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NIMCET 2008 Question Paper with solutions
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© examsnet.com
Question : 30
Total: 120
Let
A
=
2
i
+
j
−
2
k
and
B
=
i
+
j
.
If
C
is a vector such that
A
⋅
C
=
|
C
|
,
|
C
−
A
|
=
2
√
2
and the angle between
A
×
B
and
C
is
30
∘
then
|
(
A
×
B
)
×
C
|
is equal to
2
3
3
2
2
3
Validate
Solution:
Given,
|
C
−
A
|
=
2
√
2
⇒
∣
C
−
A
2
∣
=
8
⇒
C
2
+
|
A
|
2
−
2
C
⋅
A
=
8
(
∵
C
⋅
A
=
|
C
|
=
C
)
⇒
C
2
+
9
−
2
C
=
8
⇒
C
2
−
2
C
+
1
=
0
⇒
(
C
−
1
)
2
=
0
C
=
1
⇒
|
C
|
=
1
Now,
|
(
A
×
B
)
×
C
|
=
|
|
A
×
B
|
|
C
|
sin
30
∘
⋀
n
|
=
1
2
|
A
×
B
|
....(I)
A
×
B
=
|
i
j
k
2
1
−
2
1
1
0
|
=
i
(
0
+
2
)
−
j
(
0
+
2
)
+
k
(
2
−
1
)
=
2
i
−
2
j
+
k
⇒
|
A
×
B
|
=
√
4
+
4
+
1
=
3
....(ii)
From Eqs. (i) and (ii). we get
|
(
A
×
B
)
×
C
|
=
3
2
© examsnet.com
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