3.04 Two points in 3D
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 | $=$= | $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$√(x2−x1)2+(y2−y1)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).
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 | $=$= | $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$√(x2−x1)2+(y2−y1)2+(z2−z1)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 | $=$= | $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$√(x2−x1)2+(y2−y1)2+(z2−z1)2 |
| $=$= | $\sqrt{(4-3)^2+(5-0)^2+(-6-(-2))^2}$√(4−3)2+(5−0)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 | $=$= | $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$√(x2−x1)2+(y2−y1)2+(z2−z1)2 |
| $=$= | $\sqrt{(3-2)^2+(3-(-5))^2+(-2-3)^2}$√(3−2)2+(3−(−5))2+(−2−3)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}$3√10 |
So the exact distance between the two points is $3\sqrt{10}$3√10 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.
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) |