If #x/a=y/b=z/c#; calculate: #(ax+by+cz)/(a^2+b^2+c^2)- (x^2+y^2+z^2)/(ax+by+cz)# ?

If #x/a=y/b=z/c#; calculate: #(ax+by+cz)/(a^2+b^2+c^2)- (x^2+y^2+z^2)/(ax+by+cz)#

2 Answers

zero

Explanation:

#y = (bx)/a ; z = (cx)/a#

#Psi = frac{ax + (b^2x)/a + (c^2x)/a}{a^2 + b^2 + c^2} - frac{x^2 + (b^2x^2)/a^2 + (c^2x^2)/a^2}{ax + (b^2x)/a + (c^2x)/a}#

#Psi = frac{a^2x + b^2x + c^2x}{a(a^2 + b^2 + c^2)} - frac{a^2x^2 + b^2x^2 + c^2x^2}{a^3x + ab^2x + ac^2x}#

#Psi = frac{x(a^2 + b^2 + c^2)}{a(a^2 + b^2 + c^2)} - frac{x^2(a^2 + b^2 + c^2)}{ax(a^2 + b^2 + c^2)}#

#Psi = x/a - x/a = 0#

Jun 19, 2018

# 0#.

Explanation:

Prerequisite :

#"If "x/a=y/b=z/c," then Each Ratio"=(lx+my+nz)/(la+mb+nc), i.e.,#

#x/a=y/b=z/c=(lx+my+nz)/(la+mb+nc)............(star)#.

Using #(star)" with "l=a, m=b, n=c#, we have,

#"Each Ratio"=(ax+by+cz)/(a^2+b^2+c^2).........(star_1)#.

Reusing #(star)#, but now with #l=x, m=y, n=z#, we get,

#"Each Ratio"=(x^2+y^2+z^2)/(ax+by+cz)...........(star_2)#.

#:." By "(star_1) and (star_2)#

#(ax+by+cz)/(a^2+b^2+c^2)=(x^2+y^2+z^2)/(ax+by+cz)#.

#rArr (ax+by+cz)/(a^2+b^2+c^2)-(x^2+y^2+z^2)/(ax+by+cz)=0#,

as Respected Vinícius Ferraz has already derived!