Question

Find the shortest distance between the lines whose vector equations are

Answer 1

The vector equations are,

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

And

r → =( 4 i ^ +5 j ^ +6 k ^ )+μ( 2 i ^ +3 j ^ + k ^ )(2)

Comparing equation (1) and (2) with (3) and (4) respectively.

r → = a → 1 +λ b → 1 (3)

And,

r → = a → 2 +μ b → 2 (4)

We get,

a → 1 =( i ^ +2 j ^ +3 k ^ )(5)

b → 1 =( i ^ −3 j ^ +2 k ^ )(6)

a → 2 =( 4 i ^ +5 j ^ +6 k ^ )(7)

And,

b → 2 =( 2 i ^ +3 j ^ + k ^ )(8)

Subtract equation (5) from (7),

a → 2 − a → 1 =( 4 i ^ +5 j ^ +6 k ^ )−( i ^ +2 j ^ +3 k ^ ) a → 2 − a → 1 =( 3 i ^ +3 j ^ +3 k ^ ) (9)

Take cross product of the vectors, b → 1 and b → 2 .

b → 1 × b → 2 =( i ^ −3 j ^ +2 k ^ )×( 2 i ^ +3 j ^ + k ^ ) =| i ^ j ^ k ^ 1 −3 2 2 3 1 | =( ( −3 )×1−2×3 ) i ^ −( 1×1−2×2 ) j ^ +( 1×3−( −3 )×2 ) k ^ =( −3−6 ) i ^ −( 1−4 ) j ^ +( 3+6 ) k ^

Further solve the above equation.

b → 1 × b → 2 =−9 i ^ +3 j ^ +9 k ^ (10)

The magnitude of the vector in equation (10) is,

| b → 1 × b → 2 |=| −9 i ^ +3 j ^ +9 k ^ | = ( −9 ) 2 + 3 2 + 9 2 = 81+9+81 = 9( 9+1+9 )

The magnitude of the cross product is,

| b → 1 × b → 2 |=3 19 (11)

Hence, the shortest distance between the given lines is given by,

d=| ( b → 1 × b → 2 )⋅( a → 2 − a → 1 ) | b → 1 × b → 2 | |(12)

Substitute the values from equation (9), (10) and (11) in equation (12).

d=| ( −9 i ^ +3 j ^ +9 k ^ )⋅( 3 i ^ +3 j ^ +3 k ^ ) 3 19 | =| −9×3+3×3+9×3 3 19 | =| 3( −9+3+9 ) 3 19 | =| 3 19 |

Thus, the shortest distance between the given lines is 3 19  units.