How do you find vector a and b given the cross product of a x b?
@hero @angle would you mind taking a look at this one? I'm not sure tbh
you need the dot product and cross product, it's a classic intro multivariate calc problem
and iirc there's two possible solutions
Suppose\[a^ \rightarrow\] and \[b^ \rightarrow\] are vectors such that \[a^ \rightarrow \times b^ \rightarrow\] = (1,0,2). What is \[b^ \rightarrow \times (-2a^ \rightarrow )\]
Sorry, I'm not sure why it turned out like that.
what are the properties of the cross product that you know of?
We didn't learn the properties of the cross product. We were just taught how to do it.
you know of the a x b = a*b sin theta right?
Ah yes, we did talk about that.
is that enough of a hint, or do i take it further?
So we have: (1,0,2) = a*bsin theta But what else could I do without an angle or anything?
ok first off, when you switch the order of the vectors in a cross product, what happens
Changes sign?
would that cancel out with the negative in the a?
The -2a part?
yeah
and think about what the 2 does
So the -2a becomes 2a and the 2 double the value of a
yup, but due to that formula i gave you double the value of a is twice the length (but the same angle) so you just multiply your answer by 2
So we multiply the cross product by 2? Sorry, I'm sorta confused.
yup b x (-2 a) = -(-2 a) x b = 2a x b = 2(a x b)
(2,0,4) ?
yup
But how does that give me the value of vector a and b?
why do you want to know what a and b are
if I have 2 cars and I sell one, I have one car left. I don't need to know how much the cars cost or how much I sold them for.
another way to put it, a and b are not unique vectors. Let me come up with an analogy if a+b=2, what is 2a+2b? we know it's 4. What are a and b? there are an infinite number of possibilities, but we still get 4 if we multiply their sum by 2.
Ohh, right!! What am I thinking? I get it now. Tysm
yup, gl!
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