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JEE Main 9 Apr 2024 Shift 1 Solved Paper
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© examsnet.com
Question : 15
Total: 90
The shortest distance between the line
x
−
3
4
=
y
+
7
−
11
=
z
−
1
5
and
x
−
5
3
=
y
−
9
−
6
=
z
+
2
1
is:
[9 Apr 2024 Shift 1]
187
√
563
178
√
563
185
√
563
179
√
563
Validate
Solution:
👈: Video Solution
→
n
=
→
p
×
→
q
→
n
=
|
^
i
^
j
^
k
4
−
11
5
3
−
6
1
|
=
19
^
i
+
11
^
j
+
9
^
k
S.d.
=
projection of
→
AB
on
→
n
=
|
→
AB
⋅
→
n
|
→
n
|
|
=
|
(
2
^
i
+
16
^
j
−
3
^
k
)
⋅
(
19
^
i
+
11
^
j
+
9
^
k
)
∣
√
361
+
121
+
81
|
=
38
+
176
−
27
√
563
S.d.
=
187
√
563
© examsnet.com
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