hat i\times\hat j$ share the same chirality rule (either left hand or right hand). By direct check, we see that $\hat i\times\hat j=\hat k$ and since $\hat i,\hat j,\hat k$ form a right basis, we conclude that right hand thumb rule is applicable to every $\mathbf a$ and $\mathbf b$

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The Euclidean $3$-space alone isn't enough. You need to fix an orientation to define the vector product this way.

Now there are various ways to define an orientation. One way is to declare $\mathbf{i},\mathbf{j},\mathbf{k}$ is right-handed (a priori you could have chosen the opposite direction for $z$ and still call it $\mathbb{R}^3$, for example).

Then "right-handedness" of the choice of $\mathbf{n}$ (or rather, of the triple $\mathbf{a},\mathbf{b},\mathbf{n}$) means there is a rotation of $\mathbb{R}^3$ that brings $\mathbf{i},\mathbf{j},\mathbf{k}$ to $\hat{\mathbf{a}},\mathbf{b'},\mathbf{n}$ where $\hat{\mathbf{a}}=\mathbf{a}/|\mathbf{a}|$ and $\mathbf{b}'$ is the unit vector in the direction $\mathbf{b}-(\mathbf{b}\cdot\hat{\mathbf{a}})\hat{\mathbf{a}}$.

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