# Vector calculus question

1. Mar 13, 2013

### petertheta

Hi - I'm totally stuck with this question: how to interpret it and tackle it. Any advice woiuld be greatly received!! We've not covered anything like this in classes...

Let
$$A = \left( x_{A}, y_{A}, z_{A} \right)$$
$$B = \left( x_{B}, y_{B}, z_{B} \right)$$
be two given distinct points in the Euclidean space. By finding the cartesian equation, descibe the surface representing the location of points M which are solutions of the equation
$$\vec{AM}.\vec{MB} = 0$$

Thanks, PT

2. Mar 13, 2013

### joeblow

Okay. I'm now confused. What is M? The only thing that would make sense would be if M is a scalar, but then the locus of M would not be a surface, since it lives in the straight line of the real numbers.

Last edited: Mar 13, 2013
3. Mar 13, 2013

### Staff: Mentor

Pretty obviously, M is a point. It would not make sense that M would be a scalar.

4. Mar 13, 2013

### joeblow

If A, B, and M are all position vectors, then AM and BM have no readily apparent definition.

5. Mar 13, 2013

### LCKurtz

You will have more fun with this problem if you simplify your notation. Say $A=(a,b,c)$ and $B=(p,q,r)$. Then let your variable point $M=(x,y,z)$. Write out your dot product and simplify it to see what you get.

6. Mar 13, 2013

### joeblow

Does MA mean the vector connecting M and A? If so, what's wrong with using subtraction so that the problem is clear?

7. Mar 13, 2013

### Staff: Mentor

It's given in the first post that A and B are points in space. One can reasonably infer that M is also a point.

8. Mar 13, 2013

### LCKurtz

The original post indicated they were points. The notation $\vec{AB}$ is pretty common notation for the vector from $A$ to $B$.

9. Mar 13, 2013

### joeblow

That annoys me. To me, that notation defines a ray and not a vector. Apparently, I'm not in the loop on that one.

10. Mar 13, 2013

### petertheta

OK. So what I get when I take the inner product as you say:

(a-x)(x-p) +(b-y)(y-q) +(c-z)(z-r) = 0

Do I expacnd further and rearrange to something??

Thanks

11. Mar 13, 2013

### petertheta

Ah, If I expand further I get a sphere centred at the origin with radius =
(ax-ap+by-bq+cz-cr)^2

12. Mar 13, 2013

### LCKurtz

What do you think? Do you recognize the surface defined by that equation as it sits? Or do you think you should work on it some more?

13. Mar 13, 2013

### LCKurtz

Looks like you have gone off the rails. The radius wouldn't have variables in it. You need to show your work. If you think it might be a sphere, think about how you would tell if a second degree equation in x,y, and z was a sphere.

14. Mar 13, 2013

### petertheta

I could factor into:

a(x-p) +b(y-q) +c(z-r) which looks like an equation of a plane...

15. Mar 13, 2013

### LCKurtz

That isn't the equation of anything because there is no equals sign. The last correct thing you have posted was your equation in post #11:

(a-x)(x-p) +(b-y)(y-q) +(c-z)(z-r) = 0

You need to simplify it into a form you recognize and show your steps doing it. I have to go now. Maybe someone else will pick it up here or I will look again later today.

16. Mar 13, 2013

### petertheta

I'm lost. could use parameterization to check something like the volume or surface area to confirm it was a sphere?!?

17. Mar 13, 2013

### petertheta

the above staement is an equation though isn't it? You are given:
$$\vec{AM}.\vec{MB} = 0$$ in the question.

using your sugested A=(a,b,c) B=(p,q,r) and M=(x,y,z) for the points isn't

$$\vec{AM} = (a-x, b-y, c-z)$$
$$\vec{BM} = (x-p, y-q, z-r)$$

Hence the inner product is:

(a-x)(x-p) +(b-y)(y-q) +(c-z)(z-r) = 0 as given

???

18. Mar 13, 2013

### slider142

There are many volumes that are enclosed by surfaces that have the same volume and surface area as a given sphere, so that would not identify a surface as a sphere.
The usual way that we identify a collection of points as a sphere is that every point is the same distance from some central point.
In 3-dimensional Euclidean space, the distance between two arbitrary points is defined by the Pythagorean theorem applied to the Cartesian coordinates of the points. That is the distance d between the two points with Cartesian coordinates (x, y, z) and (x', y', z') is given by the equation $d^2 = (x - x')^2 + (y - y')^2 + (z - z')^2$.
Treating (x, y, z) as three variables now, if a sphere is the collection of points that are the same distance from a central point (x', y', z'), then every point on the sphere must satisfy the equation above for a constant distance d, which we now call the radius of the sphere. Therefore, if we can rearrange an equation to look exactly like the one above, where x, y, and z are variables and d, x', y' and z' are constants, those points (x, y, z) must lie on a sphere of radius d around the point (x', y', z').
For example, the set of points (x, y, z) that satisfy the equation $x^2 + y^2 - 2y + z^2 + 4z = 2$ lie on the surface of a sphere of radius $\sqrt{7}$ around the point (0, 1, -2). The reason we know this is because we can rearrange this equation algebraically to see that the same points must satisfy the equation $(x - 0)^2 + (y - 1)^2 + (z - (-2))^2 = (\sqrt{7})^2$. We compare this to the standard equation of the sphere above to find out the details.

Last edited: Mar 13, 2013
19. Mar 13, 2013

### LCKurtz

@petertheta: You still haven't shown any work simplifying (a-x)(x-p) +(b-y)(y-q) +(c-z)(z-r) = 0.

You need to look in your calculus or algebra book to remind yourself how you can tell if
$$Ax^2 +By^2+Cz^2 +Dx + Ey +Fz + H = 0$$is a sphere. Then put your equation in that form and decide.

20. Mar 13, 2013

### petertheta

The only way I know how to show something is a sphere is that all points are equal distance from the centre.

i think I need more help?