Show that $\frac{x}{\sqrt{1-a^2}}=\frac{y}{\sqrt{1-b^2}}=\frac{z}{\sqrt{1-c^2}}$

Question

Given $3$ equations:

• $ x-cy-bz=0$
  • $-cx+y-az=0$
  • $-bx-ay+z=0$

Show that $\dfrac{x}{\sqrt{1-a^2}}=\dfrac{y}{\sqrt{1-b^2}}=\dfrac{z}{\sqrt{1-c^2}}$

Now solving the 3 equations I got:

$\dfrac{x}{{1-a^2}}=\dfrac{y}{{ab+c}}=\dfrac{z}{ac+b}$

$\dfrac{x}{ac+b}=\dfrac{y}{a+bc}=\dfrac{z}{1-c^2}$

$\dfrac{x}{a+bc}=\dfrac{y}{1-b^2}=\dfrac{z}{a+bc}$

How to prove the required fact?

Any way to prove it ?

Answer 1

Hint:

Relating to geometry of triangles, looks much like sine rule to me, where $x,y$ and $z$ are sides of triangle and $a, b, c$ are the cosines of the corresponding opposite angles to the respective sides.

But here we restrict that $-1\le a, b, c\le 1$ because for the denominator to be meaningful in the "real number" sense we need to find the range of $a, b, c$

Edit:

Consider triangle $XYZ$ with sides $x, y, z$ and corresponding opposite angles $X, Y, Z$ . It is quite well known from a little trigonometry of triangles that $$x=y\cos Z +z\cos Y$$ $$y=x\cos Z +z\cos X$$ $$z=y\cos X +x\cos Y$$

And from sine rule we have $$\frac {x}{\sin X}=\frac {y}{\sin Y}=\frac {z}{\sin Z}$$

Now in your given equations just substitute $a=\cos X$,$b=\cos Y$,$c=\cos Z$

On doing this the given three equations transform to the three equations I have given above. While the equation you need to prove may be simply written as the Sine rule. I think this clarification might get you visualised what I really mean in my answer.

Answer 2

HINT:

Here you have a linear system of 3 equations and 3 unknowns. This is known as a balanced linear system. To verify that your solutions are correct you can either

(1) Solve the system of equations to derive the values given (Gaussian Elimination, Craymers Rule, LU Decomposition, etc etc (there are many ways)).

(2) Substitute in your sols of $x,y,z$ into each of your equations to see that they are solutions

Or

(3) Give us some background on your mathematical toolkit so we can suggest a method that can be used to solve this system of equations.

Answer 3

Hint:

By $(1),(2)$ $$0=x-cy-bz+c(-cx+y-az)=x(1-c^2)-z(b+ca)$$

By $(2),(3)$ $$0=a(-cx+y-az)-bx-ay+z=z(1-a^2)-x(ca+b)$$

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

What if $b+ca\ne0$