Question

The equation of the square inscribed in a tetrahedron, whose faces are x = 0, y = 0, z = 0 and x + 2y + 2z = 1 is
(a) $32({{x}^{2}}+{{y}^{2}}+{{z}^{2}})+8\left( x+y+z \right)+1=0$
(b) $32({{x}^{2}}+{{y}^{2}}+{{z}^{2}})-8\left( x+y+z \right)-1=0$
(c) $32({{x}^{2}}+{{y}^{2}}+{{z}^{2}})-8\left( x+y+z \right)+1=0$
(d) None of these

Answer 1

Hint:For solving this problem, we have to find the centre and radius of a sphere. By using the tetrahedron we evaluate the centre of the sphere and its radius also. After obtaining both the values we get the final equation.

Complete step-by-step answer:
Let (a, b, c) be the centre and r be the radius of the sphere inscribed in the tetrahedron.
The sphere is inscribed in the tetrahedron, hence the length of the perpendicular from the centre (a, b, c) upon each of the faces is equal to the radius of the sphere. So, for the tetrahedron the mathematically expression will be:
$\therefore \dfrac{a}{1}=\dfrac{b}{1}=\dfrac{c}{1}=r\ldots (1)$
The distance formula of a point (l, m, n) from a plane ax + by + cz = d is:
$\dfrac{d-ax-by-cz}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$
Now, by using the above formula, point (a, b, c) from the plane x + 2y +2z = 1, we get
$\begin{aligned} & \dfrac{1-a-2b-2c}{\sqrt{1+{{2}^{2}}+{{2}^{2}}}}=\dfrac{1-a-2b-2c}{\sqrt{1+4+4}} \\\ & \therefore \dfrac{1-a-2b-2c}{\sqrt{9}}=\dfrac{1-a-2b-2c}{3}\ldots (2) \\\ \end{aligned}$
Now, equating equation (1) and (2) we get,
$a=b=c=r=\dfrac{1-a-2b-2c}{3}$
Using the first and the last part of equality,
$\begin{aligned} & 3a=1-a-2b-2c\text{ and }a=b=c \\\ & \therefore 3a=1-a-2a-2a \\\ & 8a=1 \\\ & \Rightarrow a=\dfrac{1}{8} \\\ \end{aligned}$
Also, a = r and hence $r=\dfrac{1}{8}$.
The general equation of a sphere having centre (l, m, n) and radius a is given as:
${{(x-l)}^{2}}+{{(y-m)}^{2}}+{{(z-n)}^{2}}={{a}^{2}}$
Hence, the equation of a sphere with centre $\left( \dfrac{1}{8},\dfrac{1}{8},\dfrac{1}{8} \right)$ and radius $r=\dfrac{1}{8}$ would be:

& {{\left( x-\dfrac{1}{8} \right)}^{2}}+{{\left( y-\dfrac{1}{8} \right)}^{2}}+{{\left( z-\dfrac{1}{8} \right)}^{2}}={{\left( \dfrac{1}{8} \right)}^{2}} \\\ & \Rightarrow {{\left( \dfrac{1}{8} \right)}^{2}}\left\\{ {{\left( 8x-1 \right)}^{2}}+{{\left( 8y-1 \right)}^{2}}+{{\left( 8z-1 \right)}^{2}} \right\\}={{\left( \dfrac{1}{8} \right)}^{2}} \\\ & 64{{x}^{2}}+1-16x+64{{y}^{2}}+1-16y+64{{z}^{2}}-16z+1=1 \\\ & 32{{x}^{2}}+32{{y}^{2}}+32{{z}^{2}}-8x-8y-8z+2=0 \\\ & \therefore 32({{x}^{2}}+{{y}^{2}}+{{z}^{2}})-8\left( x+y+z \right)+1=0 \\\ \end{aligned}$$ Hence the correct option is option (c). Note: The key step for solving this problem is the knowledge of the equation of a sphere inscribed in a tetrahedron. To evaluate the final answer, we also require the knowledge of distance of a point from the given plane. By using the correct equations and formulas we obtain the desired equation of the sphere.