Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 7612 by Tawakalitu. last updated on 06/Sep/16

If  ax + by + cz = 0,   and,  a^2 x + b^2 y + c^2 z = 0  Find the ratio  x:y:z

$${If}\:\:{ax}\:+\:{by}\:+\:{cz}\:=\:\mathrm{0},\: \\ $$$${and},\:\:{a}^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} {y}\:+\:{c}^{\mathrm{2}} {z}\:=\:\mathrm{0} \\ $$$${Find}\:{the}\:{ratio}\:\:{x}:{y}:{z} \\ $$

Commented by Rasheed Soomro last updated on 06/Sep/16

ax + by + cz = 0.................(i)   a^2 x + b^2 y + c^2 z = 0.............(ii)  x=((−by−cz)/a)  x=((−b^2 y−c^2 z)/a^2 )  ((−by−cz)/a)=((−b^2 y−c^2 z)/a^2 )  −aby−acz=−b^2 y−c^2 z  b^2 y−aby=−c^2 z+acz  b(b−a)y=c(a−c)z  (y/z)=((c(a−c))/(b(b−a)))  y : z = c(a−c) : b(b−a)  ..........(iii)  Again from (i) & (ii)   z=((−ax−by)/c)  z=(( −a^2 x −b^2 y)/c^2 )  ((−ax−by)/c)=(( −a^2 x −b^2 y)/c^2 )  −acx−bcy=−a^2 x −b^2 y  a^2 x−acx=bcy−b^2 y  a(a−c)x=b(c−b)y  (x/y)=((b(c−b))/(a(a−c)))  x : y = b(c−b) : a(a−c)...................(iv)  By combining (iv) & (iii)   ((x,:,y,:,z),((b(c−b)),:,(a(a−c)),,),(,,(c(a−c)),:,(b(b−a))),((bc(c−b)),:,(ca(a−c)),:,(ab(b−a))) )

$${ax}\:+\:{by}\:+\:{cz}\:=\:\mathrm{0}.................\left({i}\right) \\ $$$$\:{a}^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} {y}\:+\:{c}^{\mathrm{2}} {z}\:=\:\mathrm{0}.............\left({ii}\right) \\ $$$${x}=\frac{−{by}−{cz}}{{a}} \\ $$$${x}=\frac{−{b}^{\mathrm{2}} {y}−{c}^{\mathrm{2}} {z}}{{a}^{\mathrm{2}} } \\ $$$$\frac{−{by}−{cz}}{{a}}=\frac{−{b}^{\mathrm{2}} {y}−{c}^{\mathrm{2}} {z}}{{a}^{\mathrm{2}} } \\ $$$$−{aby}−{acz}=−{b}^{\mathrm{2}} {y}−{c}^{\mathrm{2}} {z} \\ $$$${b}^{\mathrm{2}} {y}−{aby}=−{c}^{\mathrm{2}} {z}+{acz} \\ $$$${b}\left({b}−{a}\right){y}={c}\left({a}−{c}\right){z} \\ $$$$\frac{{y}}{{z}}=\frac{{c}\left({a}−{c}\right)}{{b}\left({b}−{a}\right)} \\ $$$${y}\::\:{z}\:=\:{c}\left({a}−{c}\right)\::\:{b}\left({b}−{a}\right)\:\:..........\left({iii}\right) \\ $$$${Again}\:{from}\:\left({i}\right)\:\&\:\left({ii}\right) \\ $$$$\:{z}=\frac{−{ax}−{by}}{{c}} \\ $$$${z}=\frac{\:−{a}^{\mathrm{2}} {x}\:−{b}^{\mathrm{2}} {y}}{{c}^{\mathrm{2}} } \\ $$$$\frac{−{ax}−{by}}{{c}}=\frac{\:−{a}^{\mathrm{2}} {x}\:−{b}^{\mathrm{2}} {y}}{{c}^{\mathrm{2}} } \\ $$$$−{acx}−{bcy}=−{a}^{\mathrm{2}} {x}\:−{b}^{\mathrm{2}} {y} \\ $$$${a}^{\mathrm{2}} {x}−{acx}={bcy}−{b}^{\mathrm{2}} {y} \\ $$$${a}\left({a}−{c}\right){x}={b}\left({c}−{b}\right){y} \\ $$$$\frac{{x}}{{y}}=\frac{{b}\left({c}−{b}\right)}{{a}\left({a}−{c}\right)} \\ $$$${x}\::\:{y}\:=\:{b}\left({c}−{b}\right)\::\:{a}\left({a}−{c}\right)...................\left({iv}\right) \\ $$$${By}\:{combining}\:\left({iv}\right)\:\&\:\left({iii}\right) \\ $$$$\begin{pmatrix}{{x}}&{:}&{{y}}&{:}&{{z}}\\{{b}\left({c}−{b}\right)}&{:}&{{a}\left({a}−{c}\right)}&{}&{}\\{}&{}&{{c}\left({a}−{c}\right)}&{:}&{{b}\left({b}−{a}\right)}\\{{bc}\left({c}−{b}\right)}&{:}&{{ca}\left({a}−{c}\right)}&{:}&{{ab}\left({b}−{a}\right)}\end{pmatrix} \\ $$

Commented by Tawakalitu. last updated on 06/Sep/16

Wow, thank you so much sir.

$${Wow},\:{thank}\:{you}\:{so}\:{much}\:{sir}. \\ $$

Commented by Tawakalitu. last updated on 06/Sep/16

Thanks so much sir. God bless you

$${Thanks}\:{so}\:{much}\:{sir}.\:{God}\:{bless}\:{you} \\ $$

Answered by Yozzia last updated on 06/Sep/16

Answer in comments.

$${Answer}\:{in}\:{comments}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com