Question

V2
51. The distance of the plane 2x-3y+oz+7=0 from the point (2,-3,-1) is​

Answer 1

Distance of plane from point(x,y,z)=

| ax+by+cz+d / √[a^2+b^2+c^2] |

here a = 2

b= -3

c= 0

d= 7

x,y,z= 2,-3,-1

Distance = 2(2)+(-3)(-3)+0(-1)+7/√13

= 4+9+7/√13

= 20/√13