Find the vector and cartesian equation of a sphere with center having position vector $2 \hat{i} - \hat{j} + 3 \hat{k}$ and radius 4 units.
Position vector of center of sphere $= \overrightarrow{c} = 2 \hat{i} - \hat{j} + 3 \hat{k}$
Radius of sphere $= a = 4 \;$ units
Position vector of any point on the sphere $= \overrightarrow{r}$
Vector equation of sphere is: $\;$ $\left|\overrightarrow{r} - \overrightarrow{c}\right| = a$
i.e. $\;$ $\left|\overrightarrow{r} - \left(2 \hat{i} - \hat{j} + 3 \hat{k}\right)\right| = 4$ $\;\;\; \cdots \; (1)$
Equation $(1)$ is the required vector equation of the sphere.
Put $\;$ $\overrightarrow{r} = x \hat{i} + y \hat{j} + z \hat{k}$ $\;$ in equation $(1)$. We have,
$\left|\left(x \hat{i} + y \hat{j} + z \hat{k}\right) - \left(2 \hat{i} - \hat{j} + 3 \hat{k}\right) \right| = 4$
i.e. $\;$ $\left|\left(x - 2\right) \hat{i} + \left(y + 1\right) \hat{j} + \left(z - 3\right) \hat{k}\right| = 4$
i.e. $\;$ $\left(x - 2\right)^2 + \left(y + 1\right)^2 + \left(z - 3\right)^2 = 4^2$
i.e. $\;$ $x^2 + y^2 + z^2 - 2x + 2y - 6z - 2 = 0$ $\;\;\; \cdots \; (2)$
Equation $(2)$ is the required cartesian equation of the sphere.