# Fleming's “right-hand rule” and cross-product of two vectors

I have been throwing around hand gestures for the past hour in a feeble attempt at trying to solve this question involving a cross product of two vectors $a$ x $b$. So far, I haven't found any satisfactory explanation on the internet as to how this rule is supposed to help me find out whether the components of $a$ and $b$ are positive, negative or zero. E.g., James Stewart explains it as though it is obvious (it isn't):

If the fingers of your right hand curl through the angle $\theta$ from $a$ to $b$, then your thumb points in the direction of $n$

What does he mean by "curl"? Sorry if this is a dumb question but I am getting really frustrated trying to use this "rule" to solve what is seemingly a really elementary question. I imagine it would be more intuitive if someone could just "show" me how to do it, but since that is not possible, if anyone could offer an explanation that is more helpful than the above, I would appreciate it.

(This question is specifically in reference to Exercise #4, Chapter 9.4 in Stewart's Single-variable calculus book, 4th edition.)

The figure shows a vector a in the xy-plane and a vector $b$ in the direction of $k$. Their lengths are $||a|| = 3$ and $||b|| = 2$. a) find $||a$ x $b||$ b) Use the right-hand rule to decide whether the components of a x b are positive, negative or zero

I am stuck on part b).

-
Think of the natural way your right hand would move when you want to slap someone. That's also the way fingers of your right hand curl. Sorry if this is a dumb answer. –  Shitikanth Apr 14 '12 at 5:14
The RHR is supposed to help you visualize the result of a cross product. I don't see how cross products are relevant to finding out "whether the components of $a$ and $b$ are positive, negative or zero." You also refer to "this question" but don't tell us what it is. –  anon Apr 14 '12 at 5:14
I guess I am not understanding what it means in terms of the angle. What does it mean to "curl through an angle"? –  Dylan Apr 14 '12 at 5:15
I am sorry, I guess your confusion is really a reflection of my own in that I don't have any clue how this rule is supposed to help me solve this question. I will post the question, but I don't know how to include the actual graph... hopefully it will still help. –  Dylan Apr 14 '12 at 5:18
Have you tried searching for some videos on this? E.g. at youtube. –  Martin Sleziak Apr 15 '12 at 4:46

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 $\vec{n}$ lies in (or between!), which would determine the sign of the components. I can't say much without seeing the picture of $a$ and $b$ though...

Update (sorry, MSPaint...):

$\hskip 1in$

I assume $a$ is on the $xy$ plane, so your thumb is too. This implies $z=0$. Moreover, the thumb is still close enough to the positively oriented $x$-ray (haha, pun) that it has positive $x$ component, but far enough from the positively oriented $y$-ray to have negative $y$ component.

-
Added an image above. I understand the explanation, but I don't understand how it relates exactly to the graph (i.e. how to position my hand, how it lines up with a and b vectors). –  Dylan Apr 14 '12 at 5:38
@Dylan: I've updated my answer. –  anon Apr 14 '12 at 5:51
I think I am getting closer... I see now how to orient my thumb with respect to a given vector. However, I am not sure I understand your explanation of positive versus zero versus negative. When you say "it's close enough to be positively oriented $x$-ray[...]", what does that mean precisely? Because, visually speaking, vector $a$ looks much closer to $y$ in terms of the angle between them than it is to $x$. –  Dylan Apr 14 '12 at 5:58
@Dylan: Visually the $x$-axis has a positive ray and a negative ray; same with the $y$-axis. Between the two rays is a perpendicular plane (namely, the $yz$ plane between the $x$-rays and the $xz$-plane between the $y$-rays). Which side of these planes a vector is on determines whether its component is positive or negative (and if it is actually on the plane, it has component $0$). Remember we're trying to analyze $a\times b$, which is the thumb. The thumb is closer to the negative $y$-ray (not actually shown) than the positive one, and we're not investigating $a$'s components. –  anon Apr 14 '12 at 6:03
After your explanation, it is much clearer now. Thank you. –  Dylan Apr 14 '12 at 6:14