2.2 Vectors in three dimensions  (Page 4/14)

Use the distance formula to find the radius $r$ of the sphere:

The standard equation of the sphere is

Since $PQ$ is a diameter of the sphere, we know the center of the sphere is the midpoint of $PQ.$ Then,

Furthermore, we know the radius of the sphere is half the length of the diameter. This gives

Then, the equation of the sphere is ${\left(x+1\right)}^2+{\left(y-3\right)}^2+{\left(z-1\right)}^2=21.$

We must have either $x-4=0$ or $z-2=0,$ so the set of points forms the two planes $x=4$ and $z=2$ ( [link] ).

The x - and y -coordinates form a circle in the xy -plane of radius $2,$ centered at $\left(2,1\right).$ Since there is no restriction on the z -coordinate, the three-dimensional result is a circular cylinder of radius $2$ centered on the line with $x=2\text{and}y=1.$ The cylinder extends indefinitely in the z -direction ( [link] ).