Question

Find the vector and Cartesian equations of the sphere on the join of the points $A$ and $B$ having position vectors $2\overrightarrow{i}+6\overrightarrow{j}-7\overrightarrow{k}$ and $-2\overrightarrow{i}+4\overrightarrow{j}-3\overrightarrow{k}$ respectively as a diameter. Find also the centre and radius of the sphere.

Answer 1

Vector equation of a sphere joining the points $A$ and $B$ whose P.V.S. are $\overrightarrow{a}$ and $\overrightarrow{b}$ is
$(\overrightarrow{r}-\overrightarrow{a}).(\overrightarrow{r}-\overrightarrow{b})=0$
Here $\overrightarrow{a}=2\overrightarrow{i}+6\overrightarrow{j}-7\overrightarrow{k}$ and $\overrightarrow{b}=-2\overrightarrow{i}+4\overrightarrow{j}-3\overrightarrow{k}$
$[\overrightarrow{r}-(2\overrightarrow{i}+6\overrightarrow{j}-7\overrightarrow{k}\left)\right].[\overrightarrow{r}-(2\overrightarrow{i}+4\overrightarrow{j}-3\overrightarrow{k}\left)\right]=0$
Cartesian form
Let $\overrightarrow{r}=x\overrightarrow{i}+y\overrightarrow{j}2\overrightarrow{k}$
$\overrightarrow{r}-\overrightarrow{a}=(x-2)\overrightarrow{i}+(y-6)\overrightarrow{j}+(z+7)\overrightarrow{k}$
$\overrightarrow{r}-\overrightarrow{b}=(x+2)\overrightarrow{i}+(y-4)\overrightarrow{j}+(z+3)\overrightarrow{k}$
$(\overrightarrow{r}-\overrightarrow{a}).(\overrightarrow{r}-\overrightarrow{b})=0$
$(x-2)(x+2)+(y-6)(y-4)+(z+7)(z+3)=0$
$x^2+y^2+z^2-10y+10z+41=0$
Comparing with
$x^2+y^2+z^2+2xx+2xy+2xz+d=0$
$x=0;v=-5;w=5;d=41$
centre is $(-x,-v,-w)=(0,5,-5)$
radius is $\sqrt{x^2+v^2+w^2-d}$
$=\sqrt{25+25-41}=\sqrt{9}=3$ units