UPSC NDA I 2016 Mathematics Solved Paper

We have, points P and Q on complex plane which are ω and ω2
We know ω=−‌

1

2

+‌

i√3

2

and ω2=−‌

1

2

−‌

i√3

2

For point P,ω=−‌

1

2

+‌

i√3

2

, Here real part is −‌

1

2

and imaginary part is ‌

√3

2

Similarly, for point Q,ω2=−‌

1

2

−‌

i√3

2

, Here real part is −‌

1

2

and imaginary part is ‌

−√3

2

On complex plane point P and Q will look something like (shown in image) ∴P(−‌

1

2

,‌

√3

2

) and Q(−‌

1

2

,−‌

√3

2

) And origin O(0,0)
Let's find out slope of OP and slope of OQ:
Slope of OP (m1)=‌

‌ √3 2 −0

−‌ 1 2 −0

=−√3
...(. slope (m)=‌

y2−y1

x2−x1

)
Slope of OP(m2)=‌

− √3 2 −0

−‌ 1 2 −0

=√3
Now we have slope of the lines, so we can estimate the angle between the lines OP and OQ
We have,
⇒boldsymbol‌θ=tan−1(‌

m2−m1

1+m1m2

)
⇒θ=tan−1(‌

√3−(−√3)

1+√3(−√3)

)
⇒θ=tan−1(‌

2√3

1−3

)
⇒θ=tan−1(−√3)
∴θ=120∘