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I am stuck on the following problem that says:

If $x=cy+bz,\,y=az+cx,\,z=bx+ay$ then show that

$x^2:y^2:z^2=(1-a^2):(1-b^2):(1-c^2)$ where ":" indicates ratio.

My try:

The result I got is: $$z(1-a^2)=x(b+ac), y(1-c^2) =(a+bc)z\,,x(1-b^2)=(c+ab)y$$

But I don't know how to progress from here. Can someone help? Thanks in advance for your time.

  • 0
    What does $ x^2:y^2:z^2$ mean? – 2017-01-01
  • 1
    Ratio of $x^2$ ,$y^2$ and $z^2$ – 2017-01-01

2 Answers 2

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Since $z=bx+ay$, we get $$x=cy+bz=cy+b(bx+ay)=cy+b^2x+aby\tag1$$ and $$y=az+cx=a(bx+ay)+cx=abx+a^2y+cx\tag2$$ Multiplying the both sides of $(1)$ by $x$ gives $$x^2=cxy+b^2x^2+abxy\tag3$$ Multiplying the both sides of $(2)$ by $y$ gives $$y^2=abxy+a^2y^2+cxy\tag4$$ Now $(3)-(4)$ gives $$x^2-y^2=b^2x^2-a^2y^2,$$ i.e. $$(1-b^2)x^2=(1-a^2)y^2$$ I think that you can continue from here.

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    Thanks a lot for the clarification...Got it. – 2017-01-01
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HINT:

Solving $x-cy-bz=0,cx-y+az=0$ for $x,y$

$x=\dfrac{z(ca+b)}{1-c^2},y=\dfrac{z(bc+a)}{1-c^2}$

Similarly, solve for $bx+ay-z=0,cx-y+az=0$ for $x,y$