[Math] Fleming’s “right-hand rule” and cross-product of two vectors

calculuscross productvector analysis

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).

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Best Answer

You're supposed to curl your hand like so:

$\hskip 1.3in$ rhr

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...):

$\hskip 1in$ rhr2

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.

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