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

## The Distance and Midpoint Formulas

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

${\displaystyle d={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}}}$


Exercises 1 Find the distance between two points

1) ${\displaystyle (4,2,3)}$ and ${\displaystyle (1,2,0)}$

Solution:
${\displaystyle d={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}}={\sqrt {(1-4)^{2}+(2-2)^{2}+(0-3)^{2}}}={\sqrt {18}}}$

2) ${\displaystyle (1,2,4)}$ and ${\displaystyle (2,5,1)}$

Solution:
${\displaystyle d={\sqrt {(2-1)^{2}+(5-2)^{2}+(1-4)^{2}}}={\sqrt {19}}}$

## Midpoint Formula in Space

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

${\displaystyle {\text{Midpoint}}=({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}},{\frac {z_{1}+z_{2}}{2}})}$


Exercises 2 Find the midpoint of two points below:

1) ${\displaystyle (4,2,3)}$ and ${\displaystyle (1,2,0)}$

Solution:
${\displaystyle {\text{Midpoint}}=({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}},{\frac {z_{1}+z_{2}}{2}})=({\frac {4+1}{2}},{\frac {2+2}{2}},{\frac {3+0}{2}})=({\frac {5}{2}},2,{\frac {3}{2}})}$

2) ${\displaystyle (1,2,4)}$ and ${\displaystyle (2,5,1)}$

Solution:
${\displaystyle {\text{Midpoint}}=({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}},{\frac {z_{1}+z_{2}}{2}})=({\frac {1+2}{2}},{\frac {2+5}{2}},{\frac {4+1}{2}})=({\frac {3}{2}},{\frac {7}{2}},{\frac {5}{2}})}$

## Standard Equation of a Sphere

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

${\displaystyle (x-h)^{2}+(y-k)^{2}+(z-j)^{2}=r^{2}}$


Exercises 3 Find the equation of the sphere that has:

1) Center: ${\displaystyle (1,2,0)}$ and radius: ${\displaystyle 5}$

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