Answer
(a) $x^2+y^2+(z-1)^2=9$
(b) center $(0,0,1)$, radius $3$.
(c) See graph and explanations.
(d) $(x-1)^2+(y)^2+(z-\frac{7}{2})^2=\frac{9}{4}$, center $(1,0,\frac{7}{2})$ radius $\frac{3}{2}$
Work Step by Step
(a) Based on the quantities given, we have $\vec r-\vec u=\langle x-2, y-2, z-2 \rangle$ and $\vec r-\vec v=\langle x+2, y+2, z \rangle$. The dot product is $(\vec r-\vec u)\cdot(\vec r-\vec v)=x^2-4+y^2-4+z^2-2z=x^2+y^2+(z-1)^2-9$. Let this dot product be zero, we have $x^2+y^2+(z-1)^2=9$ which represents a sphere.
(b) It can be seen that the last equation above represents a sphere centered at $(0,0,1)$ with radius $3$.
(c) The figure shows the vectors $\vec u, \vec v, \vec r$, the dashed arrows show the vectors $\vec r-\vec u, \vec r-\vec v$. The equation that the dot product of these two vectors is zero indicates that they are perpendicular to each other. Imagine that the connection of the end points of $\vec u, \vec v$ forms the diameter of a sphere, because the two dashed vectors form a right angle, the end point of $\vec r$ will be on the sphere centered at the midpoint of the diameter connection.
(d) Given the two new end points for vectors $\vec u, \vec v$, we can get the equation as $\langle x-2, y+1, z-4 \rangle\cdot\langle x-0, y-1, z-3 \rangle=0$ which leads to $x^2-2x+y^2-1+z^2-7z+12=0$ or
$(x-1)^2+(y)^2+(z-\frac{7}{2})^2=1-12+1+(\frac{7}{2})^2=\frac{9}{4}$. Thus it is a sphere centered at $(1,0,\frac{7}{2})$ with radius $\frac{3}{2}$
