What is the distance between the z intercept from x intercept in the equation ax+by+cz+d=0?

• put y=z=0 in eqn to get x-coordinate=((-d/a),0,0)
  put x=y=0 in eqn to get z-cordinate =(0,0,(-d/c))
  
  distance between the z intercept from x intercept =
  
  √((-d/a)^2 + (-d/c)^2) = |d|/√(1/a^2 + 1/c^2) = |d/ac|*√(a^2 + c^2)
• 10 years agoHelpfull: Yes(37) No(1)
• 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)
• 10 years agoHelpfull: Yes(2) No(3)
• ax + by + cz + d = 0
  At the z-intercept, 'x' and 'y' are both zero.
  cz + d = 0 --> z = -d/c --> The z-intercept is the point (0, 0, -d/c).
• 6 years agoHelpfull: Yes(0) No(0)