# 2.2 Vectors in three dimensions  (Page 4/14)

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## Definition

A sphere    is the set of all points in space equidistant from a fixed point, the center of the sphere ( [link] ), just as the set of all points in a plane that are equidistant from the center represents a circle. In a sphere, as in a circle, the distance from the center to a point on the sphere is called the radius . Each point ( x , y , z ) on the surface of a sphere is r units away from the center ( a , b , c ) .

The equation of a circle is derived using the distance formula in two dimensions. In the same way, the equation of a sphere is based on the three-dimensional formula for distance.

## Rule: equation of a sphere

The sphere with center $\left(a,b,c\right)$ and radius $r$ can be represented by the equation

${\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}+{\left(z-c\right)}^{2}={r}^{2}.$

This equation is known as the standard equation of a sphere    .

## Finding an equation of a sphere

Find the standard equation of the sphere with center $\left(10,7,4\right)$ and point $\left(-1,3,-2\right),$ as shown in [link] . The sphere centered at ( 10 , 7 , 4 ) containing point ( −1 , 3 , −2 ) .

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

$\begin{array}{cc}\hfill r& =\sqrt{{\left(-1-10\right)}^{2}+{\left(3-7\right)}^{2}+{\left(-2-4\right)}^{2}}\hfill \\ & =\sqrt{{\left(-11\right)}^{2}+{\left(-4\right)}^{2}+{\left(-6\right)}^{2}}\hfill \\ & =\sqrt{173}.\hfill \end{array}$

The standard equation of the sphere is

${\left(x-10\right)}^{2}+{\left(y-7\right)}^{2}+{\left(z-4\right)}^{2}=173.$

Find the standard equation of the sphere with center $\left(-2,4,-5\right)$ containing point $\left(4,4,-1\right).$

${\left(x+2\right)}^{2}+{\left(y-4\right)}^{2}+{\left(z+5\right)}^{2}=52$

## Finding the equation of a sphere

Let $P=\left(-5,2,3\right)$ and $Q=\left(3,4,-1\right),$ and suppose line segment $PQ$ forms the diameter of a sphere ( [link] ). Find the equation of the sphere.

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

$\begin{array}{cc}\hfill C& =\left(\frac{-5+3}{2},\frac{2+4}{2},\frac{3+\left(-1\right)}{2}\right)\hfill \\ & =\left(-1,3,1\right).\hfill \end{array}$

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

$\begin{array}{cc}\hfill r& =\frac{1}{2}\sqrt{{\left(-5-3\right)}^{2}+{\left(2-4\right)}^{2}+{\left(3-\left(-1\right)\right)}^{2}}\hfill \\ & =\frac{1}{2}\sqrt{64+4+16}\hfill \\ & =\sqrt{21}.\hfill \end{array}$

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

Find the equation of the sphere with diameter $PQ,$ where $P=\left(2,-1,-3\right)$ and $Q=\left(-2,5,-1\right).$

${x}^{2}+{\left(y-2\right)}^{2}+{\left(z+2\right)}^{2}=14$

## Graphing other equations in three dimensions

Describe the set of points that satisfies $\left(x-4\right)\left(z-2\right)=0,$ and graph the set.

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 set of points satisfying ( x − 4 ) ( z − 2 ) = 0 forms the two planes x = 4 and z = 2 .

Describe the set of points that satisfies $\left(y+2\right)\left(z-3\right)=0,$ and graph the set.

The set of points forms the two planes $y=-2$ and $z=3.$ ## Graphing other equations in three dimensions

Describe the set of points in three-dimensional space that satisfies ${\left(x-2\right)}^{2}+{\left(y-1\right)}^{2}=4,$ and graph the set.

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 $x=2\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y=1.$ The cylinder extends indefinitely in the z -direction ( [link] ). The set of points satisfying ( x − 2 ) 2 + ( y − 1 ) 2 = 4 . This is a cylinder of radius 2 centered on the line with x = 2 and y = 1 .

Describe the set of points in three dimensional space that satisfies ${x}^{2}+{\left(z-2\right)}^{2}=16,$ and graph the surface.

A cylinder of radius 4 centered on the line with $x=0\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}z=2.$ ## Working with vectors in ℝ 3

Just like two-dimensional vectors, three-dimensional vectors are quantities with both magnitude and direction, and they are represented by directed line segments (arrows). With a three-dimensional vector, we use a three-dimensional arrow.

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