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JEE Advanced 2018 Paper 2
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© examsnet.com
Question : 49
Total: 54
Consider the cube in the first octant with sides OP,OQ and OR of length 1, along the x-axis, y-axis and z-axis, respectively, where O(0, 0, 0) is the origin Let
S
(
1
2
,
1
2
,
1
2
)
be the centre of the cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal OT. If
→
p
=
→
S
P
.
→
q
=
→
S
Q
,
→
r
=
→
S
R
and
→
t
=
→
S
T
, then the value of
|
(
→
p
×
→
q
)
×
(
→
r
×
→
t
)
|
is ______.
Your Answer:
Validate
Solution:
→
p
=
→
S
P
=
(
1
2
,
1
2
,
−
1
2
)
=
1
2
(
⋀
i
−
⋀
j
−
⋀
k
)
→
q
=
→
S
Q
=
(
−
1
2
,
1
2
,
−
1
2
)
=
1
2
(
−
⋀
i
+
⋀
j
−
⋀
k
)
→
r
=
→
S
R
=
(
−
1
2
,
−
1
2
,
1
2
)
=
1
2
(
−
⋀
i
−
⋀
j
+
⋀
k
)
→
t
=
→
S
T
=
(
1
2
,
1
2
,
1
2
)
=
1
2
(
⋀
i
+
⋀
j
+
⋀
k
)
|
(
→
p
×
→
q
)
×
(
→
r
×
→
t
)
|
=
1
4
=
|
⋀
i
⋀
j
⋀
k
1
−
1
−
1
−
1
1
−
1
|
×
1
4
|
⋀
i
⋀
j
⋀
k
−
1
−
1
1
1
1
1
|
=
1
16
|
(
2
⋀
i
+
2
⋀
j
)
×
(
−
2
⋀
i
+
2
⋀
j
)
|
=
|
⋀
k
2
|
=
1
2
© examsnet.com
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