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JEE Main 24 June 2022 Shift 2 Solved Paper
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© examsnet.com
Question : 14
Total: 90
If the shortest distance between the lines
x
−
1
2
=
y
−
2
3
=
z
−
3
λ
and
x
−
2
1
=
y
−
4
4
=
z
−
5
5
is
1
√
3
, then the sum of all possible value of
λ
is :
[24-Jun-2022-Shift-2]
16
6
12
15
Validate
Solution:
👈: Video Solution
Let
→
a
1
=
^
i
+
2
^
j
+
3
^
k
→
a
2
=
2
^
i
+
4
^
j
+
5
^
k
→
p
=
2
^
i
+
3
^
j
+
λ
^
k
,
→
q
=
^
i
+
4
^
j
+
5
^
k
∴
→
p
×
→
q
=
(
15
−
4
λ
)
^
i
−
(
10
−
λ
)
^
j
+
5
^
k
→
a
2
−
→
a
1
=
^
i
+
2
^
j
+
2
^
k
∴
Shortest distance
=
|
(
15
−
4
λ
)
−
2
(
10
−
λ
)
+
10
√
(
15
−
4
λ
)
2
+
(
10
−
λ
)
2
+
25
|
=
1
√
3
⇒
3
(
5
−
2
λ
)
2
=
(
15
−
4
λ
)
2
+
(
10
−
λ
)
2
+
25
⇒
5
λ
2
−
80
λ
+
275
=
0
∴
Sum of values of
λ
=
80
5
=
16
© examsnet.com
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