Algebra Examples

$(7, 7, 7)$ , $(6, 6, 6)$

Step 1

To find the distance between two 3d points, square the difference of the x, y, and z points. Then, sum them and take the square root.

$\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$

Step 2

Replace $x_{1}$ , $x_{2}$ , $y_{1}$ , $y_{2}$ , $z_{1}$ , and $z_{2}$ with the corresponding values.

$D i s t a n c e = \sqrt{{(6 - 7)}^{2} + {(6 - 7)}^{2} + {(6 - 7)}^{2}}$

Step 3

Simplify the expression $\sqrt{{(6 - 7)}^{2} + {(6 - 7)}^{2} + {(6 - 7)}^{2}}$ .

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Step 3.1

Subtract $7$ from $6$ .

$D i s t a n c e = \sqrt{{(- 1)}^{2} + {(6 - 7)}^{2} + {(6 - 7)}^{2}}$

Step 3.2

Raise $- 1$ to the power of $2$ .

$D i s t a n c e = \sqrt{1 + {(6 - 7)}^{2} + {(6 - 7)}^{2}}$

Step 3.3

Subtract $7$ from $6$ .

$D i s t a n c e = \sqrt{1 + {(- 1)}^{2} + {(6 - 7)}^{2}}$

Step 3.4

Raise $- 1$ to the power of $2$ .

$D i s t a n c e = \sqrt{1 + 1 + {(6 - 7)}^{2}}$

Step 3.5

Subtract $7$ from $6$ .

$D i s t a n c e = \sqrt{1 + 1 + {(- 1)}^{2}}$

Step 3.6

Raise $- 1$ to the power of $2$ .

$D i s t a n c e = \sqrt{1 + 1 + 1}$

Step 3.7

Add $1$ and $1$ .

$D i s t a n c e = \sqrt{2 + 1}$

Step 3.8

Add $2$ and $1$ .

$D i s t a n c e = \sqrt{3}$

Step 4

The distance between $(7, 7, 7)$ and $(6, 6, 6)$ is $\sqrt{3}$ .

$\sqrt{3} \approx 1.7320508$

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