Precalculus Examples

$(5, 10, - 3)$ , $(- 1, 4, - 2)$

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.

$D i s t a n c e = \sqrt{{(- 1 - 5)}^{2} + {(4 - 10)}^{2} + {(- 2 - (- 3))}^{2}}$

Step 3

Simplify the expression $\sqrt{{(- 1 - 5)}^{2} + {(4 - 10)}^{2} + {(- 2 - (- 3))}^{2}}$ .

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Step 3.1

Simplify the expression.

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Step 3.1.1

Subtract $5$ from $- 1$ .

$D i s t a n c e = \sqrt{{(- 6)}^{2} + {(4 - 10)}^{2} + {(- 2 - (- 3))}^{2}}$

Step 3.1.2

Raise $- 6$ to the power of $2$ .

$D i s t a n c e = \sqrt{36 + {(4 - 10)}^{2} + {(- 2 - (- 3))}^{2}}$

Step 3.1.3

Subtract $10$ from $4$ .

$D i s t a n c e = \sqrt{36 + {(- 6)}^{2} + {(- 2 - (- 3))}^{2}}$

Step 3.1.4

Raise $- 6$ to the power of $2$ .

$D i s t a n c e = \sqrt{36 + 36 + {(- 2 - (- 3))}^{2}}$

Step 3.2

Simplify each term.

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Step 3.2.1

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

$D i s t a n c e = \sqrt{36 + 36 + {(- 2 + 3)}^{2}}$

Step 3.2.2

Multiply $- 1$ by $- 3$ .

$D i s t a n c e = \sqrt{36 + 36 + {(- 2 + 3)}^{2}}$

Step 3.3

Simplify the expression.

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Step 3.3.1

Add $- 2$ and $3$ .

$D i s t a n c e = \sqrt{36 + 36 + 1^{2}}$

Step 3.3.2

One to any power is one.

$D i s t a n c e = \sqrt{36 + 36 + 1}$

Step 3.3.3

Add $36$ and $36$ .

$D i s t a n c e = \sqrt{72 + 1}$

Step 3.3.4

Add $72$ and $1$ .

$D i s t a n c e = \sqrt{73}$

Step 4

The distance between $(5, 10, - 3)$ and $(- 1, 4, - 2)$ is $\sqrt{73}$ .

$\sqrt{73} \approx 8.54400374$