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Unit vector help

I have three unit vectors x,y,z and I am given x+y+z=0.

I am then asked to find (x.y)+(y.z)+(z.a) where the dots represent the scalar/dot/inner product.

I am not really sure how to do this as I know no relation between the unit vectors (they are not the standard x=(1,0,0)).




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Original post by maths01
I have three unit vectors x,y,z and I am given x+y+z=0.

I am then asked to find (x.y)+(y.z)+(z.a) where the dots represent the scalar/dot/inner product.

I am not really sure how to do this as I know no relation between the unit vectors (they are not the standard x=(1,0,0)).


I'd first check that you actually mean that bit in bold, particularly the "a".

Then looking at your original equation x+y+z=0

The obvious thing to do is see what happens if you dot the entire equation with "x" say. Does that give you something useful in relation to what you're trying to show?

If not, can it be adapted?
Reply 2
Original post by ghostwalker
I'd first check that you actually mean that bit in bold, particularly the "a".

Then looking at your original equation x+y+z=0

The obvious thing to do is see what happens if you dot the entire equation with "x" say. Does that give you something useful in relation to what you're trying to show?

If not, can it be adapted?


Got that, thank you. :smile:

Can you tell me how I could simplify 2a x 2b ? I understand that you could write 2a x b as a x (2b) or 2(a x b) but that is only when there is a scalar attached to one term?
Original post by maths01
Got that, thank you. :smile:

Can you tell me how I could simplify 2a x 2b ? I understand that you could write 2a x b as a x (2b) or 2(a x b) but that is only when there is a scalar attached to one term?


You can pull scalars to the front, regardless of how many terms they are attached to.

If you look at the definition of the cross product, we have absinθ  n|a||b|\sin\theta \;\mathbf{n} , and it should be clear that we can extract scalar multiples to the front with impunity.
Reply 4
Original post by ghostwalker
You can pull scalars to the front, regardless of how many terms they are attached to.

If you look at the definition of the cross product, we have absinθ  n|a||b|\sin\theta \;\mathbf{n} , and it should be clear that we can extract scalar multiples to the front with impunity.


So I could take the 2 to the front of 2a x 2b and that would be fine? So 2(a X b) = 2a X 2b
Reply 5
Original post by maths01
So I could take the 2 to the front of 2a x 2b and that would be fine? So 2(a X b) = 2a X 2b


Haven't you lost a factor 2? 2a X 2b = 4(a X b)

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