Question

What is the distance between `(–2, 1, 3)` and `(8, 6, 0)`?

Answer 1

`"Distance" = 11.6" units to 3 significant figures"`

Explanation:

First, calculate your distance per dimension:

• `x : 8+2 = 10`
• `y : 6-1 = 5`
• `z : 3+-0 = 3`

Next, apply 3D Pythagoras' theorem:

`h^2 = a^2 + b^2 + c^2`

Where:

• `h^2` is the square of the distance between two points
• `a^2`, `b^2`, and `c^2` are the calculated dimensional distances

We can adjust the theorem to solve directly for `h`:

`h = sqrt(a^2 + b^2 + c^2)`

Finally, substitute your values into the equation and solve:

`h = sqrt(10^2 + 5^2 + 3^2)`
`h = sqrt(100 + 25 + 9)`
`h = sqrt(134)`
`h = 11.5758369028 = 11.6" to 3 significant figures"`

`:. "Distance" = 11.6" units to 3 significant figures"`

Answer 2

`sqrt(134)`

Explanation:

The distance formula for Cartesian coordinates is

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2`
Where `x_1, y_1,z_1`, and`x_2, y_2,z_2` are the Cartesian coordinates of two points respectively.
Let `(x_1,y_1,z_1)` represent `(-2,1,3)` and `(x_2,y_2,z_2)` represent `(8,6,0)`.
`implies d=sqrt((8-(-2))^2+(6-1)^2+(0-3)^2`
`implies d=sqrt((10)^2+(5)^2+(-3)^2`
`implies d=sqrt(100+25+9`
`implies d=sqrt(134`

Hence the distance between the given points is `sqrt(134)`.