Question

Associative double cross products

Answer 1

for vectors A, B and C when is this true?

\[A \times (B \times C) = (A \times B) \times C\]

Answer 2

Wouldn't one of the vectors have to be zero?

Answer 3

Yeah, I guess that would work but I am not 100% sure if that's like everything or what

Answer 4

I believe order is important here, so that as long as you keep A, B and C in that order. Why not create three vectors A, B and C and determine whether this is in fact what happens.

Answer 5

Yeah, that's actually no a bad idea I should just make the vector components and solve to make sure

Answer 6

Yes. Notice that all three vectors A, B and C are 3-dimensional.

Answer 7

weird it just flipped the chat order around for me. Yeah, geometrically it seems to make sense that the cross product making a vector normal to the plane spanned by the vectors it came from should sorta not work out.

Answer 8

Ah a little algebra makes the geometric proof simpler for me to see now.

Just looking at one side of the equation,
\[(A \times B) \times C = -C \times (A \times B) = C \times (B \times A)\]
That means:
\[A \times ( B \times C) = C \times (B \times A)\]
So it's gotta be the 0 vector I think

Answer 9

I think if A and C are parallel we can get a nontrivial solution actually

Answer 10

Hmm, I'm having fun with the triple product expansion

\[\large A \times (B \times C) = (A \times B) \times C\]
\[\large \cancel{\color \red{(a \cdot c)b}} - (a\cdot b)c = -(c\cdot b)a + \cancel{\color \red{(c \cdot a)b}}\]

Giving us
\[\large (a \cdot b)c = (c\cdot b)a\]

\[\large \frac{a}{a \cdot b} = \frac{c}{c\cdot b}\]

Solve this guy for 'c' and we find that by letting \(\large \frac{c\cdot b}{a \cdot b} = k\) *since they will just be a constant*
we can see that \(\large c = ka\) showing us that is vector c is some multiple of vector a, this would work out

Answer 11

Lol literally just beat me to the punch >.<

Answer 12

Hahaha sorry didn't mean to steal your thunder haha buuuuut your stuff is more thorough so it shows that we can't have other possibilities too I think

Answer 13

That was fun, basically I came to asking this question cause I had heard that octonion multiplication is not associative and they gave this as an example that non-associativity isn't that uncommon.

Answer 14

\[A \times (B \times C) = (A \times B) \times C\]

lef side is parallel to BC plane
right side is parallel to AB plane
so...