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Find the locus of the point which moves so that its distance from the line x=y=z is twice its distance from the plane x + y+z=1.

I know the distance of point (x,y,z) from given plane will be mod(x+y+z-1)/3^(1/2).How to find distance that from given line?I tried to find distance of (0,0,0) and (x,y,z) and then taking its component along normal to line, but failed.

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    $\begingroup$ Direction cosines $(1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3} ) $ of cone axis with semi-vertical angle $\tan^{-1}2$ $\endgroup$
    – Narasimham
    Jun 5 '20 at 20:58
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Let $P(a,b,c)$ be any such point. As you said, it’s distance from the plane is $$D=\frac{|a+b+c-1|}{\sqrt 3}$$ Now, any general point on the line is of the form $(k,k,k)$, whose distance from $P$ is $$d(k)=\sqrt{(a-k)^2 +(b-k)^2+(c-k)^2} $$ We want to minimize this distance. Differentiate $d^2(k)$ to do this. $$(d^2(k))’ =-2(a-k)-2(b-k)-2(c-k)=0 \implies k=\frac{a+b+c}{3}$$ Then we want $$\sqrt{\left(\frac{2a-b-c}{3}\right)^2 +\left(\frac{2b-a-c}{3}\right)^2+\left(\frac{2c-a-b}{3}\right)^2}=2\times \frac{|a+b+c-1|}{\sqrt 3} $$ I’ll leave the simplification to you.

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  • $\begingroup$ I am sorry but it's not at all an achieved answer : you should square the relationship and then explain that it is the equation of a cone that @Narasimhan has observed, and which is evident if you consider the situation in a section plane passing through the straight line with equation $x=y=z$. $\endgroup$
    – Jean Marie
    Jun 6 '20 at 22:00
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    $\begingroup$ @JeanMarie I thought I’d let the OP figure out what this equation represents, instead of making it a complete solution. $\endgroup$
    – Tavish
    Jun 7 '20 at 9:40
  • $\begingroup$ Now, with a closer look at your solution, I don't understand what you have done: 1) why is this problem a minimization issue and 2) where you take into account factor $2$ in "distance of $M$ from the line is 2 times its distance from the plane". $\endgroup$
    – Jean Marie
    Jun 7 '20 at 14:14
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    $\begingroup$ @Jean Marie $1)$ The distance of a point from a line is the shortest distance between the point and the line, $d(k)$ gives the distance from $P$ to any point $(k,k,k)$ on the line, and we want the minimum of that. $2)$ I have multiplied the RHS of my last equation with $2$. $\endgroup$
    – Tavish
    Jun 7 '20 at 14:29
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    $\begingroup$ @JeanMarie That is all fine, but I do like differentiation! $\endgroup$
    – Tavish
    Jun 7 '20 at 19:58
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Here is a geometrical proof (see Fig. below)

This question is easy to tackle when you consider what happens in a section plane passing through line (D) (with equation $x=y=z$) : the locus is clearly a pair of half lines passing through point $A$ (with 3D coordinates $(0,0,1)$).

enter image description here

Revolving this 2D locus around line (D) gives the 3D locus : a half cone with axis (D), apex $A$ and aperture angle $\arctan 2$.

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