If `| vec a|= a\ a n d\ | vec b|=b,` prove that `( vec a/(a^2)- vec b/(b^2))^2=(( vec a- vec b)/(a b))^2`

Explanation

Given that,
`| vec a|= a\ a n d\ | vec b|=b,`
`LHS=( vec a/(a^2)- vec b/(b^2))^2`
`=|vec a|^2/a^4-(2(vec a * vec b))/(a^2 b^2)+|vec b|^2/b^4`
`=1/a^2+1/b^2-(2(vec a * vec b))/(a^2 b^2)`


`RHS=(( vec a- vec b)/(a b))^2`
`=(( vec a- vec b)/(a b))(( vec a- vec b)/(a b))`
`=1/(a^2b^2)[|vec a|^2- 2vec a* vec b + |vec b|^2]`
`=1/(a^2b^2)[a^2+b^2-2(vec a * vec b)]`
`=1/a^2+1/b^2-(2(vec a * vec b))/(a^2 b^2)`
so,`LHS=RHS`(proved)