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Question

Find the shortest distance between lines and .

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Solution

The given plane lines are,

r =6 i ^ +2 j ^ +2 k ^ +λ( i ^ 2 j ^ +2 j ^ ) (1)

r =4 i + k +μ( 3 i 2 j 2 k )(2)

We know that shortest distance between two lines, r = a 1 +λ b 1 and r = a 2 +λ b 2 is given by,

d=| ( b 1 × b 2 )( a 2 a 1 ) | b 1 × b 2 | |(3)

Comparing r = a 1 +λ b 1 and r = a 2 +λ b 2 to equation (1) and (2).

a 1 =6 i ^ +2 j ^ +2 k ^ b 1 = i ^ 2 j ^ +2 k ^ a 2 =4 i ^ k ^ b 2 =3 i ^ 2 j ^ 2 k ^

So,

( a 2 a 1 )=( 4 i ^ k ^ )( 6 i ^ +2 j ^ +2 k ^ ) =10 i ^ 2 j ^ 3 k ^

b 1 × b 2 = i ^ j ^ k ^ 1 2 2 3 2 2 =( 4+4 ) i ^ ( 26 ) j ^ +( 2+6 ) k ^ =8 i ^ +8 j ^ +4 k ^

| b 1 × b 2 |= ( 8 ) 2 + ( 8 ) 2 + ( 4 ) 2 = 64+64+16 = 144 =12

( b 1 × b 2 )( a 2 a 1 )=( 8 i ^ +8 j ^ +4 k ^ )( 10 i ^ 2 j ^ 3 k ^ ) =801612 =108

Substitute all the values in equation (3),

d=| 108 12 | =9

Thus, the shortest distance between the two given lines is 9 units.


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