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Byju's Answer
Standard XII
Mathematics
Distance Formula
If the line, ...
Question
If the line,
x
−
3
1
=
y
+
2
−
1
=
z
+
λ
−
2
lies on the plane
2
x
−
4
y
+
3
z
=
2
,
then the shortest distance between this line
and the line
x
−
1
12
=
y
9
=
z
4
is
A
0
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B
2
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C
1
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D
3
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Solution
The correct option is
C
0
x
−
3
1
=
y
+
2
−
1
=
z
+
λ
−
2
=
k
Let P be any point on the line,
P
=
(
k
+
3
,
−
k
−
2
,
−
2
k
−
λ
)
P lies on the plane
2
x
−
4
y
+
3
z
=
2
2
(
k
+
3
)
−
4
(
−
k
−
2
)
+
3
(
−
2
k
−
λ
)
=
2
2
k
+
6
+
4
k
+
8
−
6
k
−
3
λ
=
2
14
−
3
λ
=
2
3
λ
=
12
λ
=
4
∴
Line
1
is
x
−
3
1
=
y
+
2
−
1
=
z
+
4
−
2
Another line is
x
−
1
12
=
y
9
=
z
4
(line
2
)
Shortest distance be d.
d
=
∣
∣ ∣ ∣
∣
x
1
−
x
2
y
1
−
y
2
z
1
−
z
2
a
1
b
1
c
1
a
2
b
2
c
2
∣
∣ ∣ ∣
∣
√
∑
(
a
1
b
2
−
a
2
b
1
)
2
where
a
1
,
b
1
,
c
1
are d.r's of lines
→
∣
∣ ∣
∣
3
−
1
−
2
−
0
−
4
−
0
1
−
1
−
2
12
9
4
∣
∣ ∣
∣
=
∣
∣ ∣
∣
2
−
2
−
4
1
−
1
−
2
12
9
4
∣
∣ ∣
∣
=
|
2
(
14
)
+
2
(
28
)
−
4
(
21
)
|
=
|
28
+
56
−
84
|
=
0
∴
d
=
0
⇒
d
=
0
.
Suggest Corrections
0
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