What is the distance between point in z-axis from point in x-axis, where given plane ax+by+cz+d=0 intercept.

• sqrt{(-d/c)^2+(-d/a)^2}
• 13 years agoHelpfull: Yes(66) No(6)
• equa is ax+by+cz+d=0 put the x=0 ,y=0 for finding z co-ordinate
  now put the y=0,z=0 for finding x co-ordinate now for distance apply the distance formula..D=root(square(x1-x2)+square(y1-y2)+square(z1-z2))
• 13 years agoHelpfull: Yes(59) No(9)
• by putting y=z=0 in equation we get coordinate of x((-d/a),0,0)
  x=z=0 coordinate of y(0,(-d/b),0) and x=y=0 coordinate of z(0,0,(-d/c))
  sqrt{(-d/a)^2)+((-d/c)^2)}=d*sqrt((1/a^2)+(1/c^2))
• 13 years agoHelpfull: Yes(38) No(10)
• ax+by+cz=-d
  x/(-ad)+y/(-bd)+z/(-cd)=1
  x-intercept=ad
  z-intercept=cd
  distance between these two=(ad^2+cd^2)^(1/2)
• 13 years agoHelpfull: Yes(8) No(19)
• make the equation as ax+by+cz=-d.furter divide the terms by -d to make rhs 1.take the coefficient of x and z and subtract it.thats the ans
• 13 years agoHelpfull: Yes(4) No(6)
• The plane crosses the x axis when y = z = 0; similarly x = y = 0 when it crosses the z-axis. We have to make some assumptions about the coefficients a and c---namely that they are not zero so that the plane does actually cross these axes. (Well, there is the other degenerate case in which the plane is the xz-plane y = 0.)
  If a and c are nonzero, the intersection of the plane with the x-axis occurs at the point (-d/a,0,0) and that with the z-axis occurs at (0,0,-d/c). The distance between these points is
  √((-d/a)^2 + (-d/c)^2) = |d|/√(a^2 + c^2)
• 13 years agoHelpfull: Yes(4) No(6)
• change it in the form of...
  x/a+y/b+z/c=1....
  u wil get the ans...
• 13 years agoHelpfull: Yes(3) No(7)
• for the answer we divide equation with d then we get individual terms
  then answer is sqrt(d^2/(a^2+c^2))
• 12 years agoHelpfull: Yes(1) No(6)
• d/((x-a)^2+(z-c)^2)1/2
• 7 years agoHelpfull: Yes(0) No(0)