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:




How to prove the required fact?

Any way to prove it ?

  • 1
    $\begingroup$ You should also not that in the event of linear independence $x,y,z = 0$. Your solution profile does meet that criterion. To repeat my early question, what is known about $a,b,c$?? $\endgroup$ – user150203 Dec 25 '18 at 3:22
  • 1
    $\begingroup$ Here you have a balanced system of linear equations of $x,y,z$. There is a wide spectrum of methods to take your linear system and solve for $x,y.z$. What methods have you been working with at present? Also, there hasn't been any conditions put on $a,b,c$. Here we see that if $a,b,c = \pm 1$ that $x,y,z$ do not exist. $\endgroup$ – user150203 Dec 25 '18 at 3:24
  • $\begingroup$ @DavidG,only thing known is $x,y,z\neq \pm 1$ $\endgroup$ – user596656 Dec 25 '18 at 3:24
  • $\begingroup$ Ok, there are a number of ways you can approach this. Have you learnt about solving Linear Systems using Gaussian Elimination? en.wikipedia.org/wiki/Gaussian_elimination $\endgroup$ – user150203 Dec 25 '18 at 3:25
  • $\begingroup$ @DavidG,it wont help $\endgroup$ – user596656 Dec 25 '18 at 3:32


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 $a, b, c$


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.

  • $\begingroup$ I am sorry,I am not getting it really $\endgroup$ – user596656 Dec 25 '18 at 3:33
  • $\begingroup$ How can I solve this from above $\endgroup$ – user596656 Dec 25 '18 at 3:34
  • $\begingroup$ Digamma correctly notes that $$\sin x =\sqrt{1-\cos^2(x)}$$ $\endgroup$ – Rhys Hughes Dec 25 '18 at 3:36
  • $\begingroup$ @RhysHughes;if that is the solution,why are we given three equations $\endgroup$ – user596656 Dec 25 '18 at 3:45
  • $\begingroup$ If we assume that they are the sides of the triiangle then there is nothing to prove and no use of the equations $\endgroup$ – user596656 Dec 25 '18 at 3:46


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

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


What if $b+ca\ne0$



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


(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.

  • $\begingroup$ I have already told you that I know all the tools that you mentioned but the problem is the required solution is not coming,I dont know what background you are talking about,I have completed my undergraduate degree in Maths $\endgroup$ – user596656 Dec 25 '18 at 3:57
  • $\begingroup$ Using Cramers Rule i got 0 as the solution,Gauss is not giving too,are there any other ways $\endgroup$ – user596656 Dec 25 '18 at 3:58
  • $\begingroup$ If possible do give a way by which you find this solution can be arrived at $\endgroup$ – user596656 Dec 25 '18 at 3:58
  • $\begingroup$ If Gaussian Elimination is not giving you the given values then they are either wrong OR your use of Gaussian Elimination is wrong. $\endgroup$ – user150203 Dec 25 '18 at 3:59
  • $\begingroup$ Btw, $x,y,z = 0$ is a solution that matches the form given... $\endgroup$ – user150203 Dec 25 '18 at 4:01

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