Interpreting and Solving a Vector Calculus Question

[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

 

[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.

 

[Mark44]

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

Okay. I'm now confused. What is M? The only thing that would make sense would be if M is a scalar

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

, but then the locus of M would not be a surface, since it lives in the straight line of the real numbers.

 

[joeblow]

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

 

[LCKurtz]

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

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.

 

[joeblow]

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

 

[Mark44]

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

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.

 

[LCKurtz]

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

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

 

[joeblow]

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

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

 

[petertheta]

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.

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

 

[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

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

 

[LCKurtz]

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

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?

 

[LCKurtz]

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

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.

 

[petertheta]

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?

I could factor into:

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

 

[LCKurtz]

I could factor into:

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

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.

 

[petertheta]

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.

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

 

[petertheta]

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.

I appreciate your help.

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

?

 

[slider142]

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

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.

 

[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.

 

[petertheta]

@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.

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?

 

[Mark44]

You can tell from the equation that LCKurtz showed whether the surface is a sphere, paraboloid, or other conic section.

You need to look in your calculus or algebra book...

 

[petertheta]

You can tell from the equation that LCKurtz showed whether the surface is a sphere, paraboloid, or other conic section.

I see. Just revising conics but how do i relate this to th points i have? theyre algebraic so none of them "disappear" to form one of the possible conics?

 

[Mark44]

What do you get when you expand this?
(a-x)(x-p) +(b-y)(y-q) +(c-z)(z-r) = 0

 

[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.

I see. Just revising conics but how do i relate this to th points i have? theyre algebraic so none of them "disappear" to form one of the possible conics?

Everything in that equation but x,y,z are constants. Until you actually do what I asked above I can't help you further.

 

[petertheta]

What do you get when you expand this?
(a-x)(x-p) +(b-y)(y-q) +(c-z)(z-r) = 0

ax - ap -x^2 +xp + by - bq - y^2+yq + cz-cr-z^2+zr = 0

then rearrange to give:
x^2 + y^2 + z^2 = a(x-p) + b(y-g) + c(z-r)

but I was told I couldn't do this??

 

[Mark44]

ax - ap -x^2 +xp + by - bq - y^2+yq + cz-cr-z^2+zr = 0

then rearrange to give:
x^2 + y^2 + z^2 = a(x-p) + b(y-g) + c(z-r)

This is the equation of a sphere. I can say this because the coefficients of the second degree terms are all equal.

but I was told I couldn't do this??

Who told you that? I don't think anyone said that in this thread.

 

[petertheta]

This is the equation of a sphere. I can say this because the coefficients of the second degree terms are all equal.

Who told you that? I don't think anyone said that in this thread.

post #13 and #15 suggwsted that this was not possible but anyway...

Ok so we have a sphere. Is it this surface then that satisfies the question? No matter what values of A or B. A point M sat on the sphere... will satisfy the orthognality of the two vectors when dotted??

 

[Mark44]

post #13 and #15 suggwsted that this was not possible but anyway...

You're misinterpreting what LCKurtz said.

Ok so we have a sphere. Is it this surface then that satisfies the question?

Yes.

No matter what values of A or B. A point M sat on the sphere... will satisfy the orthognality of the two vectors when dotted??

Yes.

It might help to think about it in two dimensions. Take two distinct points A and B in the plane. The line segment that joins them defines a circle whose center is midway between the two points. The center point can be used to construct a half circle.

Take any point on the half circle, and call it M. Draw segments from A to M and from B to M. Segment AM will always be perpendicular to segment BM.

This is essentially what you've done, except with another dimension thrown it, so you get a sphere instead of a circle.

 

[petertheta]

You're misinterpreting what LCKurtz said.
Yes.

Yes.

It might help to think about it in two dimensions. Take two distinct points A and B in the plane. The line segment that joins them defines a circle whose center is midway between the two points. The center point can be used to construct a half circle.

Take any point on the half circle, and call it M. Draw segments from A to M and from B to M. Segment AM will always be perpendicular to segment BM.

This is essentially what you've done, except with another dimension thrown it, so you get a sphere instead of a circle.

I get it. so I could descibe the surface as a sphere with a centre... and radius... which I deduce algebraically?

 

[LCKurtz]

ax - ap -x^2 +xp + by - bq - y^2+yq + cz-cr-z^2+zr = 0

then rearrange to give:
x^2 + y^2 + z^2 = a(x-p) + b(y-g) + c(z-r)

but I was told I couldn't do this??

Who told you that? I don't think anyone said that in this thread.

post #13 and #15 suggwsted that this was not possible ...

You have got to be kidding!