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 $\frac{2\sqrt{r^2}}{3} = \frac{2r}{3}$
Best Answer
As $A_1$ is the midpoint of $BC$,
$2\vec{OA_1} = \vec{OB} + \vec{OC}$
As $M$ is the centroid of the triangle, it divides median $AA_1$ into ratio of $AM:MA_1 = 2:1$
$\vec{OM} = \frac{\vec{OA}+2\vec{OA_1}}{3} = \frac{\vec{OA}+\vec{OB} + \vec{OC}}{3}$
That is all the proof is.
But if you need to further show how $2\vec{OA_1} = \vec{OB} + \vec{OC}$,
$\vec{OA_1} = \vec{OB} + \vec{BA_1}$
But also, $\vec{OA_1} = \vec{OC} + \vec{CA_1} = \vec{OC} - \vec{A_1C}$
As $\vec{BA_1} = \vec{A_1C}$, adding both you have
$2\vec{OA_1} = \vec{OB} + \vec{OC}$
Similarly you can show $\vec{OM} = \frac{\vec{OA}+2\vec{OA_1}}{3}$.
$\vec{OM} = \vec{OA}+\vec{AM}$
As $\vec{AM} = \frac{2}{3} \vec{AA_1}$,
$\vec{OM} = \frac{3\vec{OA}+2\vec{AA_1}}{3} = \frac{\vec{OA}+2(\vec{OA}+\vec{AA_1})}{3} = \frac{\vec{OA}+2\vec{OA_1}}{3}$