Find the equation for the set of points equidistant from the y-axis and the plane z=6
1,432
I think it's $$\sqrt{x^2+z^2}=|z-6|$$
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Michael Rozenberg
Updated on September 22, 2020Comments
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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?
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Jihoon Kang about 6 yearsBe careful! $x=0$ is a plane and not a line!
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Admin about 6 yearsdo we have to use parametric methods to get the equation of the y-axis?
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Jihoon Kang about 6 yearsFor 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 $\sqrt{x^2 +z^2}$
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Admin about 6 yearsOk, 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 $\sqrt{y^2+z^2}$ ?
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Misha Lavrov about 6 yearsGiven 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.
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Admin about 6 yearsWhy 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?
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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.
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Admin about 6 yearsbut the hypotenuse of a right triangle is not perpendicular to any base of the triangle.
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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.
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Admin about 6 yearsHow 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 ?
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Admin about 6 yearsI 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.
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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 $\perp AC$.
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Admin about 6 yearshow did you get the left hand side?
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Michael Rozenberg about 6 years@Got By the Pythagoras theorem. Just draw it!
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Admin about 6 yearsI 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.
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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.
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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 $AC=\sqrt{x^2+z^2}$.
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Admin about 6 yearsI'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
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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.