The line passing through ( 1 , 1 , - 1 ) and parallel to the vector i ^ + 2 j ^ - k ^ meets the line x - 3 - 1 = y + 2 5 = z - 2 - 4 at A and the plane 2 x - y + 2 z + 7 = 0 at B . Then A B =

Step-by-step solution

The line passing through ( 1 , 1 , - 1 ) and parallel to the vector i ^ + 2 j ^ - k ^ , will be, x - 1 1 = y - 1 2 = z + 1 - 1 = k   (let) So, x = k + 1   ,   y = 2 k + 1   ,   z = - k - 1 To find the point of intersection of x - 1 1 = y - 1 2 = z + 1 - 1 and x - 3 - 1 = y + 2 5 = z - 2 - 4 , A , put the values of x ,   y ,   z in the second equation ∴ k + 1 - 3 - 1 = 2 k + 1 + 2 5 ⇒ k = 1 Hence, A = ( 1 + 1 ,   2 × 1 + 1 ,   - 1 - 1 ) = ( 2 , 3 , - 2 )