what is the distance between x=(x(sub)1, x(sub)2, x(sub)3) and y= (y(sub)1, y(sub)2, y(sub)3)?

Question

my teacher doesn't simply want the distance formula. He wants to prove the formula using a proof.

Answer 1

Well, the distance formula is just the Pythagorean Theorem in disguise.

I'm not going to prove the Pythagorean Theorem; if you want that there are lots of good websites out there with nice pictures and explanations.

Instead, I'll give an "intuitive idea" of why the distance formula works in three dimensions as well as two.

First, pick one of the points and imagine a plane going through it that is parallel to the familiar x-y plane, that is, the set of all points that have a z-coordinate of x₃.

Now imagine taking a flashlight and shining it perpendicular to the plane. The other point (y₁, y₂, y₃) would cast a shadow on the plane that would have coordinates (y₁, y₂, x₃), i.e. we're "projecting" the point into the plane by drawing a line through it perpendicular to the plane, and noting where that line "pierces" the plane.

Now if you draw a line from the first point (x₁, x₂, x₃) to the point (y₁, y₂, x₃), you should notice that this line would be the hypotenuse of a right triangle whose side lengths can be calculated as the difference in the x-values and the difference in the y-values. By the Pythagorean Theorem, therefore, the distance between (x₁, x₂, x₃) and (y₁, y₂, x₃) is given by:

d = √((x₁ - y₁)² + (x₂ - y₂)²)

So that's the 2-D version of the distance formula.

But if you've drawn the picture, now you should also be able to see that the real distance between (x₁, x₂, x₃) and (y₁, y₂, y₃) is ALSO the hypotenuse of a right triangle, where the base is the distance we just calculated, and the height is the difference between the z-coordinates of the two points.

d = √( [√((x₁ - y₁)² + (x₂ - y₂)²)]² + (x₃ - y₃)²)

Simplify and you get the distance formula in amazing 3D!

d = √( (x₁ - y₁)² + (x₂ - y₂)² + (x₃ - y₃)² )

If you're having trouble visualizing it, try this diagram:

http://www.ltcconline.net/greenl/Courses/107/Vectors/vect.h9.gif

Answer 2

These are two points in three dimensions. Each point has three co-ordinates.

Consider the horizontal plane through x. This is the plane where all points have x3 as their third co-ordinate.

Now drop a perpendicular line from y to this plane. Call the point where it meets the plane z. Calculate the co-ordinates of z. Calculate the distance between z and x using pythagoras's theorem. Calculate the distance between z and y. (it's obvious). Calculate the distance between x and y using the previous two distances and pythagoras's theorem.

Answer 3

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Answer 4

Notation X(sub)1 = x_1 etc.

distance = sqrt( (x_1 - y_1)^2 + (x_2-y_2)^2 + (x_3-y_3)^2)