A plane moves so that its distance from the origin is a constant $p$. Write down the equation of the locus of the centroid of the triangle formed by its intersection with the three coordinates plane.

I was thinking about this question many times, but i could not get any hint to find the locus of the centroid of the triangle . But i know that how to find the centroid of triangle we have to add (x+y+z/3,X+Y+Z/3), i don't know the locus of the centroid of the triangle.

if anbody help me i would be very thankful to him

  • $\begingroup$ So the plane is always tangent to the sphere? $\endgroup$ – Michael Hoppe Aug 9 '17 at 5:53

The equation of a variable plane in intercept form is $$\frac xa+\frac yb+ \frac zc=1$$

Here, $a,b$ and $c$ are intercept on $x,y$ and $z$ axes.

Given, it's distance from origin is constant $$\frac{\Big|\Big(\frac xa+\frac yb+ \frac zc-1\Big)_{x=y=z=0}\Big|}{\sqrt{\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2}}}=p(\text{constant})$$

And centroid of triangle $$(x,y,z)=\left(\frac a3,\frac b3,\frac c3\right)$$

Therefore you get $$\frac{1}{p^2}=\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2}$$

Also, $a=3x,b=3y$ and $c=3z$

$$\implies \frac{1}{p^2}=\frac {1}{9x^2}+\frac {1}{9y^2}+\frac {1}{9z^2}$$

Hence the desired locus is : $$\color{blue}{\frac {1}{x^2}+\frac {1}{y^2}+\frac {1}{z^2}=\frac{9}{p^2}}$$

  • $\begingroup$ thanks a lot@ jaideep khare,, but i don't know where i haveto put this centroid coordinates and make a equation $\endgroup$ – lomber Aug 9 '17 at 5:57
  • $\begingroup$ What about the plane $z=p$? $\endgroup$ – Michael Hoppe Aug 9 '17 at 5:57
  • $\begingroup$ @MichaelHoppe I don't understand what you want to say. $\endgroup$ – Jaideep stands with Monica Aug 9 '17 at 6:00
  • $\begingroup$ @lomberlego Check my edit. $\endgroup$ – Jaideep stands with Monica Aug 9 '17 at 6:00
  • $\begingroup$ The plane given by $z=p$ doesn't have an $x$- or a $y$-intercept. $\endgroup$ – Michael Hoppe Aug 9 '17 at 6:01

Let $ac+by+cz+d=0$ be an equation of the plain.

Thus, $abc\neq0$ and $$\frac{|d|}{\sqrt{a^2+b^2+c^2}}=p,$$ which gives $$|d|=p\sqrt{a^2+b^2+c^2}.$$

Let $A\left(\frac{-d}{a},0,0\right)$, $B\left(0,-\frac{d}{b},0\right)$ and $C\left(0,0,-\frac{d}{c}\right)$ and for the centroid $M$ we obtain $$M\left(\frac{-d}{3a},\frac{-d}{3b},\frac{-d}{3c}\right),$$ which gives $$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{9(a^2+b^2+c^2)}{d^2}=\frac{9}{p^2}$$

  • $\begingroup$ thanks a lots @ michael Rozenberg $\endgroup$ – lomber Aug 9 '17 at 13:18
  • $\begingroup$ @lomber lego You are welcome! $\endgroup$ – Michael Rozenberg Aug 9 '17 at 14:26

Let $\vec s$ be a vector of length $p$. Then the plane's equation is $\langle\vec s,\vec x\rangle=p^2$. Provided that no coordinate of $\vec s$ is zero, the intercepts are obviously $$\frac{p^2}{s_k}\vec e_k\quad\text{for $k=1,2,3$}$$ where $\vec e_k$ denote the $k^\text{th}$ unit vector.

The result may be easily generalized to arbitrary dimensions.


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