The shortest distance between the line x = y = z and the line of intersection of 2 x + y + z - 1 = 0 and 3 x + y + 2 z - 2 = 0 is

Step-by-step solution

P 1 = 2 x + y + z - 1 = 0 P 2 = 3 x + y + 2 z - 2 = 0 The line of intersection is parallel to i ^ j ^ k ^ 2 1 1 3 1 2 = i ^ - j ^ - k ^ For the point on the line, putting z = 0 , we get, 2 x + y = 1 3 x + y = 2 x = 1 , y = - 1 So, the required point is 1 , - 1,0 Hence, the equation of the line of intersection is x - 1 1 = y + 1 - 1 = z - 0 - 1 Let, a → = i ^ - j ^ b → = i ^ - j ^ - k ^ c → = 0 → d → = i ^ - j ^ - k ^ c → - a → = - i ^ + j ^ b → × d → = - 2 j ^ + 2 k ^ Shortest distance = c → - a → ⋅ b → × d → b → ×