Question

If $\left(\frac{9}{4},\frac{5}{4},\frac{15}{4}\right)$ is the centroid of a tetrahedron whose vertices are $(a,2,1),(1,b,4),(4,0,c)$ and $(1,1,7)$, then

• A. $a=b=c$
• B. $a=b=c+1$
• C. $b=c=a+1$
• D. $a=c=b+1$

Answer 1

Answer: (D)

Vertices of tetrahedron
$(a,2,1),(1,b,4),(4,0,c),(1,1,7)$
Centroid
$\left(\frac{a+1+4+1}{4},\frac{2+b+0+1}{4},\frac{1+4+c+7}{4}\right)$
$=\left(\frac{9}{4},\frac{5}{4},\frac{15}{4}\right)$
$\Rightarrow \frac{a+6}{4}=\frac{9}{4},\frac{b+3}{4}=\frac{5}{4},\frac{c+12}{4}=\frac{15}{4}$
$\Rightarrow a=3,b=2c=3$
$\Rightarrow a=c=b+1$