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NDA Math Complex Numbers
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© examsnet.com
Question : 42
Total: 66
Suppose ω is a cube root of unity with
ω
≠
1
.Suppose P and q are on the complex plane defined by ω and
ω
2
.If O is the origin,ten what is the angle between OP and OQ ?
[NDA I 2016]
60°
90°
120°
150°
Validate
Solution:
We have, points
P
and
Q
on complex plane which are
ω
and
ω
2
We know
ω
=
−
1
2
+
i
√
3
2
and
ω
2
=
−
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
,
ω
2
=
−
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
(
m
1
)
=
√
3
2
−
0
−
1
2
−
0
=
−
√
3
.
.
.
(
.
slope
(
m
)
=
y
2
−
y
1
x
2
−
x
1
)
Slope of
OP
(
m
2
)
=
−
√
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,
⇒
boldsymbol
θ
=
tan
−
1
(
m
2
−
m
1
1
+
m
1
m
2
)
⇒
θ
=
tan
−
1
(
√
3
−
(
−
√
3
)
1
+
√
3
(
−
√
3
)
)
⇒
θ
=
tan
−
1
(
2
√
3
1
−
3
)
⇒
θ
=
tan
−
1
(
−
√
3
)
∴
θ
=
120
∘
© examsnet.com
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