@AkivaWeinberger Makes sense, thanks. So in general, if you have an equation with vectors and you take the magnitude of both sides, in cases where there is more than one vector on one side you only take the magnitude of one vector?
That first equation doesn't make sense. (a x c)(e x f) is a type error. a x c and e x f are both vectors; you've written it like you're trying to multiply them like they're scalars
It can be confusing keeping track of the type (vector vs scalar) of everything. In your notes, it can be useful to add annotations pointing out what bits of your equations have what types
I guess that means we technically should've had |sine(theta)|. (But two vectors in space will always have an angle between them between 0 and 180 degrees so the sine will be positive anyway)
@AkivaWeinberger If say you have 1 vector (a) on one side and you have an operation involving 2 vectors (b and c) on the other side. Does the "take magnitude" operation involve taking the magnitude of both b and c? e.g. a = b some operation c. Is that equal to |a| = |b| same operation c or do you have to take the magnitude of c as well?
So if you have a x b = |a||b|sin(theta)nhat, that equals to |a x b| = ||a|| ||b|| |sin(theta)| |nhat|, which is equal to |a x b| = |a| |b| sin(theta) |nhat|, since ||x|| = |x| and |sin(theta)| = sin(theta) for 0 <= theta <= 180?
So if you have a x b = |a||b|sin(theta)nhat, that equals to |a x b| = ||a|| ||b|| |sin(theta)| |nhat|, which is equal to |a x b| = |a| |b| sin(theta) |nhat|, since ||x|| = |x| and |sin(theta)| = sin(theta) for 0 <= theta <= 180?
I'm getting confused about taking the magnitude of both sides of the equation. How does a x b = |a||b| sine(theta) nhat the same as |a x b| = |a||b| sine(theta) |nhat|?
@TedShifrin A scalar is a variable rather than an actual vector, right?
So if you have a x b = |a||b|sin(theta)nhat, that equals to |a x b| = ||a|| ||b|| |sin(theta)| |nhat|, which is equal to |a x b| = |a| |b| sin(theta) |nhat|, since ||x|| = |x| and |sin(theta)| = sin(theta) for 0 <= theta <= 180?
This is all fine
|| means absolute value of absolute value there. He's using the fact that $|(|x|)|=|x|$, but writing it out in full
(in fancy words, absolute value is idempotent, meaning doing it twice is the same as doing it once)
You're being confused by the fact that people often use ||.|| to mean magnitude, I think, but in the context of the conversation that's not the notation we're using
It’s absolute value of magnitude. And one needs to understand what is magnitude and what is absolute value. Along with understanding what’s a vector and what’s a scalar. But throw symbols around without understanding. Fine.
And magnitude of a scalar isn't substantially different from absolute value, because they are both the distance from 0. (Or, if you like, the scalar field may be naturally identified with the one-dimensional vector field.)
xnero this is actually an instance where it would be helpful to use chatjax. some of these questions do appear to touch on type-checking issues and it helps to use notation that reflects that.
So if you have a x b = |a||b|sin(theta)nhat, that equals to |a x b| = ||a|| ||b|| |sin(theta)| |nhat|, which is equal to |a x b| = |a| |b| sin(theta) |nhat|, since ||x|| = |x| and |sin(theta)| = sin(theta) for 0 <= theta <= 180?
if you write |x| for the magnitude of the vector x, then you could write this as ||x|| = |x| but now we have | |s that mean different things and maybe it's better to stay away from that sort of thing until later.
you can denote it however you like. i'm pointing out that when you do that, the two | |s have different meanings, as do the two different uses of the word "magnitude" when you say "magnitude of the magnitude"
if you write ||x|| = |x| with your convention, the inner | |'s around x refer to the magnitude of whatever type of vector x is. the outer | |s refer to the absolute value (or magnitude, if you will) of a real number.
if your vectors aren't all just real numbers, you might want to mentally flag this different status assigned to the same symbol.
So if you have a x b = |a||b|sin(theta)nhat, that equals to |a x b| = ||a|| ||b|| |sin(theta)| |nhat|, which is equal to |a x b| = |a| |b| sin(theta) |nhat|, since ||x|| = |x| and |sin(theta)| = sin(theta) for 0 <= theta <= 180?
i don't know. it's fine. i'm pointing out that | | is playing two roles here, as is your use of the term 'magnitude.'
it's fine to do that, but be aware that you're doing it, particularly if you're asking questions that may turn upon people being aware of what you're doing.
@leslietownes I am taking the magnitude of both sides. The magnitude of a x b is |a x b|, the magnitude of |a||b| sin(theta) n cap = ||a|| ||b|| |sin(theta)| |ncap|, simplifying to |a| |b| sin(theta) |ncap|. Correct?
I am finding the sine of the acute angle between the vectors a = 2i + j + 2k, b = -3j + 4k using the cross product. The textbook says to substitute |nhat| = 1 into the formula a x b = |a||b|sin(theta)nhat. Firstly, the formula uses nhat which is a vector, not |nhat|. Secondly, why can you substite |ncap| = 1?
|a x b| = |a| |b| sin(theta), then solve for theta?
OK. you're not substituting anything into the vector formula. you're taking the norm of the vector formula, getting a formula with numbers, and then simplifying that by noting that one of the numbers is 1.
yeah. there's a little bit of ambiguity about recovering theta from sine theta some of the time, but if there's an acute angle here i guess there isn't.
it seemed for a moment like the problem was just asking for the sine of the angle and not the angle itself. that may just be how you worded it as an intermediate step. it is not clear to me.
Those have very different angles, but the cross product is the same. This is why you cannot tell the angle from the cross product, unless you know whether it is acute or obtuse
