A sphere of radius $r$ passes through the origin and the other points where it meets the coordinate axes are A, B and C. Find the distance of the centroid of the triangle ABC from the origin (in terms of $r$).

i was taking the coordinate of the centroid ($\frac{x+y+z}{3}$, $\frac{X+Y+Z}{3}$). but i don't know the distance of the centroid of triangle ABC from the origin. From my point of view i was using the distance formula from the origin,, but i didn't get it

if anybody help me me i would be very thankful to him thank you

  • $\begingroup$ Are we in two or three dimensions ? I guess in two ; a point with coordinates $(u,v)$ has distance $\sqrt{u^2+v^2}$ from the origin. In three dimensions, it is not much harder : The distcnae of a point with coordinates $(u,v,w)$ from the origin is $\sqrt{u^2+v^2+w^2}$ $\endgroup$
    – Peter
    Aug 9 '17 at 15:04
  • $\begingroup$ You considered that entering something like $5+4+7/3$ into a calculator does not give the correct result $(5+4+7)/3$ , didn't you ? $\endgroup$
    – Peter
    Aug 9 '17 at 15:12

If the sphere intersects the three axes at the points A, B and C, it means that the sphere passes through these three points:

$(A,0,0)$; $(0,B,0)$; $(0,0,C)$

i.e., the sphere intersects x-axis at $(A,0,0)$ and y-axis at $(0,B,0)$ and z-axis at $(0,0,C)$

Now, coordinates of the centroid of triangle formed by these three co-ordinates will be:

$(\frac{A+0+0}{3},\frac{0+B+0}{3},\frac{0+0+C}{3})$ = $(\frac{A}{3},\frac{B}{3},\frac{C}{3})$

To find distance of this centroid from origin $(0,0,0)$ we simply use the two point distance formula in 3d:

Distance between $(\frac{A}{3},\frac{B}{3},\frac{C}{3})$ and $(0,0,0)$ =

$\sqrt{(\frac{A}{3}-0)^2 + (\frac{B}{3}-0)^2 + (\frac{C}{3}-0)^2}$ = $\frac{\sqrt{A^2 + B^2 + C^2}}{3}$

In our case, the radius of sphere is given as $r$.

Let's assume the centre of sphere to be at $(x_0,y_0,z_0)$. Then the equation of the sphere can be written as:

$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2 = r^2$

Since it's also given that the sphere passes through the origin $(0,0,0)$, this point must satisfy the above equation of sphere.

Or, $(0-x_0)^2+(0-y_0)^2+(0-z_0)^2 = r^2$

Or, $x_0^2 + y_0^2 + z_0^2 = r^2$ --- $(1)$

Now, the sphere also passes through the point $(A,0,0)$. Substituting this point in the equation of sphere, we have:

$(A-x_0)^2+(0-y_0)^2+(0-z_0)^2 = r^2$

Or, $(A-x_0)^2+y_0^2+z_0^2 = r^2$

From $(1)$, let's also substitute $r^2$ with $x_0^2 + y_0^2 + z_0^2$

$(A-x_0)^2+y_0^2+z_0^2 = x_0^2 + y_0^2 + z_0^2$

Or, $(A-x_0)^2 = x_0^2$

Or, $A-x_0 = x_0$

Thus we have $A = 2x_0$.

Similarly by substituting $(0,B,0)$ and $(0,0,C)$ in the equation of sphere we get $B = 2y_0$ and $C= 2z_0$.

Hence, the three points A, B and C are $(2x_0,0,0)$, $(0,2y_0,0)$ and $(0,0,2z_0)$ respectively.

Thus, by using the formula we arrived at in section 1, we can say the distance of centroid of triangle ABC from origin =

$\frac{\sqrt{A^2 + B^2 + C^2}}{3}$ = $\frac{\sqrt{(2x_0)^2 + (2y_0)^2 + (2z_0)^2}}{3}$ = $\frac{2\sqrt{x_0^2 + y_0^2 + z_0^2}}{3}$

But from equation $(1)$ we know that $x_0^2 + y_0^2 + z_0^2 = r^2$.

Substituting this we have the required distance as $\frac{2\sqrt{r^2}}{3} = \frac{2r}{3}$

  • $\begingroup$ What is $(A,0,0)$ and the others ?, they shall be $(R,0,0)$ etc. (it,s a sphere not an ellipsoid)! the remaining is ok $\endgroup$
    – G Cab
    Aug 9 '17 at 16:18
  • 1
    $\begingroup$ @GCab you're confused. The centre of sphere is not the origin. The sphere is passing through the origin. So the three points A, B and C will be at distances a, b and c respectively. None of a, b or c necessarily need be equal to R. I assumed the three distances as A, B and C (since it's not clear in the question whether they're just the labels or distances). Ellipsoid/sphere is irrelevant here. We can also replace A, B, C with X, Y, Z respectively.. $\endgroup$
    – yathish
    Aug 9 '17 at 16:34
  • $\begingroup$ Oh yes, you are right, I misunderstood the sphere as centered at the origin. $\endgroup$
    – G Cab
    Aug 9 '17 at 17:51
  • $\begingroup$ thanks a lot @ yathish but my teacher said that answer is 2r/3,, i don't how 2r/3 comes.......... $\endgroup$
    – user469754
    Aug 10 '17 at 7:07
  • $\begingroup$ @legohlegoh can you please share the exact question that your teacher asked you? $\endgroup$
    – yathish
    Aug 10 '17 at 8:03

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