Question

Eliminate x, y, z from the equations
$a_{1}x+b_{1}y+c_{1}z=0...\left(1\right),$
$a_{2}x+b_{2}y+c_{2}z=0...\left(2\right),$
$a_{3}x+b_{3}y+c_{3}z=0...\left(3\right)$

Answer 1

Given equations are

$a_{1}x+b_{1}y+c_{1}z=0...\left(1\right),$
$a_{2}x+b_{2}y+c_{2}z=0...\left(2\right),$

$a_{3}x+b_{3}y+c_{3}z=0...\left(3\right),$

From (2) and (3), by cross multiplication

$\frac{x}{b_2{c}_3-c_2{b}_3}=\frac{y}{c_2{a}_3-c_3{a}_2}=\frac{z}{a_2{b}_3-a_3{b}_2}=k$

$\Rightarrow x=(b_2{c}_3-c_2{b}_3)k$, $y=(c_2{a}_3-a_2{c}_3)k$ and $z=(a_2{b}_3-b_2{a}_3)k$

Substituting in $\left(1\right)$ and dividing out by $k,$ we obtain

$a_1(b_2{c}_3-b_3{c}_2)+b_1(c_2{a}_3-c_3{a}_2)+c_1(a_2{b}_3-a_3{b}_2)=0.$

This relation is called the eliminant of the given equations.