Finding the length between two points in 3 Dimensions with Pythagorean's Theorem

Question

I am a grade 11 student and I have to learn vectors for the IB exam. I know that to find the distance of a vector between two points in a 3 dimensional space for lets say point A and B, then you would use the formula \begin{align*} |AB|=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}. \end{align*} I am having difficulty visualizing and understanding why the coordinates of the points are being subtracted here. I really appreciate the feedback. Sorry for the low quality of the link, but I guess I don't have enough reputation yet to insert images directly. Also, in the formula all the subscripts should be reversed. So x1 becomes x2 and vice versa.

Answer 1

I'm assuming you know why the distance between $(x_1, y_1)$ and $(x_2,y_2)$ is $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Refer to the picture below

Clearly, the vecctor representing $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ lies in the $XY$ plane.

$z_2-z_1$ lies on the $Z$ axis.

The normal vector to the $XY$ plane is $(0,0,k)$, indicating that the vector representing $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ and $z_2-z_1$ are perpendicular.

Now, you have two vectors i.e $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ and $z_2-z_1$ which are mutually perpendicular and hence you can apply the Pythagorean's Theorem here.

This gives distance between $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ = $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_1)^2}$