 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# The Midpoint and Distance Formulas in 3D - Problem 3

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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The distance formula for vectors can be used to derive equations. Let's take a look at an example. It says find an equation that must be satisfied by the points equidistant from O, the origin, and point A; 3, 2, 6.

So I've drawn a little picture here. Here is the origin, here is point 3, 2, 6, and here is the midpoint between them. It's probably the first one you think about that's equidistant from the two points. In space, or even in two dimensions, there are lots of other points that are equidistant from A, and O.

So how do we find all of them? Well I'm going to start by calling this vector AP, and this vector OP. I'm going to start with the condition that defines the set. That vector OP's length has to equal vector AP's length.

Now in order to analyze that a little bit further, I need to come up with components for these two vectors. So first OP, note that that's a position vector, because O is the origin, so vector OP is going to be x, y, z. What about vector AP? AP will be x minus 3, y minus 2, and z minus 6. So now I have to take these two vectors, and set their lengths equal to each other.

So the length of OP is the square root of x² plus y² plus z². The length of vector AP, x minus 3², plus y minus 2², plus z minus 6². Now to make any sense out of this equation, I'm going to have to square both sides. Let me do that up here. I have x², plus y², plus z² on the left and on the right x minus 3², plus 1 minus 2² plus z minus 6².

Now let me expand each of these binomials, because there's going to be a lot of cancellation here. These guys are all going to cancel. X², plus y² plus z² equals x² minus 6x plus 9. Y² minus 4x plus 4, 4y plus 4. Z² minus 12z plus 36. And now the fun part, the cancellation. So you can see that x² will cancel, y² will cancel and z² cancels. Now what are we left with? Well we're left with 0 in the left, and on the right we've got -6x. Let me pull all the variable coefficients together, variable terms; minus 4y minus 12z. Then I have 9 plus 4 plus 36.

Now just to neaten things up, I'll pull all the variable terms to the left. I have 6x plus 4y plus 12z, and then 40 plus 9, 49. That's a rectangular equation for the set of points that are equidistant from point O, and point A. It turns out to be a plane. This is the equation of a plane in space.

So the set of all points equidistant from these two points is a plane. In fact in space, the set of all points equidistant from any two points, is a plane that perpendicularly bisects the segment connecting the two points.