Algebra Examples

$(- 1, - 3, 6)$ , $(- 1, 6, - 4)$

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.

$D i s t a n c e = \sqrt{{(- 1 - (- 1))}^{2} + {(6 - (- 3))}^{2} + {(- 4 - 6)}^{2}}$

Step 3

Simplify the expression $\sqrt{{(- 1 - (- 1))}^{2} + {(6 - (- 3))}^{2} + {(- 4 - 6)}^{2}}$ .

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Step 3.1

Simplify each term.

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Step 3.1.1

Multiply $- 1$ by each element of the matrix.

$D i s t a n c e = \sqrt{{(- 1 + 1)}^{2} + {(6 - (- 3))}^{2} + {(- 4 - 6)}^{2}}$

Step 3.1.2

Multiply $- 1$ by $- 1$ .

$D i s t a n c e = \sqrt{{(- 1 + 1)}^{2} + {(6 - (- 3))}^{2} + {(- 4 - 6)}^{2}}$

Step 3.2

Simplify the expression.

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Step 3.2.1

Add $- 1$ and $1$ .

$D i s t a n c e = \sqrt{0^{2} + {(6 - (- 3))}^{2} + {(- 4 - 6)}^{2}}$

Step 3.2.2

Raising $0$ to any positive power yields $0$ .

$D i s t a n c e = \sqrt{0 + {(6 - (- 3))}^{2} + {(- 4 - 6)}^{2}}$

Step 3.3

Simplify each term.

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Step 3.3.1

Multiply $- 1$ by each element of the matrix.

$D i s t a n c e = \sqrt{0 + {(6 + 3)}^{2} + {(- 4 - 6)}^{2}}$

Step 3.3.2

Multiply $- 1$ by $- 3$ .

$D i s t a n c e = \sqrt{0 + {(6 + 3)}^{2} + {(- 4 - 6)}^{2}}$

Step 3.4

Simplify the expression.

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Step 3.4.1

Add $6$ and $3$ .

$D i s t a n c e = \sqrt{0 + 9^{2} + {(- 4 - 6)}^{2}}$

Step 3.4.2

Raise $9$ to the power of $2$ .

$D i s t a n c e = \sqrt{0 + 81 + {(- 4 - 6)}^{2}}$

Step 3.4.3

Subtract $6$ from $- 4$ .

$D i s t a n c e = \sqrt{0 + 81 + {(- 10)}^{2}}$

Step 3.4.4

Raise $- 10$ to the power of $2$ .

$D i s t a n c e = \sqrt{0 + 81 + 100}$

Step 3.4.5

Add $0$ and $81$ .

$D i s t a n c e = \sqrt{81 + 100}$

Step 3.4.6

Add $81$ and $100$ .

$D i s t a n c e = \sqrt{181}$

Step 4

The distance between $(- 1, - 3, 6)$ and $(- 1, 6, - 4)$ is $\sqrt{181}$ .

$\sqrt{181} \approx 13.45362404$

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