Question

Find the shortest distance between the lines

Answer 1

The given equations of lines are,

r → =( i ^ +2 j ^ + k ^ )+λ( i ^ − j ^ + k ^ ) r → ′ =( 2 i ^ − j ^ − k ^ )+μ( 2 i ^ + j ^ +2 k ^ )

Two lines in vector form are,

r → = a → 1 +λ b → 1 (1)

And,

r → = a → 2 +μ b → 2 (2)

The formula for the shortest distance d → between these lines is,

d → =| ( b → 1 × b → 2 ).( a → 2 − a → 1 ) | b → 1 × b → 2 | | (3)

Compare the given equation of lines with equation (1) and (2) respectively.

a → 1 =1 i ^ +2 j ^ +1 k ^  ,  b → 1 =1 i ^ −1 j ^ +1 k ^ a → 2 =2 i ^ −1 j ^ −1 k ^  ,  b → 2 =2 i ^ +1 j ^ +2 k ^

The values of b → 1 × b → 2 and a → 2 − a → 1 are,

( b → 1 × b → 2 )=| i ^ j ^ k ^ 1 −1 1 2 1 2 | = i ^ [ ( −1×2 )−( −1×1 ) ]− j ^ [ ( 1×2 )−( 2×1 ) ]+ k ^ [ ( 1×1 )−( 2×−1 ) ] = i ^ ( −3 )− j ^ ( 2−2 )+( 1+2 ) k ^ =−3 i ^ −0 j ^ +3 k ^

And,

( a → 2 − a → 1 )=( 2 i ^ −1 j ^ −1 k ^ )−( 1 i ^ +2 j ^ +1 k ^ ) = i ^ −3 j ^ −2 k ^

The value of | b → 1 × b → 2 | is,

| b → 1 × b → 2 |= ( −3 ) 2 + 0 2 + 3 2 = 9+0+9 = 18 =3 2

Substitute these values in equation (3).

d → =| ( −3 i ^ −0 j ^ +3 k ^ )⋅( i ^ −3 j ^ −2 k ^ ) 3 2 | =| −3+0−6 3 2 | =| −9 3 2 | = 3 2

Thus, the shortest distance between the two given lines is 3 2 units.