Question

If bcx=cay=abz, then proof that (ax+by)/(a²+b²)=(by+cz)/(b²+c²)=(cz+ax)/(c²+a²).


please solve this..​

Answer 1

Answer:

Given, ax+by+cz=0→(1)

bcx+cay+abz=0→(2)

xyz+abc(a3x+b3y+c3z)=0→(3)

(1)×bc−(2)×a⇒abcx+b2cy+bc2z−abcx−ca2y−a2b2=0

c(b2−a2)y=b(a2−c2)z

(1)×ac−(2)×b⇒a2cx+abcy+ac2z−b2cx−abcy−ab2z=0

c(a2−b2)x=a(b2−c2)z

(1)×ab−(2)×c⇒a2bx+ab2y+Given, ax+by+cz=0→(1)

bcx+cay+abz=0→(2)

xyz+abc(a3x+b3y+c3z)=0→(3)

(1)×bc−(2)×a⇒abcx+b2cy+bc2z−abcx−ca2y−a2b2=0

c(b2−a2)y=b(a2−c2)z

(1)×ac−(2)×b⇒a2cx+abcy+ac2z−b2cx−abcy−ab2z=0

c(a2−b2)x=a(b2−c2)z

(1)×ab−(2)×c⇒a2bx+ab2y+