Let $n =1$ then,
if $a$ $=$ $\left \{ 2 \right \}$, $b$ $=$ $\left \{ 1 \right \}$
< $a,b$ > $=$ $2$
$||$ $b$ $||$ $=$ $1$
Hence, i is incorrect.
Let a = {-1} , b ={1}
$< a,b > = -1$
$||$ $a$ $||$ $||$ $b$ $||$ $=$ $1$ x $1$ = $1$
Hence, iii and iv is incorrect, if iv is incorrect $d$ and $e$ both can't be right.
So, by elimination $(a)$ is correct.