Question

$ABCD$ is a tetrahedron where $A(2,0,0),$ $B(0,4,0)$ and $CD=\sqrt{14}.$ The edge $CD$ lies on the line $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}.$ If locus of centroid of tetrahedron is $\frac{x-\frac{3}{2}}{1}=\frac{y-y_1}{a}=\frac{z-z_1}{b},$ then

• A. $a+b=5$
• B. $y_1+z_1=6$
• C. $y_1-z_1=1$
• D. $a+b+y_1=8$

Answer 1

Answer: (A), (B), (D)

$a+b+y_1=8$

Given : $A(2,0,0),B(0,4,0)$ and $CD$ lies on the line $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$
Let $C\equiv ({\lambda }_1+1,2{\lambda }_1+2,3{\lambda }_1+3)$
and $D\equiv ({\lambda }_2+1,2{\lambda }_2+2,3{\lambda }_2+3)$
$CD=\sqrt{14}$
$\Rightarrow ({\lambda }_1-{\lambda }_2{)}^2+4({\lambda }_1-{\lambda }_2{)}^2+9({\lambda }_1-{\lambda }_2{)}^2=14$
$\Rightarrow {\lambda }_1-{\lambda }_2=\pm 1$
Taking negative sign, we get
${\lambda }_2=1+{\lambda }_1$
$\therefore C\equiv ({\lambda }_1+1,2{\lambda }_1+2,3{\lambda }_1+3)$ and
$D\equiv ({\lambda }_1+2,2{\lambda }_1+4,3{\lambda }_1+6)$

Let centroid of tetrahedron be $G\equiv (\alpha ,\beta ,\gamma )$
$\therefore \alpha =\frac{2+0+{\lambda }_1+1+{\lambda }_1+2}{4}$
$\Rightarrow 4\alpha =5+2{\lambda }_1$
Similarly, $4\beta =10+4{\lambda }_1$
and $4\gamma =9+6{\lambda }_1$
Thus, locus of the centroid of tetrahedron is $\frac{4x-5}{2}=\frac{4y-10}{4}=\frac{4z-9}{6}$
$\Rightarrow \frac{x-\frac{5}{4}}{\frac{1}{2}}=\frac{y-\frac{5}{2}}{1}=\frac{z-\frac{9}{4}}{\frac{3}{2}}$
$\Rightarrow \frac{x-\frac{5}{4}}{\frac{1}{2}}-\frac{1}{2}=\frac{y-\frac{5}{2}}{1}-\frac{1}{2}=\frac{z-\frac{9}{4}}{\frac{3}{2}}-\frac{1}{2}$
$\Rightarrow \frac{x-\frac{3}{2}}{1}=\frac{y-3}{2}=\frac{z-3}{3}$
$\therefore y_1=3=z_1,a=2$ and $b=3$

Now, $a+b=2+3=5$
$y_1+z_1=3+3=6$
$y_1-z_1=3-3=0$
and $a+b+y_1=2+3+3=8$