# Difference between revisions of "Math 22 The Three-Dimensional Coordinate System"

## The Distance and Midpoint Formulas

 The distance $d$ between the points $(x_{1},x_{2},x_{3})$ and $(x_{2},y_{2},z_{2})$ is

$d={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}}$ Exercises 1 Find the distance between two points

1) $(4,2,3)$ and $(1,2,0)$ Solution:
$d={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}}={\sqrt {(1-4)^{2}+(2-2)^{2}+(0-3)^{2}}}={\sqrt {18}}$ 2) $(1,2,4)$ and $(2,5,1)$ Solution:
$d={\sqrt {(2-1)^{2}+(5-2)^{2}+(1-4)^{2}}}={\sqrt {19}}$ ## Midpoint Formula in Space

 The midpoint of the line segment joining the points $(x_{1},x_{2},x_{3})$ and $(x_{2},y_{2},z_{2})$ is

${\text{Midpoint}}=({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}},{\frac {z_{1}+z_{2}}{2}})$ Exercises 2 Find the midpoint of two points below:

1) $(4,2,3)$ and $(1,2,0)$ Solution:
${\text{Midpoint}}=({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}},{\frac {z_{1}+z_{2}}{2}})=({\frac {4+1}{2}},{\frac {2+2}{2}},{\frac {3+0}{2}})=({\frac {5}{2}},2,{\frac {3}{2}})$ 2) $(1,2,4)$ and $(2,5,1)$ Solution:
${\text{Midpoint}}=({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}},{\frac {z_{1}+z_{2}}{2}})=({\frac {1+2}{2}},{\frac {2+5}{2}},{\frac {4+1}{2}})=({\frac {3}{2}},{\frac {7}{2}},{\frac {5}{2}})$ ## Standard Equation of a Sphere

 The standard equation of a sphere with center at $(h,k,j)$ and radius $r$ is:

$(x-h)^{2}+(y-k)^{2}+(z-j)^{2}=r^{2}$ Exercises 3 Find the equation of the sphere that has:

1) Center: $(1,2,0)$ and radius: $5$ Solution:
$(x-h)^{2}+(y-k)^{2}+(z-j)^{2}=r^{2}$ So, $(x-1)^{2}+(y-2)^{2}+(z-0)^{2}=5^{2}$ Therefore, $(x-1)^{2}+(y-2)^{2}+z^{2}=25$ 2) Center: $(-1,2,4)$ and radius: $2$ Solution:
$(x-h)^{2}+(y-k)^{2}+(z-j)^{2}=r^{2}$ So, $(x-(-1))^{2}+(y-2)^{2}+(z-4)^{2}=2^{2}$ Therefore, $(x+1)^{2}+(y-2)^{2}+(z-4)^{2}=4$ 