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JEE Main Math Class 12 Three Dimensional Geometry Part 3 Questions
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© examsnet.com
Question : 9
Total: 134
The shortest distance between the lines
x
−
3
2
=
y
+
15
−
7
=
z
−
9
5
and
x
+
1
2
=
y
−
1
1
=
z
−
9
−
3
is
[6 Apr 2024 Shift 1]
6
√
3
4
√
3
5
√
3
8
√
3
Validate
Solution:
👈: Video Solution
x
−
3
2
=
y
+
15
−
7
=
z
−
9
5
&
x
+
1
2
=
y
−
1
1
=
z
−
9
−
3
S.D
=
|
(
a
2
⋅
a
1
)
⋅
(
b
1
⋅
b
2
)
|
|
b
1
×
b
2
|
a
1
=
3
,
−
15
,
9
b
1
=
2
,
−
7
,
5
a
2
=
−
1
,
1
,
9
b
2
=
2
,
1
,
−
3
a
2
−
a
1
=
−
4
,
16
,
0
b
1
×
b
2
=
|
^
i
^
j
^
k
2
−
7
5
2
1
−
3
|
=
^
i
(
16
)
−
^
j
(
−
16
)
+
^
k
(
16
)
16
(
^
i
+
^
j
+
^
k
)
|
b
1
×
b
2
|
=
16
√
3
∴
(
a
2
−
a
1
)
⋅
(
b
1
−
b
2
)
=
16
[
−
4
+
16
]
=
(
16
)
(
12
)
S.D.
=
(
16
)
(
12
)
16
√
3
=
4
√
3
© examsnet.com
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