Find the distance from the point (3, 1, 2) to the point (1, 4, -1).

Step-by-step solution

1. 1

   Use the three-dimensional distance formula. It is the Pythagorean theorem applied twice — once in the horizontal plane, once vertically — which is why a third squared term simply joins the other two:

   $d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$

2. 2

   Compute the three coordinate differences. Subtracting in a consistent order (second point minus first):

   $\Delta x=1-3=-2,\qquad \Delta y=4-1=3,\qquad \Delta z=-1-2=-3$

   The order does not ultimately matter, because every difference is about to be squared — but mixing the order between coordinates does.

3. 3

   Square each difference.

   $(-2)^{2}=4,\qquad 3^{2}=9,\qquad (-3)^{2}=9$

   Note $(-3)^{2}=+9$: squaring is what makes the distance independent of direction.

4. 4

   Add and take the square root.

   $d=\sqrt{4+9+9}=\sqrt{22}$

   Since $22=2\times 11$ has no square factor, $\sqrt{22}$ cannot be simplified further.

5. 5

   Give the decimal and sanity-check it. $\sqrt{22}\approx 4.6904$. As a check, the largest single coordinate gap is $3$, and the distance must exceed it but be less than $2+3+3=8$; $4.69$ sits comfortably in that window ✓.