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)#
If
2 Answers
zero
Explanation:
Explanation:
Prerequisite :
Using
Reusing
as Respected VinÃcius Ferraz has already derived!