Question

To determine the equation of a sphere if the end points of diameter are \((2,1,4)\) and\((4,3,10)\).

Short answer

The equation of a sphere is \({(x - 3)^2} + {(y - 2)^2} + {(z - 7)^2} = 11\).

Step-by-step solution

Formulae and concept used to determine the equation of sphere

1. The distance between two points\({P_1}\left( {{x_1},{y_1},{z_1}} \right)\)and\({P_2}\left( {{x_2},{y_2},{z_2}} \right)\)is\(\left| {{P_1}{P_2}} \right| = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} .\)

2. The equation of a sphere with centre\(C(h,k,l)\)and radius\(r\)is\({(x - h)^2} + {(y - k)^2} + {(z - l)^2} = {r^2}\).

Calculate the radius and centre of sphere (with given endpoints of diameter of that sphere)

The given two end points of the diameter of a sphere are \((2,1,4)\) and \((4,3,10)\).

In order to find the radius of a sphere, use the distance formula.

Let the two end points of diameter be \(P(2,1,4)\) and \(Q(4,3,10)\) as follows.

Substitute \(P\left( {{x_1},{y_1},{z_1}} \right) = P(2,1,4)\) and \(Q\left( {{x_2},{y_2},{z_2}} \right) = Q(4,3,10)\) in the distance formula and find \(PQ\).

\(\begin{aligned}{l}|PQ| = \sqrt {{{(4 - 2)}^2} + {{(3 - 1)}^2} + {{(10 - 4)}^2}} \\ = \sqrt {4 + 4 + 36} \\ = \sqrt {44} \\ = 2\sqrt {11} \end{aligned}\)

Thus, the length of the diameter \(PQ\) in the sphere is\(2\sqrt {11} \).

Then, the radius of a sphere is\(\frac{{2\sqrt {11} }}{2} = \sqrt {11} \).

Note that, mid points of two end points is the centre of the sphere that can be obtain by\(\left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2},\frac{{{z_1} + {z_2}}}{2}} \right)\).

Hence, the centre of a sphere with two end points \(P\left( {{x_1},{y_1},{z_1}} \right) = P(2,1,4)\) and \(Q\left( {{x_2},{y_2},{z_2}} \right) = Q(4,3,10)\) is

\(\left( {\frac{{2 + 4}}{2},\frac{{1 + 3}}{2},\frac{{4 + 10}}{2}} \right) = (3,2,7)\).

Determine the equation of the sphere

Apply the formula (2) and obtain the equation of sphere with centre \((3,2,7)\) and radius \(\sqrt {11} \).\(\begin{aligned}{l}{(x - 3)^2} + {(y - 2)^2} + {(z - 7)^2} = {(\sqrt {11} )^2}\\{(x - 3)^2} + {(y - 2)^2} + {(z - 7)^2} = 11\end{aligned}\)

Thus, the equation of a sphere is\({(x - 3)^2} + {(y - 2)^2} + {(z - 7)^2} = 11\).