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Test Index
JEE Main Math Class 12 Three Dimensional Geometry Part 3 Questions
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© examsnet.com
Question : 3
Total: 134
Let
P
be the point of intersection of the lines
x
−
2
1
=
y
−
4
5
=
z
−
2
1
and
x
−
3
2
=
y
−
2
3
=
z
−
3
2
.
Then, the shortest distance of
P
from the line
4
x
=
2
y
=
z
is
[4 Apr 2024 Shift 2]
5
√
14
7
√
14
7
3
√
14
7
6
√
14
7
Validate
Solution:
👈: Video Solution
L
1
≡
x
−
2
1
=
y
−
4
5
=
z
−
2
1
=
λ
P
(
λ
+
2
,
5
λ
+
4
,
λ
+
2
)
L
2
≡
x
−
3
2
=
y
−
2
3
=
z
−
3
2
P
(
2
µ
+
3
,
3
µ
+
2
,
2
µ
+
3
)
λ
+
2
=
2
µ
+
3
3
µ
+
2
=
5
λ
+
4
λ
=
2
µ
+
1
3
µ
=
5
λ
+
2
3
µ
=
5
(
2
µ
+
1
)
+
2
3
µ
=
10
µ
+
7
µ
=
−
1
λ
=
−
1
Both satisfies (P)
P
(
1
,
−
1
,
1
)
L
3
≡
x
1
∕
4
=
y
1
∕
2
=
z
1
L
3
=
x
1
=
y
2
=
z
4
=
k
Coordinates of
Q
(
k
,
2
k
,
4
k
)
DR
's of
PQ
=
⟨
k
−
1
,
2
k
+
1
,
4
k
−
1
>
PQ
⟂
to
L
3
(
k
−
1
)
+
2
(
2
k
+
1
)
+
4
(
4
k
−
1
)
=
0
k
−
1
+
4
k
+
2
+
16
k
−
4
=
0
k
=
1
7
Q
(
1
7
,
2
7
,
4
7
)
PQ
=
√
(
1
−
1
7
)
2
+
(
−
1
−
2
7
)
2
+
(
1
−
4
7
)
2
=
√
36
49
+
81
49
+
9
49
=
√
126
7
PQ
=
3
√
14
7
Option-3 will satisfy
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