Q7 of 181 Page 1196

Show that the point (1, 2, 1) is equidistant from the planes and

Formula :

where () is point from which distance is to be calculated

Therefore ,

First Plane can be written in cartesian form as

x + 2y - 2z = 5

x + 2y - 2z – 5 = 0

Point = ( 1 , 2 , 1 )

Distance for first plane

Second Plane can be written in cartesian form as

2x - 2y + z + 3 = 0

Point = ( 1 , 2 , 1 )

Distance for second plane

Hence proved.

Find the distance of the point (2, 1, 0) from the plane 2x + y + 2z + 5 = 0.

Find the distance of the point (2, 1, - 1) from the plane x – 2y + 4z = 9.

Show that the points ( - 3, 0, 1) and (1, 1, 1) are equidistant from the plane 3x + 4y – 12z + 13 = 0.

Find the distance between the parallel planes 2x + 3y + 4 = 4 and 4x + 6y + 8z = 12.