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**mathdrop****Member**- Registered: 2022-03-07
- Posts: 72

I would like to intersect the function y=(a^n-x^n-z^n)^(1/n),

which describes a sphube, being a rounded cube with its center in the origin,

with the line, going trough the origin and P=(1|1|1).

I tried it with y=x+z and y=3-x-z,

but the computer says I'm giving planes.

Is it possible to somehow get the point, where they intersect ?

For numericals, one could take a=1 and n=4.

Edit: Being unsure what the correct equation for the line is.

*Last edited by mathdrop (2023-01-18 07:09:41)*

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,631

hi mathdrop

It looks like you've got

The line joining the origin to (1,1,1) is

So every point on the line has x = y = z

So it will intersect the shape at

Hope that helps,

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**mathdrop****Member**- Registered: 2022-03-07
- Posts: 72

3*x^4=1 where x=y=z

x=y=z=+/- 1/3^(1/4)≈+/- 0.75984

Your point hits the shape :-)

How have you gotten 3*x^4=1 ?

I moved everything to one side:

x^4+y^4+z^4-1=0

f*x+f*y+f*z-r=0

Set them equal:

x^4+y^4+z^4-1=f*x+f*y+f*z-r

And tried to solve the above for x,

but this gave me wild formulas.

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,631

If the line goes through (0,0,0) and (1,1,1) then its equation is

and so for a given lambda x=y=z.

Replace y and z with x and you get

Trying to solve with x, y, and z being maybe different makes the algebra horrible but as they are equal it's enough to solve 3x^4 = 1 which you have done.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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