Find the distance between the points (1802, 139, -4524) and (1885, 132, -4592).

Step-by-step solution

1. 1

   Write the three-dimensional distance formula. It is the Pythagorean theorem applied twice:

   $d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}$

   Only the differences matter, so the large absolute values of the coordinates are irrelevant to the answer.

2. 2

   Compute the three differences.

   $\Delta x=1885-1802=83,\qquad \Delta y=132-139=-7,\qquad \Delta z=-4592-(-4524)=-68$

   The $z$ difference is the one to watch: subtracting a negative gives $-4592+4524=-68$.

3. 3

   Square and add.

   $83^2=6889,\qquad (-7)^2=49,\qquad (-68)^2=4624$

   $6889+49+4624=11562$

   Squaring removes the signs, so the direction of travel does not affect the distance.

4. 4

   Take the square root and try to simplify.

   $d=\sqrt{11562}$

   Factoring, $11562=2\cdot 3\cdot 41\cdot 47$ — every prime appears to the first power, so there is no perfect-square factor to pull out. The radical is already in simplest form.

5. 5

   Give the decimal value and sanity-check it.

   $d=\sqrt{11562}\approx 107.53$

   The largest single difference is $83$, and the distance must exceed it but be less than the sum $83+7+68=158$; $107.53$ sits comfortably inside that window $\checkmark$. Squaring back, $107.53^2=11562.7$, consistent to rounding.