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JEE Main Math Class 12 Three Dimensional Geometry Part 2 Questions
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© examsnet.com
Question : 99
Total: 100
The perpendicular distance from the origin to the plane containing the two lines,
x
+
2
3
=
y
−
2
5
=
z
+
5
7
and
x
−
1
1
=
y
−
4
4
=
z
+
4
7
,
is :
[Jan. 12, 2019 (I)]
11
√
6
11
∕
√
6
11
6
√
11
Validate
Solution:
∵
plane containing both lines.
D.R. of plane
=
|
^
i
^
j
^
k
3
5
7
1
4
7
|
=
7
^
i
−
14
^
j
+
7
^
k
Now, equation of plane is,
7
(
x
−
1
)
−
14
(
y
−
4
)
+
7
(
z
+
4
)
=
0
⇒
x
−
1
−
2
y
+
8
+
z
+
4
=
0
⇒
x
−
2
y
+
z
+
11
=
0
Hence, distance from (0,0,0) to the plane,
=
11
√
1
+
4
+
1
=
11
√
6
© examsnet.com
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