Find the equation for the set of points equidistant from the y-axis and the plane z=6

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I think it's $$\sqrt{x^2+z^2}=|z-6|$$

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Michael Rozenberg
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Michael Rozenberg

Updated on September 22, 2020

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  • Michael Rozenberg
    Michael Rozenberg about 3 years

    I want to solve this question:

    Find the equation for the set of points equidistant from the $y$-axis and the plane $z=6$.

    My attempt:

    The $y$-axis has the equation $x=0$. The distance from any point to the plane $z=6$ is the absolute value of that point's z-coordinate minus $6$ or $|z-6|$.

    Is the equation $|x-0|=|z-6|$? And is the distance from any point to the $y$-axis $|x'-0|$ where $x'$ is the $x$ coordinate of that arbitrary point?

    Even If I'm correct, I really don't understand why or have an intuitive sense on how to find points equidistant from axis, planes, lines, etc. Can I get help on how to develop intuition for knowing equidistance?

    • Jihoon Kang
      Jihoon Kang about 6 years
      Be careful! $x=0$ is a plane and not a line!
    • Admin
      Admin about 6 years
      do we have to use parametric methods to get the equation of the y-axis?
    • Jihoon Kang
      Jihoon Kang about 6 years
      For any point $(x,y,z)$, the distance to the plane is $|z-6|$ as you have pointed out. The distance to the y-axis is $\sqrt{x^2 +z^2}$
    • Admin
      Admin about 6 years
      Ok, so I should just memorize that the distance from any point to the z-axis is consequently $\sqrt{x^2+y^2}$ and the equation from any point to the x-axis is $\sqrt{y^2+z^2}$ ?
    • Misha Lavrov
      Misha Lavrov about 6 years
      Given a point $(x,y,z)$, the closest point to it on the $y$-axis is $(0,y,0)$. The distance between $(x,y,z)$ and $(0,y,0)$ is $\sqrt{x^2+z^2}$ by the distance formula.
    • Admin
      Admin about 6 years
      Why is the distance between $ (x,y,z)$ and $ (0,y,0)$ equal to the length of the hypotenuse of the right triangle with one base equal to $ x$ , one base equal to$ z$ , and the third side of the triangle equal to the distance between $ (x,y,z)$ and $ (0,y,0)$ ? Why is the distance between (x,y,z) and (0,y,0) the length of the longest side of the right triangle? Why can't the distance from that arbitrary point (x,y,z) not equal the length of the hypotenuse of the right triangle?
    • Michael Rozenberg
      Michael Rozenberg about 6 years
      @Got Because the distance between a point and a straight line it's a length of the perpendicular from the point to the straight line.
    • Admin
      Admin about 6 years
      but the hypotenuse of a right triangle is not perpendicular to any base of the triangle.
    • Michael Rozenberg
      Michael Rozenberg about 6 years
      @Got $AC\perp$ $y$-axis and $AB\perp BC$. From here $AC^2=AB^2+BC^2$. See my comment under my post.
    • Admin
      Admin about 6 years
      How do you prove that the hypotenuse which is the distance between some arbitrary point and the y-axis is orthogonal to the y-axis of the right triangle with one base parallel to x-axis and the other base parallel to the z-axis ?
    • Admin
      Admin about 6 years
      I don't believe that the the side of the triangle represented by $\sqrt{x^2+z^2}$ is orthogonal to the y-axis. I'm looking for a proof.
    • Michael Rozenberg
      Michael Rozenberg about 6 years
      @Got $y$-axis$\perp BC$ and $y$-axis$\perp AB$. Thus, $y$-axis$\perp(ABC)$, which says that $y$-axis $\perp AC$.
  • Admin
    Admin about 6 years
    how did you get the left hand side?
  • Michael Rozenberg
    Michael Rozenberg about 6 years
    @Got By the Pythagoras theorem. Just draw it!
  • Admin
    Admin about 6 years
    I don't have the visualization skills to draw it. The left hand side looks like the equation of a circle or the euclidean distance formula, or a right triangle.
  • Michael Rozenberg
    Michael Rozenberg about 6 years
    @Got Draw the $x$-axis, the $y$-axis, the $z$-axis and the point $(x,y,z)$. Do it and see now.
  • Michael Rozenberg
    Michael Rozenberg about 6 years
    @Got Let $A(x,y,z)$, $AB$ be a perpendicular to the plane $xy$ and $AC$ be a perpendicular to the $y$-axis. Thus, $AB=|z|$, $BC=|x|$ and $AC=\sqrt{x^2+z^2}$.
  • Admin
    Admin about 6 years
    I'm confused on why the distance from some arbitrary point to the y-axis is the hypotenuse of a right triangle. I don't think the distance is the longest side of a right triangle. I mean when we compute the distance from some arbitrary point to the x-axis the distance is also the hypotenuse of some arbitrary triangle X although a different triangle. The distance from some arbitrary point to the z-axis is also the hypotenuse of some arbitrary triangle Y. Can you draw all these different triangles in for me? I don't think the triangle is unique even though it must be
  • Michael Rozenberg
    Michael Rozenberg about 6 years
    @Got The distance between a point and a straight line it's a length of the perpendicular from the point to the straight line. I don't know to draw in the net. I am sorry.