Help to solve this:

If x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1. Prove: xy+yz+zx=0

Help to solve this:

If x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1. Prove: xy+yz+zx=0

If x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1. Prove: xy+yz+zx=0

- Martingale
**Posts:**335**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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japiga wrote:Help to solve this:

If x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1. Prove: xy+yz+zx=0

hint :

therefore

hence

And then? How to include this in the first equation x/a=y/b=z/c ?

- Martingale
**Posts:**335**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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japiga wrote::confused: And then? How to include this in the first equation x/a=y/b=z/c ?

solve for x in terms of z and solve for y in terms of z

then plug those into

But what is a purpose of plugging into the equation of xy + yz + zx? Actually, how to reach this equation?

- Martingale
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japiga wrote:But what is a purpose of plugging into the equation of xy + yz + zx? Actually, how to reach this equation?

Don't you want to show that xy+yz+xz=0? ....Yes!!

so start with xy+yz+xz and show that is has to be zero using the hints I provided.

You mean that this equation x/a=y/b=z/c should be split in two equations such as x/a=y/b and y/b=z/c. So x= (az)/c and y=(bz)/c. Does it what you mean?

- Martingale
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japiga wrote:... So x= (az)/c and y=(bz)/c. Does it what you mean?

yes

Sorry, but it is not so helpful!

Please give me step by step procedure.

Please give me step by step procedure.

- Martingale
**Posts:**335**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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japiga wrote:...So x= (az)/c and y=(bz)/c. Does it what you mean?

japiga wrote:Sorry, but it is not so helpful!

Please give me step by step procedure.

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