Find the vector and cartesian equation of the sphere on the join of the points $A$ and $B$ having position vectors $\;$ $2 \hat{i} + 6 \hat{j} - 7 \hat{k}$ $\;$ and $\;$ $-2 \hat{i} + 4 \hat{j} - 3 \hat{k}$ $\;$ respectively as diameter. Find also the center and radius of the sphere.
Position vector of point $A = \overrightarrow{a} = 2 \hat{i} + 6 \hat{j} - 7 \hat{k}$
Position vector of point $B = \overrightarrow{b} = -2 \hat{i} + 4 \hat{j} - 3 \hat{k}$
Let position vector of any point on the required sphere be $= \overrightarrow{r}$
Vector equation of the required sphere is
$\left(\overrightarrow{r} - \overrightarrow{a}\right) \cdot \left(\overrightarrow{r} - \overrightarrow{b}\right) = 0$
i.e. $\;$ $\left[\overrightarrow{r} - \left(2 \hat{i} + 6 \hat{j} - 7 \hat{k}\right)\right] \cdot \left[\overrightarrow{r} - \left(-2 \hat{i} + 4 \hat{j} - 3 \hat{k}\right)\right] = 0$ $\;\;\; \cdots \; (1)$
Equation $(1)$ is the required vector equation of the sphere.
Put $\;$ $\overrightarrow{r} = x \hat{i} + y \hat{j} + z \hat{k}$ $\;$ on equation $(1)$. We have,
$\left[\left(x \hat{i} + y \hat{j} + z \hat{k}\right) - \left(2 \hat{i} + 6 \hat{j} - 7 \hat{k}\right)\right] \cdot \left[\left(x \hat{i} + y \hat{j} + z \hat{k}\right) - \left(-2 \hat{i} + 4 \hat{j} - 3 \hat{k}\right)\right] = 0$
i.e. $\;$ $\left[\left(x - 2\right) \hat{i} + \left(y - 6\right) \hat{j} + \left(z + 7\right) \hat{k}\right] \cdot \left[\left(x + 2\right) \hat{i} + \left(y - 4\right) \hat{j} + \left(z + 3\right) \hat{k}\right] = 0$
i.e. $\;$ $x^2 - 4 + y^2 - 10 y + 24 + z^2 + 10 z + 21 = 0$
i.e. $\;$ $x^2 + y^2 + z^2 -10 y + 10z + 41 = 0$ $\;\;\; \cdots \; (2)$
Equation $(2)$ is the required cartesian equation of the sphere.
Comparing equation $(2)$ with the general equation of sphere
$x^2 + y^2 + z^2 + 2ux + 2 vy + 2 wz + d = 0$ $\;$ gives
$u = 0, \;\; v = -5, \;\; w = 5, \;\; d = 41$
Center of the sphere $= \left(-u, -v, -w\right) = \left(0, 5, -5\right)$
$\begin{aligned}
\text{Radius of sphere} & = \sqrt{u^2 + v^2 + w^2 -d} \\\\
& = \sqrt{\left(0\right)^2 + \left(-5\right)^2 + \left(5\right)^2 - 41} \\\\
& = \sqrt{9} = 3 \; units
\end{aligned}$