CBSE Class 12 Math 2018 Solved Paper

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Question : 21
Total: 29
Find the shortest distance between the lines
r
= (4
^
i
^
j
)
+ λ (
^
i
+2
^
i
3
^
k
)
and
r
= (
^
i
^
j
+2
^
k
)
+ µ (2
^
i
+4
^
j
5
^
k
)

Solution:  
The shortest distance is given by,
= |
(A2A1).(B1×B2)
|B1×B2|
|

A2A1 = (
^
i
^
j
2
^
k
)
- (4
^
i
^
j
)
= - 3
^
i
2
^
k

B1×B2 = |
^
i
^
j
^
k
123
245
|
= 2
^
i
^
j

(A2A1) . (B1×B2) = (3
^
i
2
^
k
)
. (2
^
i
^
j
)
= - 6
|B1×B2| = 22+12 = 5
so shortest distance between two lines = |
6
5
|
=
6
5
units
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