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 ^ −( −2−6 ) 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 ^ ) =−80−16−12 =−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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