Explanation
A zero vector can never be a unit vector
$$\therefore $$ option $$A$$ is not correct
Let as assume four unit vectors in the $$xy$$ plane as shown in the figure
$$\underset { { V }_{ 1 } }{ \rightarrow } =\dfrac { i }{ \sqrt { 2 } } +\dfrac { j }{ \sqrt { 2 } } \\ \underset { { V }_{ 2 } }{ \rightarrow } =-\dfrac { i }{ \sqrt { 2 } } +\dfrac { j }{ \sqrt { 2 } } \\ \underset { { V }_{ 3 } }{ \rightarrow } =-\dfrac { i }{ \sqrt { 2 } } -\dfrac { j }{ \sqrt { 2 } } \\ \underset { { V }_{ 4 } }{ \rightarrow } =\dfrac { i }{ \sqrt { 2 } } -\dfrac { j }{ \sqrt { 2 } } $$
If we add up any of the two vectors it will either lies on the $$x$$ aisx or the $$y$$ axis, none of them lies in the first quardrant
$$\therefore $$ option $$B$$ is not correct
Sum of any two vectors less than or greater than zero does not make any sense
$$\therefore $$ options $$C$$ and $$D$$ are incorrect
So none of the options are correct.
Scalar projection of $$\vec { a } $$ on $$\vec { b } $$ $$=\dfrac { \left| \vec { a } .\vec { b } \right| }{ \left| \vec { b } \right| } $$
let $$ \vec { a } \ =xi-j+k$$ and $$\vec { b } =2i-j+5k$$
Then projection of $$\vec { a } $$ on $$\vec { b } $$ $$=\dfrac { (xi-j+k).(2i-j+5k) }{ \left| 2i-j+5k \right| } $$
Given scalar projection $$=\dfrac { 1 }{ \sqrt { 30 } } $$
$$\Rightarrow \dfrac { 1 }{ \sqrt { 30 } } =\dfrac { 2x+1+5 }{ \sqrt { { 2 }^{ 2 }+{ (-1) }^{ 2 }+{ 5 }^{ 2 } } } \\ \Rightarrow \dfrac { 1 }{ \sqrt { 30 } } =\dfrac { 2x+1+5 }{ \sqrt { 30 } } \\ \Rightarrow 2x+1+5=1\\ \Rightarrow x=-\dfrac { 5 }{ 2 } $$
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