JEE Main 26-Aug-2021 Shift 1 Solved Paper

In the given figure, the magnitude of vectors OA, OB and OC areequal. Let it be m.
From figure, we can write   OA=m(cos‌30°‌

^

i

+sin‌30°‌

^

j

) =m(

√3

2

^

i

+

1

2

^

j

)
OB=m(cos‌60°‌

^

i

−sin‌60°‌

^

j

) =m(

1

2

^

i

−

√3

2

^

j

)
OC=m(cos‌45°(−

^

i

)+sin‌45°‌

^

j

) =m{

−1

√2

^

i

+

1

√2

^

j

} −m(−

1

√2

^

i

+

1

√2

^

j

)
=m

^

i

(

√3

2

+

1

2

+

1

√2

)+m

^

j

(

1

2

−

√3

2

−

1

√2

)
∴ Angle with X - axis is given by
tan‌θ=

y

x

=

m( 1 2 − √3 2 − 1 √2 )

m( √3 2 + 1 2 + 1 √2 )

⇒tan‌θ=

1 2 − √3 2 − √2 2

√3 2 + 1 2 + √2 2

⇒tan‌θ=

1−√3−√2

√3+1+√2

⇒θ=tan−1(

1−√3−√2

1+√3+√2

)