Find the shortest distance between the lines r → = ( 4 i ^ - j ^ ) + λ ( i ^ + 2 j ^ - 3 k ^ ) , and r → = ( i ^ - j ^ + 2 k ^ ) + μ ( 2 i ^ + 4 j ^ - 5 k ^ )

We know that the shortest distance between the lines r → = a → 1 + λ b → 1 and r → = a → 2 + λ b → 2 is given by d = a → 2 - a → 1 · b → 1 × b → 2 b → 1 × b → 2 Here, we have, a → 1 = 4 i ^ - j ^ , a → 2 = i ^ - j ^ + 2 k ^ , b → 1 = i + 2 j - 3 k ^ and b → 2 = 2 i ^ + 4 j ^ - 5 k ^ Now, a → 2 - a → 1 = i ^ - j ^ + 2 k ^ - 4 i ^ + j ^ = - 3 i ^ + 0 j ^ + 2 k ^ And b → 1 × b → 2 = i ^ j ^ k ^ 1 2 - 3 2 4 - 5 = - 10 + 12 i ^ - - 5 + 6 j ^ + 4 - 4 k ^ = 2 i ^ - j ^ + 0 k ^ Therefore, a → 2 - a → 1 · b → 1 × b → 2 = ( - 3 i ^ + 0 j ^ + 2 k ^ ) · ( 2 i ^ - j ^ + 0 k ^ ) = - 6 + 0 + 0 = - 6 And b → 1 × b → 2 = 4 + 1 + 0 = 5 Therefore, d = a → 2 - a → 1 · b → 2 × b → 1 b → 2 × b → 1 = - 6 5 = 6 5 5 units. Hence, the shortest distance between the lines is 6 5 5 units.

Let l 0 be the line defined by the vector (equation) i ^ + 2 j ^ + 3 k ^ + λ ( i ^ + j ^ + k ^ ) , with λ real. Which of the following vector equations, with μ real, defines a line which intersects l 0 ?

i ^ + 3 j ^ + 5 k ^ + μ ( 2 i ^ + 3 j ^ + 4 k ^ )

- i ^ - 2 j ^ - 3 k ^ + μ ( - i ^ - j ^ - k ^ )

3 i ^ - 2 j ^ + k ^ + μ ( - i ^ + j ^ )

2 i ^ + 3 j ^ + μ ( i ^ - j ^ )

Vector equations of two straight lines are r → = ( 3 - t ) i ^ + ( 4 + 2 t ) j ^ + ( t - 2 ) k ^ and r → = ( 1 + s ) i ^ + ( 3 s - 7 ) j ^ + ( 2 s - 2 ) k ^ . Find the shortest distance between them.

Let standard vector equation of a line is r ¯ = a ¯ + λ b ¯ for the line r ¯ = ( i ^ - 2 j ^ + 3 k ^ ) + t ( - i ^ + j ^ - 2 k ^ ) and standard equation to be r ¯ = c ¯ + λ d ¯ for the line r ¯ = ( i ^ - j ^ - k ^ ) + s ( i ^ + 2 j ^ - 2 k ^ ) These two vectors are not parallel because these two vectors are not collinear. As they are not parallel lines, it can be either intersecting or skew lines. Intersecting lines is a special case of skew lines, where distance between two lines is zero. Distance between two skew lines is = ( c ¯ - a ¯ ) · ( b ¯ × d ¯ ) | b ¯ × d ¯ | Now, b ¯ × d ¯ = i ^ j ^ k ^ - 1 + 1 - 2 + 1 + 2 - 2 = i ^ ( 1 ( - 2 ) - ( - 2 ) ( 2 ) ) - j ^ ( ( - 1 ) ( - 2 ) - ( 1 ) ( - 2 ) ) + k ^ ( ( - 1 ) ( 2 ) - ( 1 ) ( 1 ) ) ⇒ b ¯ × d ¯ = 2 i ^ - 4 j ^ - 3 k ^ | b ¯ × d ¯ | = 2 2 + ( - 4 ) 2 + ( - 3 ) 2 = 29 Now, ( c ¯ - a ¯ ) = ( i ^ - 4 k ^ ) Distance = ( j ^ - 4 k ^ ) · ( i ^ - 4 j ^ 3 k ^ ) 29 = 0 - 4 + 12 29 = 8 29 As the distance between the skew lines is not zero, so they are not intersecting lines but the skew lines. Thus, the shortest distance between vector equations of two straight lines is 8 29 .

Find the shortest distance between the lines r → = 6 i ^ + 2 j ^ + 2 j ^ + λ ( i ^ - 2 j ^ + 2 k ^ ) and r → = − 4 i ^ − k ^ + μ ( 3 i ^ − 2 j ^ − 2 k ^ ) .

Given lines are r → = 6 i ^ + 2 j ^ + 2 j ^ + λ ( i ^ - 2 j ^ + 2 k ^ ) and r → = − 4 i ^ − k ^ + μ ( 3 i ^ − 2 j ^ − 2 k ^ ) . We have to find the shortest distance between the two lines. Let a → = 6 i ^ + 2 j ^ + 2 j ^ , b → = ( i ^ - 2 j ^ + 2 k ^ ) , c → = − 4 i ^ − k ^ and d → = 3 i ^ − 2 j ^ − 2 k ^ . Shortest distance between the given lines is ∣ [ ( a → - c → ) b → d → ] | | b → × d → | [ ( a → - c → ) b → d → ] = | 10 2 3 1 − 2 2 3 − 2 − 2 | = 10 ( 4 + 4 ) − 2 ( − 2 − 6 ) + 3 ( − 2 + 6 ) = 108 b → × d → = | i ^ j ^ k ^ 1 − 2 2 3 − 2 − 2 | = i ^ ( 4 + 4 ) − j ^ ( − 2 − 6 ) + k ^ ( − 2 + 6 ) = 8 i ^ + 8 j ^ + 4 k ^ | b → × d → | = 64 + 64 + 16 = 12 ∴ shortest distance = 108 12 = 9 Thus, the shortest distance between the given two lines is 9 units.