Vector Algebra — Class 12 Maths Solution

Let $OP$ lie in $XY -$plane so that$\angle XOP = {30^o},\angle POY = {60^o}$ and $\angle POZ = {90^o}$.

Direction cosines of $OP$ are $< \cos {30^o},\cos {60^o},\cos {90^o} >$

i.e., $< \cfrac{{\sqrt 3 }}{2},\cfrac{1}{2},0 >$

$\therefore$ $\overrightarrow {OP} = \cfrac{{\sqrt 3 }}{2}\hat i + \cfrac{1}{2}\hat j$

Now, $|\overrightarrow {OP} | = \sqrt {{{\left( {\cfrac{{\sqrt 3 }}{2}} \right)}^2} + {{\left( {\cfrac{1}{2}} \right)}^2}} = \sqrt {\cfrac{3}{4} + \cfrac{1}{4}} = \sqrt 1 = 1$

Hence the required vector is $\cfrac{{\sqrt 3 }}{2}\hat i + \cfrac{1}{2}\hat j$