find the distance between the planes r.(2i-3j+6k)-4=0 and r.(6i-9j+18k)+30=0 Share with your friends Share 5 Manbar Singh answered this The given equation of planes are : r→ . 2i^ - 3j^ + 6k^ - 4 = 0⇒xi^ + yj^ + zk^ . 2i^ - 3j^ + 6k^ - 4 = 0⇒2x - 3y + 6z - 4 = 0 .....1r→ . 6i^ - 9j^ + 18k^ +30 = 0⇒xi^ + yj^ + zk^ .6i^ - 9j^ + 18k^ +30 = 0⇒6x - 9y +18z +30= 0 ⇒2x - 3y + 6z + 10 = 0 .....2Now, the given 2 planes are parallel.Let Px1, y1, z1 be any point on the plane 2x - 3y + 6z - 4 = 0 Then 2x1 - 3y1 + 6z1 = 4 Now, distance p of the point Px1, y1, z1 from the plane 2x - 3y + 6z + 10 = 0 is given byp = 2x1 - 3y1 + 6z1 + 1022 + -32 + 62=4+104+9+36=147=2 units 12 View Full Answer
