Find the Distance (5,4,3) , (2,7,4)

$(5, 4, 3)$ , $(2, 7, 4)$

Step 1

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

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

Step 2

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

$D i s t a n c e = \sqrt{{((2) - (5))}^{2} + {((7) - (4))}^{2} + {((4) - (3))}^{2}}$

Step 3

Multiply $- 1$ by $5$ .

$D i s t a n c e = \sqrt{{(2 - 5)}^{2} + {((7) - (4))}^{2} + {((4) - (3))}^{2}}$

Subtract $5$ from $2$ .

$D i s t a n c e = \sqrt{{(- 3)}^{2} + {((7) - (4))}^{2} + {((4) - (3))}^{2}}$

Raise $- 3$ to the power of $2$ .

$D i s t a n c e = \sqrt{9 + {((7) - (4))}^{2} + {((4) - (3))}^{2}}$

Multiply $- 1$ by $4$ .

$D i s t a n c e = \sqrt{9 + {(7 - 4)}^{2} + {((4) - (3))}^{2}}$

Subtract $4$ from $7$ .

$D i s t a n c e = \sqrt{9 + 3^{2} + {((4) - (3))}^{2}}$

Raise $3$ to the power of $2$ .

$D i s t a n c e = \sqrt{9 + 9 + {((4) - (3))}^{2}}$

Multiply $- 1$ by $3$ .

$D i s t a n c e = \sqrt{9 + 9 + {(4 - 3)}^{2}}$

Subtract $3$ from $4$ .

$D i s t a n c e = \sqrt{9 + 9 + 1^{2}}$

One to any power is one.

$D i s t a n c e = \sqrt{9 + 9 + 1}$

Add $9$ and $9$ .

$D i s t a n c e = \sqrt{18 + 1}$

Add $18$ and $1$ .

$D i s t a n c e = \sqrt{19}$

Step 4

The distance between $(5, 4, 3)$ and $(2, 7, 4)$ is $\sqrt{19}$ .

$\sqrt{19} \approx 4.35889894$