finding locus of a point in 3d.

Question

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.

Answer 1

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 $(0,0,1)$).

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

Answer 2

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.