Q8 of 181 Page 1196

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


Formula :


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


Therefore ,


Plane = 3x + 4y - 12z + 13 = 0


First Point = ( - 3 , 0 , 1 )


Distance for first point





Plane = 3x + 4y - 12z + 13 = 0


Second Point = ( 1 , 1 , 1 )


Distance for first point





Hence proved.


More from this chapter

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6

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

 7

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

9

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

 10

Find the distance between the parallel planes x + 2y - 2z + 4 = 0 and x + 2y – 2z – 8 = 0.