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JEE Main Math Class 12 Three Dimensional Geometry Part 3 Questions
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© examsnet.com
Question : 50
Total: 134
If the lines
x
−
1
2
=
2
−
y
−
3
=
z
−
3
α
and
x
−
4
5
=
y
−
1
2
=
z
β
intersect, then the magnitude of the minimum value of
8
α
β
is _______.
[6-Apr-2023 shift 2]
Your Answer:
Validate
Solution:
If the lines
x
−
1
2
=
2
−
y
−
3
=
z
−
3
α
and
x
−
4
5
=
y
−
1
2
=
z
β
intersect Point of first line
(
1
,
2
,
3
)
and point on second line
(
4
,
1
,
0
)
.
Vector joining both points is
−
3
^
i
+
^
j
+
3
k
Now vector along second line is
2
^
i
+
3
^
j
+
α
k
Also vector along second line is
5
^
i
+
2
^
j
+
β
k
Now these three vectors must be coplanar
⇒
|
2
3
α
5
2
β
−
3
1
3
|
⇒
2
(
6
−
β
)
−
3
(
15
+
3
β
)
+
α
(
11
)
=
0
⇒
α
−
β
=
3
Now
α
=
3
+
β
Given expression
8
(
3
+
β
)
⋅
β
=
8
(
β
2
+
3
β
)
=
8
(
β
2
+
3
β
+
9
4
−
9
4
)
=
8
(
β
+
3
2
)
2
−
18
So magnitude of minimum value
=
18
© examsnet.com
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