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JEE Main Math Class 12 Three Dimensional Geometry Part 3 Questions
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© examsnet.com
Question : 51
Total: 134
The shortest distance between the lines
x
−
4
4
=
y
+
2
5
=
z
+
3
3
and
x
−
1
3
=
y
−
3
4
=
z
−
4
2
is
[8-Apr-2023 shift 1]
2
√
6
3
√
6
6
√
3
6
√
2
Validate
Solution:
S
d
=
|
(
→
a
−
→
b
)
×
(
→
n
1
×
→
n
2
)
|
→
n
1
×
→
n
2
|
|
a
=
(
4
,
−
2
,
−
3
)
b
=
(
1
,
3
,
4
)
n
1
=
(
4
,
5
,
3
)
n
2
=
(
3
,
4
,
2
)
n
1
×
n
2
=
|
i
j
k
4
5
3
3
4
2
|
=
^
i
(
−
2
)
−
^
j
(
−
1
)
+
^
k
(
1
)
=
(
−
2
,
1
,
1
)
S
d
=
(
3
,
−
5
,
−
7
)
⋅
(
−
2
,
1
,
1
)
√
6
=
|
−
6
−
5
−
7
√
6
|
=
3
√
6
© examsnet.com
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