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AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper
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© examsnet.com
Question : 33
Total: 160
If
−
a
,
−
b
,
−
c
are three non-coplanar vectors, then match the items of List-I with those of List-II.
List-I
List-II
A)
[
−
b
×
−
c
−
c
×
−
a
−
a
×
−
b
]
I)
[
−
a
−
b
−
c
]
2
B)
[
−
a
×
−
b
−
a
×
−
c
−
b
]
II) 2
[
−
a
−
b
−
c
]
C)
[
−
a
+
−
b
−
b
+
−
c
−
c
+
−
a
]
III)
[
−
a
−
b
−
c
]
D)For three mutually perpendicular unit vectors
−
a
,
−
b
,
−
c
:
[
(
−
a
+
−
b
+
−
c
)
−
b
×
−
c
−
c
×
−
a
]
=
IV)
[
−
a
−
b
−
c
]
[
−
a
.
−
b
]
V) 0
A - III , B - IV , C - V , D - II
A - IV , B - V , C - II , D - III
A - I , B - IV , C - V , D - III
A - I , B - IV , C - II , D - III
Validate
Solution:
Consider the solutions of the vector expression as,
(A)
[
b
×
c
c
×
a
a
×
b
]
=
(
b
×
c
)
×
(
c
×
a
)
×
(
a
×
b
)
Let
d
=
b
×
c
=
{
d
×
(
c
×
a
)
×
(
a
×
b
)
}
=
{
(
d
.
a
)
c
−
(
d
.
c
)
a
}
.
(
a
×
b
)
=
{
[
a
b
c
]
c
−
0
}
.
(
a
×
b
)
=
[
a
b
c
]
[
c
b
a
]
=
[
a
b
c
]
2
(
B
)
[
a
×
b
a
×
c
b
]
=
{
(
a
×
b
)
×
(
a
×
c
)
}
.
b
Let
d
=
a
×
b
=
{
d
×
(
a
×
c
)
}
.
b
=
{
(
d
.
c
)
×
a
−
(
d
.
a
)
×
c
}
.
b
=
{
(
a
×
b
)
.
c
}
a
.
b
−
{
(
a
×
b
)
.
a
}
c
.
b
=
[
a
b
c
]
[
a
.
b
]
−
0
=
[
a
b
c
]
[
a
.
b
]
(C)
[
a
+
b
b
+
c
c
+
a
]
=
(
a
+
b
)
{
(
b
+
c
)
×
(
c
+
a
)
}
=
(
a
+
b
)
.
(
b
×
c
+
b
×
a
+
c
×
c
+
c
×
a
)
=
[
a
b
c
]
+
[
b
c
a
]
=
2
[
a
b
c
]
(
D
)
a
,
b
,
c
are three mutually perpendicular unit vectors
a
.
b
=
b
.
c
=
c
.
a
=
0
[
(
a
+
b
+
c
)
b
×
c
c
×
a
]
=
(
a
+
b
+
c
)
.
(
b
×
c
)
(
c
×
a
)
=
[
c
a
b
]
=
[
a
b
c
]
Thus, option (4) is correct.
© examsnet.com
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