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Vectors

The vectors a, b and c are such that bxa= (2 1 0) and axc= (0,3, -2). Find the vector (-2b-c)x(-3a-4b+2c). I used the distributive rule to get (-2bx-3a) + (-2bx-4b) + (-2bx2c) + (-cx-3a) + (-cx-4b) + (-cx2c). I understand that -cx-2c =0 and -2bx-4b =0. However, I was unsure about the other values. Also, I have no idea how to work out what -cx-4b is equal to. Thanks in advance!
Original post by A1ana2003
The vectors a, b and c are such that bxa= (2 1 0) and axc= (0,3, -2). Find the vector (-2b-c)x(-3a-4b+2c). I used the distributive rule to get (-2bx-3a) + (-2bx-4b) + (-2bx2c) + (-cx-3a) + (-cx-4b) + (-cx2c). I understand that -cx-2c =0 and -2bx-4b =0. However, I was unsure about the other values. Also, I have no idea how to work out what -cx-4b is equal to. Thanks in advance!


Recall the properties of the vector product:

Scalars can be pulled out to the front. And bxa = - axb.

That's enough to simplify your expression to something that can be evaluated.
Original post by ghostwalker
Recall the properties of the vector product:

Scalars can be pulled out to the front. And bxa = - axb.

That's enough to simplify your expression to something that can be evaluated.

Thanks, I understand that. However, how do I find out what bxc is equal to??
Original post by A1ana2003
Thanks, I understand that. However, how do I find out what bxc is equal to??


Sorry, I miscalculated. Thought the bxc would vanish, but it doesn't; though I suspect it's meant to given the coefficients.

Do you have an image of the original question, and given answer?

Edit: As it stands there is insufficient information to evaluate bxc, so I suspect there is a sign error in the question. There are several possibilities, any of which would lead to the bxc term having a coeff of 0, and thus avoiding the need to work out that particular cross product.
(edited 3 years ago)
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deimos_xe
Original post by username5207810
The vectors a, b and c are such that bxa= (2 1 0) and axc= (0,3, -2). Find the vector (-2b-c)x(-3a-4b+2c). I used the distributive rule to get (-2bx-3a) + (-2bx-4b) + (-2bx2c) + (-cx-3a) + (-cx-4b) + (-cx2c). I understand that -cx-2c =0 and -2bx-4b =0. However, I was unsure about the other values. Also, I have no idea how to work out what -cx-4b is equal to. Thanks in advance!
i know this thread is dead but i was given this question by my math teacher and he said the vector your supposed to be finding has a typo in it instead of finding (-2b-c)x(-3a-4b+2c) you need to find (2b-c)x(3a-4b+2c)
Original post by deimos_xe
i know this thread is dead but i was given this question by my math teacher and he said the vector your supposed to be finding has a typo in it instead of finding (-2b-c)x(-3a-4b+2c) you need to find (2b-c)x(3a-4b+2c)
Good to know.

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