Question

A sphere is inscribed in a tetrahedron whose vertices are $A\equiv (6,0,0),B\equiv (0,4,0),C\equiv (0,0,2)$ and $D\equiv (0,0,0)$. If the radius of sphere is $\frac{m}{n}$ (where $mandn$ are relatively prime), then $(m+n)$ is

• A. $5$
• B. $4$
• C. $6$
• D. $7$

Answer 1

Answer: (A)

$5$

Given
A(6,0,0)
B(0,4,0)
C(0,0,2)
D(0,0,0)
let R(a,b,c) be the centre of sphere and r will be perpendicular from R to each plane
eq of plane ABC
formula to find eq of plane from three points
$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
$\left|\begin{array}{ccc}x-6& y-0& z-0\\ 0-6& 4-0& 0-0\\ 0-6& 0-0& 2-0\end{array}\right|=0$
$\left|\begin{array}{ccc}x-6& y& z\\ -6& 4& 0\\ -6& 0& 2\end{array}\right|=0$
$(x-6)\left(8\right)-y(-12)+z(+24)=0$
$8x-48+12y+24z=0$
$2x+3y+6z=12$
perpendicular distance r from point R(a,b,c)
formula
$distance=\left|\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}\right|$
Here $a=2,b=3,c=6,d=-12$ and $x_1,y_1,z_1$ are coordinates of point R
$r=\left|\frac{2a+3b+6c-12}{\sqrt{2^2+3^2+6^2}}\right|$
$r=\left|\frac{2a+3b+6c-12}{\sqrt{4+9+36}}\right|$
$r=\left|\frac{2a+3b+6c-12}{\sqrt{49}}\right|$
$r=\left|\frac{2a+3b+6c-12}{7}\right|$---------(1)
eq of plane ABD
formula to find eq of plane from three points
$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
$\left|\begin{array}{ccc}x-6& y-0& z-0\\ 0-6& 4-0& 0-0\\ 0-6& 0-0& 0-0\end{array}\right|=0$
$\left|\begin{array}{ccc}x-6& y& z\\ -6& 4& 0\\ -6& 0& 0\end{array}\right|=0$
$(x-6)\left(0\right)-y\left(0\right)+z(+24)=0$
$24z=0$
$z=0$
perpendicular distance r from point R(a,b,c)
formula
$distance=\left|\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}\right|$
$r=\left|\frac{c}{\sqrt{1^2}}\right|$
$r=\left|\frac{c}{\sqrt{1}}\right|$
$r=c$------(2)
now for plane ACD
formula to find eq of plane from three points
$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
$\left|\begin{array}{ccc}x-6& y-0& z-0\\ 0-6& 0-0& 2-0\\ 0-6& 0-0& 0-0\end{array}\right|=0$
$\left|\begin{array}{ccc}x-6& y& z\\ -6& 0& 2\\ -6& 0& 0\end{array}\right|=0$
$(x-6)\left(0\right)-y(-12)+z\left(0\right)=0$
$12y=0$
$y=0$
perpendicular distance r from point R(a,b,c)
formula
$distance=\left|\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}\right|$
$r=\left|\frac{b}{\sqrt{1^2}}\right|$
$r=\left|\frac{b}{\sqrt{1}}\right|$
$r=b$------(3)
eq of plane BCD
formula to find eq of plane from three points
$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
$\left|\begin{array}{ccc}x-0& y-4& z-0\\ 0-0& 0-4& 2-0\\ 0-0& 0-4& 0-0\end{array}\right|=0$
$\left|\begin{array}{ccc}x& y-4& z\\ 0& -4& 2\\ 0& -4& 0\end{array}\right|=0$
$\left(x\right)(-8)-(y-4)\left(0\right)+z\left(0\right)=0$
$8x=0$
$x=0$
perpendicular distance r from point R(a,b,c)
formula
$distance=\left|\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}\right|$
$r=\left|\frac{a}{\sqrt{1^2}}\right|$
$r=\left|\frac{a}{\sqrt{1}}\right|$
$r=a$------(4)

from eq (1),(2),(3) and (4)
$r=\left|\frac{2a+3b+6c-12}{7}\right|=a=b=c$
here putting value of a,b,c in eq(1)
$r=\left|\frac{2r+3r+6r-12}{7}\right|$
opening modulus with positive sign
$r=\frac{11r-12}{7}$
$7r=11r-12$
$4r=12$
$r=\frac{12}{4}$
$r=\frac{3}{1}$
here 3 and 1 are not co prime so rejected
now opening modulus with negative sign
$r=-\frac{11r-12}{7}$
$7r=-11r+12$
$18r=12$
$r=\frac{12}{18}$
$r=\frac{2}{3}$
Here 2 and 3 are co prime no so we accept them
$m+n=2+3=5$