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TS EAMCET 11-Sep-2020 Shift 1 Solved Paper
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Section:
Mathematics
© examsiri.com
Question :
30 of 160
Marks:
+1
,
-0
If the point of intersection of the lines
−
r
=
−
i
−
6
−
j
+
(
p
s
e
c
α
)
−
k
+
t
(
−
i
+
2
−
j
+
−
k
)
and
−
r
=
4
−
j
+
−
k
+
λ
(
2
−
i
+
(
p
t
a
n
α
)
−
j
+
2
−
k
)
is
8
−
i
+
8
−
j
+
9
−
k
,(where
0
<
α
<
Ï€
2
), then
p
=
√
5
√
3
√
2
0
Validate
Solution:
It is given that the lines
r
=
∧
i
−
6
∧
j
(
p
‌
sec
‌
α
)
∧
k
+
t
(
∧
i
+
2
∧
j
+
∧
k
)
and
r
=
4
∧
j
+
∧
k
+
λ
(
2
∧
i
+
(
p
‌
tan
‌
α
)
∧
j
+
2
∧
k
)
Intersected at point
8
∧
i
+
8
∧
j
+
9
∧
k
, so
1
+
t
=
8
,
−
6
+
2
t
=
8
and
p
‌
sec
‌
α
+
t
=
9
∴
‌
‌
t
=
7
and
p
‌
sec
‌
α
=
2
...(i)
and
2
λ
=
8
,
4
+
λ
p
‌
tan
‌
α
=
8
and
1
+
2
λ
=
9
∴
‌
‌
λ
=
4
and
p
‌
tan
‌
α
=
1
...(ii)
From Eqs. (i) and (ii), we get
p
2
sec
2
α
−
p
2
tan
2
α
=
4
−
1
⇒
p
2
=
3
⇒
p
=
±
√
3
© examsiri.com
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