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JEE Main 27-Aug-2021 Shift 1 Solved Paper
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Section:
Mathematics
© examsiri.com
Question :
81 of 90
Marks:
+1
,
-0
Let
a
=
^
i
+
5
^
j
+
α
^
k
,
b
=
^
i
+
3
^
j
+
β
^
k
and
c
=
−
^
i
+
2
^
j
−
3
^
k
be three vectors such that,
|
b
×
c
|
=
5
√
3
and
a
is perpendicular to
b
. Then, the greatest amongst the values of
|
a
|
2
is
[27 Aug 2021 Shift 1]
Your Answer:
Validate
Solution:
Given,
a
=
^
i
+
5
^
j
+
α
^
k
b
=
^
i
+
3
^
j
+
β
^
k
and
c
=
−
^
i
+
2
^
j
−
3
^
k
∵
a
⊥
b
⇒
a . b
= 0
⇒
(
^
i
+
5
^
j
+
α
^
k
)
.
(
^
i
+
3
^
j
+
β
^
k
)
=
0
⇒
1
+
15
+
α
β
=
0
or
α
β
=
−
16
...(i)
Now,
b
×
c
=
|
^
i
^
j
^
k
1
3
β
−
1
2
−
3
|
=
^
i
(
−
9
−
2
β
)
−
^
j
(
−
3
+
β
)
+
^
k
(
2
+
3
)
b
×
c
=
^
i
(
−
9
−
2
β
)
+
^
j
(
3
−
β
)
+
5
^
k
Given,
|
b
×
c
|
=
5
√
3
⇒
|
b
×
c
|
2
=
75
⇒
(
−
9
−
2
β
)
2
+
(
3
−
β
)
2
+
25
=
75
⇒
β
2
+
6
β
+
8
=
0
⇒
β
=
−
2
,
−
4
From Eq. (i), we get
For
β
=
−
2
,
α
=
8
For
β
=
−
4
,
α
=
4
For maximum value of
|
a
|
2
,
α
=
8
∴
|
α
|
2
=
1
+
25
+
64
=
90
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