## Distance in 3D Space

Consider points (1, 2, 3) and (4, 5, 6) exist in a three-dimensional space. What is the distance between the two points?

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__**Hint**

**Hint**

In a three-dimensional space, the distance between two points is

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

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__**Hint 2**

**Hint 2**

Since the difference between coordinates is squared, it doesn’t matter if Point 1 is assigned
$$(x_1, y_1, z_1)$$
or
$$(x_2, y_2, z_2)$$
as long as Point 2 is assigned as the other set.

In a three-dimensional space, the distance between two points is

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

Since the difference between coordinates is squared, it doesn’t matter if Point 1 is assigned
$$(x_1, y_1, z_1)$$
or
$$(x_2, y_2, z_2)$$
as long as Point 2 is assigned as the other set. Let’s arbitrarily set
$$(1, 2, 3)$$
as
$$(x_1, y_1, z_1)$$
:

$$$d=\sqrt{(4-1)^2+(5-2)^2+(6-3)^2}$$$

$$$=\sqrt{(3)^2+(3)^2+(3)^2}=\sqrt{9+9+9}=\sqrt{27}=5.2$$$

5.2