An equation of the line passing through 3 i ^ - 5 j ^ + 7 k ^ and perpendicular to the plane 3 x - 4 y + 5 z = 8 is

r → = 3 i ^ - 5 j ^ + 7 k ^ + λ 3 i ^ - 4 j ^ + 5 k ^

r → = 3 i ^ - 4 j ^ - 5 k ^ + μ 3 i ^ + 5 j ^ + 7 k ^

Show that the lines x - 5 7 = y + 2 - 5 = z 1 and x 1 = y 2 = z 3 are perpendicular to each other.

The equation of the given lines are: x - 5 7 = y + 2 - 5 = z 1 & x 1 = y 2 = z 3 The direction ratios of these lines are 7 , - 5 , 1 & 1 , 2 , 3 . Two lines with direction ratios a 1 , b 1 , c 1 and a 2 , b 2 , c 2 are perpendicular to each other if a 1 a 2 + b 1 b 2 + c 1 c 2 = 0 ∴ a 1 a 2 + b 1 b 2 + c 1 c 2 = ( 7 × 1 ) + ( - 5 × 2 ) + ( 1 × 3 ) = 7 - 10 + 3 = 10 - 10 = 0 Thus, given lines are perpendicular to each other.

Find the distance between the lines ℓ 1 and ℓ 2 given by r → = i ^ + 2 j ^ - 4 k ^ + λ 2 i ^ + 3 j ^ + 6 k ^ and r → = 3 i ^ + 3 j ^ - 5 k ^ + μ 2 i ^ + 3 j ^ + 6 k ^ .

The two lines are parallel. We have, a → 1 = i ^ + 2 j ^ - 4 k ^ , a → 2 = 3 i ^ + 3 j ^ - 5 k ^ and b → = 2 i ^ + 3 j ^ + 6 k ^ . Therefore, the distance between the lines is given by d = b → × a → 2 - a → 1 | b → | b → × a → 2 - a → 1 = i ^ j ^ k ^ 2 3 6 2 1 - 1 = - 9 i ^ + 14 j ^ - 4 k ^ b → = 4 + 9 + 36 = 49 d = | - 9 i ^ + 14 j ^ - 4 k ^ | 49 = 293 49 = 293 7 units

Find the angle between the pair of lines r → = 3 i ^ + 2 j ^ - 4 k ^ + λ i ^ + 2 j ^ + 2 k ^ and r → = 5 i ^ - 2 j ^ + μ 3 i ^ + 2 j ^ + 6 k ^ .

Given, the pair of lines is r → = 3 i ^ + 2 j ^ - 4 k ^ + λ ( i ^ + 2 j ^ + 2 k ^ ) and r → = 5 i ^ - 2 j ^ + μ ( 3 i ^ + 2 j ^ + 6 k ^ ) a 1 → = 3 i ^ + 2 j ^ - 4 k ^ and a 2 → = 5 i ^ - 2 j ^ b 1 → = i ^ + 2 j ^ + 2 k ^ and b 2 → = 3 i ^ + 2 j ^ + 6 k ^ Now, cos θ = b 1 → · b 2 → | b 1 → | | b 2 → | ⇒ cos θ = ( i ^ + 2 j ^ + 2 k ^ ) · ( 3 i ^ + 2 j ^ + 6 k ^ ) 1 2 + 2 2 + 2 2 3 2 + 2 2 + 6 2 ⇒ cos θ = 3 + 4 + 12 3 7 ⇒ cos θ = 19 21 ∴ θ = cos - 1 19 21 .