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

Dec 30, 2015

$\text{Distance" = 11.6" units to 3 significant figures}$

#### Explanation:

First, calculate your distance per dimension:

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

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:

$h = \sqrt{{10}^{2} + {5}^{2} + {3}^{2}}$
$h = \sqrt{100 + 25 + 9}$
$h = \sqrt{134}$
$h = 11.5758369028 = 11.6 \text{ to 3 significant figures}$

$\therefore \text{Distance" = 11.6" units to 3 significant figures}$

Dec 30, 2015

$\sqrt{134}$

#### 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 $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ represent $\left(- 2 , 1 , 3\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$ represent $\left(8 , 6 , 0\right)$.
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 $\sqrt{134}$.