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Mathematics
Let veca and vecb be the vectors along the diagonal of a parallelogram having area 2 √2. Let the angle between veca and vecb be acute. | veca|=1 and | veca ⋅ vecb|=| veca × vecb|. If vecc=2 √2( veca × vecb)-2 vecb, then an angle between vec b and vec c is :
Q. Let
a
and
b
be the vectors along the diagonal of a parallelogram having area
2
2
. Let the angle between
a
and
b
be acute.
∣
a
∣
=
1
and
∣
a
⋅
b
∣
=
∣
a
×
b
∣
. If
c
=
2
2
(
a
×
b
)
−
2
b
, then an angle between
b
and
c
is :
1881
160
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A
4
π
26%
B
−
4
π
20%
C
6
5
π
9%
D
4
3
π
46%
Solution:
Area
=
2
1
∣
a
×
b
∣
=
2
2
⇒
∣
a
×
b
∣
=
4
2
∣
a
∣
=
1
and
∣
a
⋅
b
∣
=
∣
a
×
b
∣
⇒
cos
θ
=
sin
θ
⇒
θ
=
4
π
∴
∣
a
×
b
∣
=
4
2
⇒
∣
a
∣∣
b
∣
sin
4
π
=
4
2
⇒
∣
b
∣
=
8
Now,
c
=
2
2
(
a
×
b
)
−
2
b
∣
c
∣
=
(
2
2
)
2
∣
a
×
b
∣
2
+
(
2∣
b
∣
)
2
=
16
2
Now,
b
⋅
c
=
−
2∣
b
∣
2
⇒
8
×
16
2
×
cos
α
=
−
2.64
⇒
cos
α
=
−
2
1
⇒
α
=
4
3
π