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Standard Level

3.04 Two points in 3D

Lesson

To find the distance between points  $(x_1,y_1)$(x1,y1) and $(x_2,y_2)$(x2,y2) in two dimensions we use the formula: 

$d$d $=$= $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$(x2x1)2+(y2y1)2

 

In three dimensions, we need to add a third variable,  $z$z. This adds a third dimension to the Cartesian plane as shown: 

3 dimensional Cartesian plane

A point on the 3D plane has coordinates: $(x,y,z)$(x,y,z).

Remember!

So to find the distance between $(x_1,y_1,z_1)$(x1,y1,z1) and $(x_2,y_2,z_2)$(x2,y2,z2) on the 3D plane we would use the following formula:

$d$d $=$= $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$(x2x1)2+(y2y1)2+(z2z1)2
 

Worked examples

Example 1

Find the distance between the points $(3,0,-2)$(3,0,2) and $(4,5,-6)$(4,5,6).

Think: We need to substitute these coordinates into the above formula. You can choose which point you let be  $(x_1,y_1,z_1)$(x1,y1,z1) and $(x_2,y_2,z_2)$(x2,y2,z2). In the working, we have let  $(x_1,y_1,z_1)=(3,0,-2)$(x1,y1,z1)=(3,0,2), and $(x_2,y_2,z_2)=(4,5,-6).$(x2,y2,z2)=(4,5,6).

Do:

$d$d $=$= $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$(x2x1)2+(y2y1)2+(z2z1)2
  $=$= $\sqrt{(4-3)^2+(5-0)^2+(-6-(-2))^2}$(43)2+(50)2+(6(2))2
  $=$= $\sqrt{1^2+5^2+(-4)^2}$12+52+(4)2
  $=$= $\sqrt{1+25+16}$1+25+16
  $=$= $\sqrt{42}$42
  $\approx$ $6.48$6.48

 

So the distance between the two points is approximately  $6.48$6.48 units.

Example 2

Find the exact distance between the two points shown in the 3D Cartesian plane:

 

Think: First we should let  $(x_1,y_1,z_1)=(2,-5,3)$(x1,y1,z1)=(2,5,3), and $(x_2,y_2,z_2)=(3,3,-2).$(x2,y2,z2)=(3,3,2).

Do:

 

$d$d $=$= $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$(x2x1)2+(y2y1)2+(z2z1)2
  $=$= $\sqrt{(3-2)^2+(3-(-5))^2+(-2-3)^2}$(32)2+(3(5))2+(23)2
  $=$= $\sqrt{1^2+8^2+(-5)^2}$12+82+(5)2
  $=$= $\sqrt{1+25+16}$1+25+16
  $=$= $\sqrt{90}$90
  $=$= $3\sqrt{10}$310

 

So the exact distance between the two points is  $3\sqrt{10}$310 units.

Midpoint of two points in 3D

 

We can also find the midpoint of two points in the 3D plane by using the  $z$z-coordinate.

Remember!

So to find the midpoint between $(x_1,y_1,z_1)$(x1,y1,z1) and $(x_2,y_2,z_2)$(x2,y2,z2) on the 3D plane we would use the following formula:

Midpoint $=$= $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$(x1+x22,y1+y22,z1+z22)

 

Example 3

Find the coordinates of the midpoint between the points shown in the diagram:

 

 

Think: First we should let  $(x_1,y_1,z_1)=(1,6,0)$(x1,y1,z1)=(1,6,0), and $(x_2,y_2,z_2)=(-2,5,4).$(x2,y2,z2)=(2,5,4).

Do:

Midpoint $=$= $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$(x1+x22,y1+y22,z1+z22)
  $=$= $\left(\frac{1+(-2)}{2},\frac{6+5}{2},\frac{0+4}{2}\right)$(1+(2)2,6+52,0+42)
  $=$= $\left(\frac{-1}{2},\frac{11}{2},2\right)$(12,112,2)

 

 

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