Question

If $\dfrac{{ay - bx}}{c} = \dfrac{{cx - az}}{b} = \dfrac{{bz - cy}}{a}$ , then how to prove that $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$ ?

Answer 1

Hint : In the question, we are given an equation that provides us with a relation between six variables and we prove a given expression such that the ratio of the variables in groups of two are equal. Therefore, we are we define an arbitrary constant k such that $\dfrac{{ay - bx}}{c} = \dfrac{{cx - az}}{b} = \dfrac{{bz - cy}}{a} = k$ so that to obtain three equations in three variables from the condition given to us and simplify them to prove the required equation.

Complete step-by-step answer :
So, in the given question, we are required to prove the condition given to us. So, we first assign a new arbitrary constant such that $\dfrac{{ay - bx}}{c} = \dfrac{{cx - az}}{b} = \dfrac{{bz - cy}}{a} = k$ .
So, $\dfrac{{ay - bx}}{c} = k$
$\Rightarrow ay - bx = ck - - - - - - \left( 1 \right)$
Now, $\dfrac{{cx - az}}{b} = k$
$\Rightarrow cx - az = bk - - - - - - \left( 2 \right)$
Also, $\dfrac{{bz - cy}}{a} = k$
$\Rightarrow bz - cy = ka - - - - - - \left( 3 \right)$
Now, we have three linear equations in three variables. So, we can solve these easily and prove the required result.
Now, multiplying the equation $\left( 1 \right)$ by c and equation $\left( 3 \right)$ by a, we get,
$acy - bcx = {c^2}k$ and $abz - acy = k{a^2}$
Adding both, we get,
$acy - bcx + abz - acy = k{c^2} + k{a^2}$
$\Rightarrow abz - bcx = k\left( {{c^2} + {a^2}} \right) - - - - \left( 4 \right)$
So, now we have two equations in x and z. Multiplying the equation by b, we get,
$bcx - abz = k{b^2}$
Now, adding the equation above with equation $\left( 4 \right)$ , we get,
$\Rightarrow abz - bcx + bcx - abz = k{c^2} + k{a^2} + k{b^2}$
$\Rightarrow k\left( {{a^2} + {b^2} + {c^2}} \right) = 0$
So, we conclude that the value of k is zero.
Hence, we get, $\dfrac{{ay - bx}}{c} = \dfrac{{cx - az}}{b} = \dfrac{{bz - cy}}{a} = k = 0$
$ay = bx$ , $cx = az$ and $bz = cy$ .
Hence, we get, $\dfrac{y}{b} = \dfrac{x}{a}$ , $\dfrac{x}{a} = \dfrac{z}{c}$ and $\dfrac{z}{c} = \dfrac{y}{b}$ .
So, $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$
Hence, Proved.

Note : Such questions requiring us to prove an algebraic result involving variables and ratios of the same can be proved in more than one way. There can be numerous ways of solving such problems and can have innovative solutions to standard problems.