171) The distance between the line x - 1 3 = y + 2 - 2 = z - 1 2 and the plane 2x + 2y - z = 6 is: (a) 9 units (b) 1 units (c) 2 units (d) 3 units Share with your friends Share 0 Manbar Singh answered this The given equation of the line is,x - 13 = y + 2-2 = z - 12Now, the above line passes through A1,-2,1 and having direction ratios 3, -2, 2Now, position vector of A = r1→ = i^ - 2j^ + k^Now, the given line is parallel to vector m→ = 3i^ - 2j^ + 2k^Now, the vector equation of the line is,r→ = r1→ + λm→⇒r→ = i^ - 2j^ + k^ + λ3i^ - 2j^ + 2k^The equation of the plane is, 2x + 2y - z = 6⇒xi^ + yj^ + zk^ . 2i^ + 2j^ - k^ = 6⇒r→ . 2i^ + 2j^ - k^ = 6 vector equation of planeWe know that line r→ = r1→ + λm→ is parallel to the plane r→ . n→ = q only when m→ . n→ = 0And the distance between the line and the plane is given bydistance = r1→ .n→ - q n→ Now, r1→ = i^ - 2j^ + k^m→ = 3i^ - 2j^ + 2k^n→ = 2i^ + 2j^ - k^ q = 6Now, m→ . n→ = 3i^ - 2j^ + 2k^ . 2i^ + 2j^ - k^ = 6-4-2 = 0So, the given line is parallel to given plane.Distance between line and plane = r1→ .n→ - q n→ =i^ - 2j^ + k^ . 2i^ + 2j^ - k^ - 62i^ + 2j^ - k^=2-4-1-622+22+-12=-94+4+1=99=93=3 units 1 View Full Answer Dishank Gupta answered this 3rd option is correct i.e. 3units 1 Ajeet Kumar Talari answered this 3 units 2