 ###### 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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# Introduction to the 3D Coordinate System - Problem 2

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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We're talking about the 3D coordinate system. One of the first kinds of shapes you learned about is a sphere. Here is the definition. Given point C in space, and r a number greater than 0, the set of all points that are a distance r from C is called a sphere. R is the radius of the sphere, and C is the center. So here is the problem. Let's find an equation of the sphere centered at the origin with radius 12.

So I've got a picture here. Sphere centered at the origin, and I want my radius to be 12. So this distance here is going to be 12. Now I've also labelled a random point on this sphere, and given it coordinates x, y, and z. So let me make the observation that if the radius is 12, then the magnitude of vector OP is 12.

Now vector OP is a position vector, and it's going to have components the same as the coordinates of its endpoint. So it's going to be x., y, z. That means that, the length of the vector is going to be the square root of the square of each of these components; x² plus y² plus z² that equals 12. So we square both sides. We get an equation, it looks very familiar.

Now if I covered up the z², this would look a lot like the equation of the circle centered at the origin. So the equation of the sphere, just has one more term. X² plus y² plus z² equals 144, and notice the radius is squared here too.

So in general, a sphere centered at the origin, with radius r, has equation x² plus y² plus z² equals r².