Question

Let $A B C D$ be a tetrahedron in which the coordinates of each of its vertices are in arithmetic progression. If the centroid $G$ of the tetrahedron is $(2,3, k )$ then the distance of $G$ from the origin is

• A. $\sqrt{38}$
• B. $7$
• C. $\sqrt{22}$
• D. $\sqrt{29}$

Answer 1

Answer: (D)

Coordinates of verticies of tetrahedron are in AP. Let
$A\left(x_{1}, y_{1}, z_{1}\right), B\left(x_{2}, y_{2}, z_{2}\right), C\left(x_{3}, y_{3}, z_{3}\right)$
and $D\left(x_{4}, y_{4}, z_{4}\right)$
Now, each coordinates are in $A P$
$\therefore y_{1}=x_{1}+d, z_{1}=x_{1}+2 d$
$y_{2}=x_{2}+d_{1} z_{2}=x_{2}+2 d$
$y_{3}=x_{3}+d, z_{3}=x_{3}+2 d$
$\therefore $ centroid of tetrahedron
$=\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}, \frac{y_{1}+y_{2}+y_{3}+y_{4}}{4}$
$\frac{z_{1}+z_{2}+z_{3}+z_{4}}{4}$
$\therefore 2=\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}$,
$3=\frac{x_{1}+x_{2}+x_{3}+x_{4}+4 d}{4}$
and $k=\frac{x_{1}+x_{2}+x_{3}+x_{4}+8 d}{4}$
$\therefore x_{1}+x_{2}+x_{3}+x_{4}=8, d=1$
$\Rightarrow k=\frac{8+8}{4}=4$
$\therefore $ Distance of $G$ from the origion
$=\sqrt{2^{2}+3^{2}+4^{2}}=\sqrt{29}$