x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

japiga · Postby **japiga** » Mon Oct 18, 2010 8:00 pm

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

Martingale · Postby **Martingale** » Mon Oct 18, 2010 8:36 pm

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

:

$(a+b+c)=1=1^2=(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$

therefore

$1=1+2ab+2ac+2bc$

hence

$ab+ac+bc=0$

japiga · Postby **japiga** » Tue Oct 19, 2010 7:47 am

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

Martingale · Postby **Martingale** » Tue Oct 19, 2010 12:45 pm

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

$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$

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

then plug those into $xy+yz+zx$

japiga · Postby **japiga** » Tue Oct 19, 2010 1:49 pm

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

Martingale · Postby **Martingale** » Tue Oct 19, 2010 2:19 pm

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.

japiga · Postby **japiga** » Tue Oct 19, 2010 2:47 pm

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 · Postby **Martingale** » Tue Oct 19, 2010 4:13 pm

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

yes

japiga · Postby **japiga** » Tue Oct 19, 2010 4:27 pm

Sorry, but it is not so helpful! :confused:

Please give me step by step procedure.

Martingale · Postby **Martingale** » Tue Oct 19, 2010 4:34 pm

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.

$xy+xz+yz=\frac{a}{c}z\frac{b}{c}z+\frac{a}{c}z\cdot z+\frac{b}{c}z\cdot z=\frac{ab}{c^2}z^2+\frac{a}{c}\cdot z^2+\frac{b}{c}\cdot z^2$

...

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x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

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Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1