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JEE Main Math Class 12 Three Dimensional Geometry Part 3 Questions
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© examsnet.com
Question : 8
Total: 134
Let the point
(
−
1
,
α
,
β
)
lie on the line of the shortest distance between the lines
x
+
2
−
3
=
y
−
2
4
=
z
−
5
2
and
x
+
2
−
1
=
y
+
6
2
=
z
−
1
0
. Then
(
α
−
β
)
2
is equal to ______.
[5 Apr 2024 Shift 2]
Your Answer:
Validate
Solution:
👈: Video Solution
P
(
−
3
λ
−
2
,
4
λ
+
2
,
2
λ
+
5
)
Q
(
−
µ
−
2
,
2
µ
−
6
,
1
)
DRS
of
PQ
=
(
3
λ
−
µ
,
2
µ
−
4
λ
−
8
,
−
2
λ
−
4
)
D
R
S
of
P
Q
=
|
^
i
^
j
^
k
−
1
2
0
−
3
4
2
|
=
(
4
^
i
+
2
^
j
+
2
^
k
)
OR
(
2
,
1
,
1
)
3
λ
−
µ
2
=
2
µ
−
4
λ
−
8
1
=
−
2
λ
−
4
1
⇒
µ
=
λ
+
2
&
7
λ
=
µ
−
8
λ
=
−
1
µ
=
1
Q
:
(
−
3
,
−
4
,
1
)
L
PQ
=
x
+
3
2
=
y
+
4
1
=
z
−
1
1
(
−
1
,
α
,
β
)
⇒
1
=
α
+
4
1
=
β
−
1
1
⇒
α
=
−
3
,
β
=
2
(
α
−
β
)
2
=
25
© examsnet.com
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