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JEE Main Math Class 12 Three Dimensional Geometry Part 1 Questions
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© examsnet.com
Question : 28
Total: 100
The shortest distance between the lines
x
2
=
y
2
=
z
1
and
x
+
2
−
1
=
y
−
4
8
=
z
−
5
4
lies in the interval:
[Online April 9, 2016]
(3, 4]
(2, 3]
[1, 2)
[0, 1)
Validate
Solution:
Shortest distance between two lines
x
−
x
1
a
1
=
y
−
y
1
b
1
=
z
−
z
1
c
1
and
x
−
x
2
a
2
=
y
−
y
2
b
2
=
z
−
z
2
c
2
is given by,
|
|
x
2
−
x
1
y
2
−
y
1
z
2
−
z
1
a
1
b
1
c
1
a
2
b
2
c
2
|
√
(
b
1
c
2
−
b
2
c
1
)
2
+
(
c
1
a
2
−
c
2
a
1
)
2
+
(
a
1
b
2
−
a
2
b
1
)
2
|
∴
The shortest distance between given lines are
|
|
−
2
4
5
2
2
1
−
1
8
4
|
√
(
8
−
8
)
2
+
(
−
1
−
8
)
2
+
(
16
+
2
)
2
|
=
|
0
−
36
+
90
√
405
|
=
54
20.1
=
2.68
© examsnet.com
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