If $straight a with rightwards arrow on top$ is a non-zero vector of magnitude ‘a’ and λ a non-zero scalar, then $straight lambda straight a with rightwards arrow on top$ is a unit vector if

$straight lambda space equals space 1$
$straight lambda equals negative 1$
$straight a space equals space open vertical bar straight lambda close vertical bar$
$straight a space equals fraction numerator 1 over denominator open vertical bar straight lambda close vertical bar end fraction$

Solution

straight a space equals fraction numerator 1 over denominator open vertical bar straight lambda close vertical bar end fraction

Since $straight a with rightwards arrow on top$ is a non-zero vector of magnitude a
$therefore space space space space space open vertical bar straight a with rightwards arrow on top close vertical bar space equals space straight a$
Now, $straight lambda space straight a with rightwards arrow on top$ is a unit vector if $open vertical bar straight lambda straight a with rightwards arrow on top close vertical bar space equals space 1$
i.e. if $open vertical bar straight lambda close vertical bar space open vertical bar straight a with rightwards arrow on top close vertical bar space equals 1$
i.e. if $open vertical bar straight lambda close vertical bar space straight a space equals space 1$
i.e. if $straight a space equals space fraction numerator 1 over denominator open vertical bar straight lambda close vertical bar end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis straight lambda space not equal to space 0 right parenthesis$
$therefore space space space left parenthesis straight D right parenthesis space is space correct space answer.$