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

[japiga]

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]

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

[japiga]

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

[Martingale]

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

[japiga]

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

[Martingale]

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]

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]

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

yes

[japiga]

Sorry, but it is not so helpful!
Please give me step by step procedure.

[Martingale]

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

Sorry, but it is not so helpful!
Please give me step by step procedure.

...