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AP EAMCET Engineering 18 Sep 2020 Shift 2 Paper
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© examsnet.com
Question : 4
Total: 160
If the magnitude of the vector product of the vector
∧
i
+
∧
j
+
∧
k
with a unit vector along the sum of the vectors
2
∧
i
+
4
∧
j
−
5
∧
k
and
λ
∧
i
+
2
∧
j
+
3
∧
k
is equal to √2, then the value of ‘ λ’ is
−
1
1
0
2
Validate
Solution:
Let
a
=
∧
i
+
∧
j
+
∧
k
b
=
2
∧
i
+
4
∧
j
−
5
∧
k
c
=
λ
∧
i
+
2
∧
j
+
3
∧
k
b
+
c
=
(
2
+
λ
)
∧
i
+
6
∧
j
−
2
∧
k
∧
d
=
unit vector along
(
b
+
c
)
=
b
+
c
|
b
+
c
|
=
(
2
+
λ
)
∧
i
+
6
∧
j
−
2
∧
k
√
(
2
+
λ
)
2
+
36
+
4
d
=
(
2
+
λ
)
∧
i
+
6
∧
j
−
2
∧
k
√
λ
2
+
4
λ
+
44
∧
a
×
∧
d
=
|
∧
i
∧
j
∧
k
1
1
1
(
2
+
λ
)
√
λ
2
+
4
λ
+
44
6
√
λ
2
+
4
λ
+
44
−
2
√
λ
2
+
4
λ
+
44
|
=
1
√
λ
2
+
4
λ
+
44
|
∧
i
∧
j
∧
k
1
1
1
2
+
λ
6
−
2
|
=
1
√
λ
2
+
4
λ
+
44
|
∧
i
(
−
8
)
−
∧
j
(
−
2
−
2
−
λ
)
+
∧
k
(
6
−
2
−
λ
)
|
∧
a
×
∧
d
=
1
√
λ
2
+
4
λ
+
44
|
−
8
∧
i
+
(
4
+
λ
)
j
+
(
4
−
λ
)
∧
k
|
|
∧
a
×
∧
d
|
=
1
√
λ
2
+
4
λ
+
44
√
64
+
(
4
+
λ
)
2
+
(
4
−
λ
)
2
√
2
=
√
64
+
16
+
λ
2
+
16
+
λ
2
√
λ
2
+
4
λ
+
44
Squaring on both sides,
2
=
2
λ
2
+
96
λ
2
+
4
λ
+
44
2
λ
2
+
8
λ
+
88
=
2
λ
2
+
96
8
λ
=
8
λ
=
1
© examsnet.com
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