Chapter 10.1, Problem 61E

To determine

Tofind:The standard form of the equation of sphere with the endpoints of a diameter as $\left(3,0,0\right)$ and $\left(0,0,6\right)$ .

Answer to Problem 61E

The standard form of the equation of sphere with endpoints of the diameter as $\left(3,0,0\right)$ and $\left(0,0,6\right)$ is ${\left(x-\frac{3}{2}\right)}^2+y^2+{\left(z-3\right)}^2={\left(\frac{3\sqrt{5}}{2}\right)}^2$ .

Explanation of Solution

Given information:

The vertices of endpoints of a diameter are $\left(3,0,0\right)$ and $\left(0,0,6\right)$ .

Calculation:

The general standard form of the equation of sphere is ${\left(x-h\right)}^2+{\left(y-k\right)}^2+{\left(z-j\right)}^2=r^2$ with center $\left(h,k,j\right)$ and $r$ radius.

The center of the sphere is the midpoint of the vertices of endpoints of diameter.

Use the formula for midpoint.

  $\text{Midpoint}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

Substitute $3$ , $0$ , $0$ for $x_1$ , $y_1$ , $z_1$ and $0$ , $0$ , $6$ for $x_2$ , $y_2$ , $z_2$ in the midpoint formula.

  $\begin{array}{c}\text{Midpoint}=\left(\frac{3+0}{2},\frac{0+0}{2},\frac{0+6}{2}\right)\\ =\left(\frac{3}{2},0,\frac{6}{2}\right)\\ =\left(\frac{3}{2},0,3\right)\end{array}$

So, the center of the sphere is $\left(\frac{3}{2},0,3\right)$ .

Use the distance formula for the length of diameter.

  $\text{Distance}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Substitute $3$ , $0$ , $0$ for $x_1$ , $y_1$ , $z_1$ and $0$ , $0$ , $6$ for $x_2$ , $y_2$ , $z_2$ in the distance formula.

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(0-3\right)}^2+{\left(0-0\right)}^2+{\left(6-0\right)}^2}\\ =\sqrt{{\left(-3\right)}^2+0+6^2}\\ =\sqrt{9+0+36}\\ =\sqrt{45}\\ =3\sqrt{5}\end{array}$

So, the length of the diameter is $3\sqrt{5}$ .

The radius of sphere is half the length of the diameter.

  $r=\frac{3\sqrt{5}}{2}$

Substitute $\frac{3}{2}$ , $0$ , $3$ for $h$ , $k$ , $j$ and $\frac{3\sqrt{5}}{2}$ for $r$ in the general equation of sphere.

  $\begin{array}{c}{\left(x-h\right)}^2+{\left(y-k\right)}^2+{\left(z-j\right)}^2=r^2\\ {\left(x-\frac{3}{2}\right)}^2+{\left(y-0\right)}^2+{\left(z-3\right)}^2={\left(\frac{3\sqrt{5}}{2}\right)}^2\\ {\left(x-\frac{3}{2}\right)}^2+y^2+{\left(z-3\right)}^2={\left(\frac{3\sqrt{5}}{2}\right)}^2\end{array}$

Therefore, the standard form of the equation of the sphere with endpoints of the diameter as $\left(3,0,0\right)$ and $\left(0,0,6\right)$ is ${\left(x-\frac{3}{2}\right)}^2+y^2+{\left(z-3\right)}^2={\left(\frac{3\sqrt{5}}{2}\right)}^2$ .