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KEAM 2010 Math Paper
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© examsnet.com
Question : 11
Total: 120
If the scalar product of the vector
∧
i
+
∧
j
+
2
∧
k
with the unit vector along
m
∧
i
+
2
∧
j
+
3
∧
k
is equal to
2
,
then one of the values of
m
is
3
4
5
6
7
Validate
Solution:
The unit vector along
m
∧
i
+
2
∧
j
+
3
∧
k
is
m
∧
i
+
2
∧
j
+
3
∧
k
√
m
2
+
4
+
9
=
m
∧
i
+
2
∧
j
+
3
∧
k
√
m
2
+
13
Now,
(
∧
i
+
∧
j
+
2
∧
k
)
⋅
(
m
∧
i
+
2
∧
j
+
3
∧
k
√
m
2
+
13
)
=
2
⇒
m
√
m
2
+
13
+
2
√
m
2
+
13
+
6
√
m
2
+
13
=
2
⇒
m
+
8
√
m
2
+
13
=
2
Squaring on both sides, we get
⇒
(
m
+
8
)
2
=
4
(
m
2
+
13
)
⇒
m
2
+
16
m
+
64
=
4
m
2
+
52
⇒
3
m
2
−
16
m
−
12
=
0
⇒
m
=
16
±
√
256
+
144
6
=
16
±
20
6
⇒
m
=
6
,
−
2
3
⇒
m
=
6
as
−
2
3
is not possible.
© examsnet.com
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