What is the distance between $(2, 3, 5)$ and $(2, 7, 4)$ ?

Question

1433 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-13T23:31:25

$sqrt(17) ~~ 4.123$

The distance between $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula:

$d = sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)$

So in our case:

$d = sqrt((2-2)^2+(7-3)^2+(4-5)^2)$

$=sqrt(0+16+1)=sqrt(17) ~~ 4.123$

Distance formula for $3$ dimensions

Here's how to derive the distance formula for $3$ dimensions from Pythagoras theorem:

Given points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ , consider the triangle with vertices:

$(x_1, y_1, z_1)$ , $(x_2, y_1, z_1)$ , $(x_2, y_2, z_1)$

This is a right angled triangle with legs:

$(x_1, y_1, z_1) (x_2, y_1, z_1)$ of length $abs(x_2-x_1)$

$(x_2, y_1, z_1) (x_2, y_2, z_1)$ of length $abs(y_2-y_1)$

and hypotenuse:

$(x_1, y_1, z_1) (x_2, y_2, z_1)$ of length $sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

Next consider the triangle with vertices:

$(x_1, y_1, z_1)$ , $(x_2, y_2, z_1)$ , $(x_2, y_2, z_2)$

This is a right angled triangle with legs:

$(x_1, y_1, z_1) (x_2, y_2, z_1)$ of length $sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$(x_2, y_2, z_1) (x_2, y_2, z_2)$ of length $abs(z_2-z_1)$

and hypotenuse:

$(x_1, y_1, z_1) (x_1, y_2, z_2)$ of length:

$sqrt((sqrt((x_2-x_1)^2+(y_2-y_1)^2))^2+(z_2-z_1)^2)$

$=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)$

What is the distance between (2, 3, 5) (2, 3, 5) and (2, 7, 4) (2, 7, 4) ?