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Vector analysis problems (Math and Physics)

dexterdexter Registered User regular
edited June 2011 in Help / Advice Forum
I know there are a whole bunch of math nerds on these forums and the problems I'm having are pretty reasonable to ask over a forum I think.

First off let me make it clear this isn't homework by any means. I'm on holidays between semesters right now and in my second year of uni in a math and physics double major. Going into complex analysis, vector calc, fourier analysis and electrodynamics this coming semester, and just wanted to polish some of my basics (and having a tougher time that I'd hoped).

I'm currently working off the text "Introduction to Electrodynamics" 3rd international edition by David J. Griffiths. I'll put all the details pertaining to this text because I think it's quite a popular one and some of you might be familiar with it.

Problem 1.3 page 7: Find the angle between the body diagonals of a cube.

This is the first time I've heard of "body diagonals" but I finally found a few pictures. I used a unit cube and used as end points the vector A=(1,1,1) and B=(0,1,1) where A begins at (0,0,0) and B at (0,0,1).
This is where I'm confused, to be honest I don't actually know how to approach this problem, and that's what I'm trying to work on this break. Surely I need to find where both these vectors intersect and use that as the origin, but then how does that change the end points?

I used the dot product straight up just to see what I got, even though it makes no sense, and found the angle equal to 35 degrees. Does the dot product take into account the vectors will intersect at some point?


The following isn't a priority since I appreciate you guys spending the time to answer the first one and don't want to get too greedy! Also, I'm not really sure how to explain the problem.

Also, Problem 1.4 page 7: Use the cross product to find the components of the unit vector n perpendicular to the plane shown in Fig. 1.11.

This is why I cited the text, since I'm not sure how to explain the plane. Basically there are the points (1,2,3) and a plane cuts through each of them. I understand how the cross product works, I'm just not sure how to find two vectors which are parallel to this plane so I can make a new perpendicular one.
Again, it's because I can't see how I can have some vectors not coming from the origin.

Thanks a lot for your time dudes, I really appreciate it. I realize my post is a little lengthy. Any tips on approaching problems in physics or math would really be great too since I have a lot of trouble in this area.. Setting things up, that is.

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dexter on

Posts

  • DemerdarDemerdar Registered User regular
    edited June 2011
    If you use three points to define a plane, then you already have the vectors that are parallel to that plane, does that make sense? Once you have two vectors, take their cross product to find the normal vector to the plane (this is another way to define a plane, and it's much more convenient than using 3 points IMO). Vectors don't necessarily have to come from the origin.

    For example, you have two points, A & B. A = (1, 2, 1) and B = (0,1,2). To find the vector pointing FROM A to B you simply subtact B-A = (0,1,2)-(1,2,1) = <-1,-1,1>.

    Does this make sense?

    Demerdar on
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  • dexterdexter Registered User regular
    edited June 2011
    I don't really understand the part using the points of the plane to find a vector. Do I find a linearly independent basis or something? I think I just need to understand how to find two vectors from the plane described by (1,2,3).

    Thanks, Demerdar.

    dexter on
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  • SmasherSmasher Starting to get dizzy Registered User regular
    edited June 2011
    dexter wrote: »
    I don't really understand the part using the points of the plane to find a vector. Do I find a linearly independent basis or something? I think I just need to understand how to find two vectors from the plane described by (1,2,3).

    Thanks, Demerdar.

    The three points are all on the plane, so any vector defined by two of them (ie, any vector that starts at one of the points and ends at another) will also be on the plane. To find such a vector just subtract the coordinates of the starting point from the coordinates of the end point as Demerdar described.

    Also, you probably shouldn't list three points as (1,2,3); it makes it look like you're talking about a single point with those coordinates.

    Smasher on
  • dexterdexter Registered User regular
    edited June 2011
    Thanks for the tip on describing the plane, I'm really not thinking properly!

    I think I know what you mean, I found a starting point at (1,0,0) and end point at (0,2,0), and then

    start - end = (1,-2,0)

    When I draw the vectors out this makes sense, but my final vector doesn't look right... Were those vectors Demerdar used in the plane I'm using? I can't see how they would be?

    I might need to look back at another text book or something where this is all explained step by step because I haven't done it for so long and I'm getting stuck on really simple things.. Thanks for your patience guys.

    dexter on
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  • SmasherSmasher Starting to get dizzy Registered User regular
    edited June 2011
    You need to subtract the start from the end rather than vice versa, so end - start.

    Smasher on
  • DemerdarDemerdar Registered User regular
    edited June 2011
    Okay, lets do this from the origin:

    Say we have three points that define a plane (this is the XY plane, BTW)

    A=(0,0,0), B=(2,1,0), C=(4,3,0)

    We want to make TWO vectors (so we can take a cross product) that are parallel to this plane. Lets make A the origin and form vectors out of points B and C. I will label vectors in lower case and points in uppercase.

    <b> = B-A = (2,1,0) - (0,0,0) = <2, 1, 0>
    <c> = C-A = (4,3,0) - (0,0,0) = <4, 3, 0>

    Now simply take the cross product of these two vectors (it doesn't really matter which way you take it, but use the right hand rule to get some sense of where this resultant vector is going to end up)

    You'll end up finding out that this cross product <b> x <c> will end up creating a vector in the Z-direction (and only the z direction).. because we defined THESE points as the plane (and it happens to be the X-Y plane).

    You can apply this in your problem by not assuming you are coming from the origin, but from some other reference point. The steps are exactly the same. And when in doubt, use the right hand rule to check to see if your cross product makes any sense at all.

    Demerdar on
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  • dexterdexter Registered User regular
    edited June 2011
    Ahh I think I have it now! I found two vectors, each originating from point (1,0,0).

    A=(1,0,0)
    B=(0,2,0)
    u=B-A=(-1,2,0)

    A=(1,0,0)
    D=(0,0,3)
    v=D-A=(-1,0,3)

    So to find a vector normal to the plane in the positive n direction, I use the cross product of u x v, which gives

    u x v=(6,-3,2)n

    This looks pretty good since it's orthogonal to both my u and v! So have I done this right?

    I imagine putting an n out the front of that vector is the same as writing (6x,3y,2z), is that right?

    dexter on
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  • dexterdexter Registered User regular
    edited June 2011
    Sorry Demerdar, I didn't see your post until just now... That's where I was getting stuck. Since these vectors aren't originating from the origin, but some other reference point, where is that reference point? So my u vector used above, where is that originating from?

    Actually, it's just originating from which ever start point we assign it right? So A was the start and B the end for the u vector, this u vector will still originate from the same starting point huh. Man, this has been WAY harder than it should have been haha.

    dexter on
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  • CptHamiltonCptHamilton Registered User regular
    edited June 2011
    Problem 1:

    The dot product doesn't 'know' anything about the vectors. But to use the dot product to find the angle between two vectors, they must share the same basis. Vectors don't remember what their origin is, they only remember their magnitude and their direction. So if you have a line segment A representing the line between points (0,0,0)->(1,1,1), the vector A is not <1,1,1>.

    To find the vector components of A, you need to first figure out the length of the line segment A. Then you need to find the projection of that length on the x, y, and z axis. I'll leave that as an exercise for you and skip to the meaty bits.

    Let's say that our vector A has component notation sqrt(3)<1/sqrt(2), 1/sqrt(2), 1/2>.

    Our vector B is going to be defined by a line from (0,0,1) to (1,1,0). Remember that we don't actually care about where those points are as far as defining our vector; we just need to work with the length of the line and its direction. To make it more obvious we can translate the line to the origin such that it points from (0,0,0) to (1,1,-1) (I just subtracted the position of the 'start' point from both points).

    Doing the same sort of math again, we can say that we have a vector with coordinates
    B = sqrt(3)<1/sqrt(2), 1/sqrt(2), -1/2>

    Now we do our dot product trick and find that A . B = 0. Therefore the vectors are orthoganol, or, the line segments bisect one another (the angle between them is 90 degrees). This should be our expectation since these are lines connecting opposite corners of a cube. They should bisect one another.

    Problem 2:
    This is sort of the same deal. We need to define vectors out of line segments. You have three points: A, B, and C. So make line segments AB and AC. It's easier if you translate the whole mess to the origin, so subtract A from A, B, and C (making A = (0,0,0)) first. Now do the same thing you did above to figure out vector components for the vectors AB and AC. Take the cross product and whalla, you have a vector perpendicular to the plane defined by A, B, and C.

    Getting the coordinates:
    So if you have a line segment in three dimensions that starts at the origin and goes to a point B located at (x,y,z), the length is just sqrt(x^2 + y^2 + z^2). Then you can get a vector along the line segment by:
    v = <d*cos(theta)*sin(phi), d*cos(theta)*cos(phi), d*sin(theta)>
    where theta is the angle between the line and the unit vector k-hat (or z-hat or whatever, the one pointing up off the xy plane) and phi is the angle between the line and the i-hat (or x-hat) unit vector (or the x axis).

    In the first problem the angles are both 45 degrees, so the sin is 1/sqrt(2) and the cos is 1/sqrt(2), making the trig stuff <d/sqrt(2), d/sqrt(2), d/(sqrt(2)*sqrt(2))>. d is the length of the line, which is sqrt(3) for the unit cube.

    Edit: Corrected my wonky second vector

    CptHamilton on
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  • DemerdarDemerdar Registered User regular
    edited June 2011
    dexter wrote: »
    Sorry Demerdar, I didn't see your post until just now... That's where I was getting stuck. Since these vectors aren't originating from the origin, but some other reference point, where is that reference point? So my u vector used above, where is that originating from?

    Actually, it's just originating from which ever start point we assign it right? So A was the start and B the end for the u vector, this u vector will still originate from the same starting point huh. Man, this has been WAY harder than it should have been haha.

    As long as your points A, B, and D are the points that define the plane in your problem, then yeah A was your "reference point". D could have been your reference point and you will still get the same answer.

    A unit vector is a vector that is divided by it's length, such that it's length is 1 (hence a unit vector).

    So, you have a vector

    <2, 5, -1>.. but that is not a unit vector. You can verify that by taking the magnitude of <2, 5, -1>. You'll find that it is greater than 1. To make that vector into a unit vector, you simply take the magnitude of <2, 5, -1> and divide it by the vector so..

    u = <2,5,-1> / magnitude(<2,5,-1>).

    So, you did everything right above, but you need to correct your vector so that it is a unit vector.

    In the example I gave you above, the unit vector will simply end up being

    <0, 0, 1>.

    edit:

    Pretty much a unit vector gives you purely direction and not necessarily magnitude.

    Demerdar on
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  • DemerdarDemerdar Registered User regular
    edited June 2011
    dexter wrote: »
    I imagine putting an n out the front of that vector is the same as writing (6x,3y,2z), is that right?

    No, absolutely not. What you just described with that is a vector field, and is probably a bit beyond the scope of what you are doing right now.

    Demerdar on
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  • dexterdexter Registered User regular
    edited June 2011
    Demerder, Smasher, thanks a lot dudes, I appreciate all the help! I always forget about making my unit vectors! When would I start playing with vector fields? I am doing vector calculus this semester, and I'm doing introductory electrodynamics too, would I use them there?

    CptHamilton, thank you for your post, I'll check it out in the morning!

    dexter on
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  • DemerdarDemerdar Registered User regular
    edited June 2011
    Electrodynamics?

    Probably very soon, once you start talking about magnetic fields (I use them all the time in fluid mechanics). In fact, you'd be surprised by the amount of overlap all of the disciplines have as you climb further up the tree....

    Demerdar on
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  • CptHamiltonCptHamilton Registered User regular
    edited June 2011
    I may have made an error in my components. I'll let you check my math :p

    Griffiths is kind of a bitch of a book, so don't feel bad. It's the standard text for undergrad E&M but it's not fun by any stretch. Also, it sounds like you have sort of a shaky grasp on what vectors are as mathematical entities. Based on your course load for next semester I'm pretty sure you'll clear that up pretty quickly. I think I had almost that exact same semester at one point (I did a physics and math double in undergrand, too). You'll likely get the short form of everything that you'd learn in the vector calc class in the first few weeks of E&M, then be really bored in vector calc.

    CptHamilton on
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  • WickedKWickedK Registered User new member
    Just want to say this forum was a life saver! Been reading your guys for years now. Anyhow, for the life of me I couldn't figure out my physic hw for an exam today because the angular momentum was taken at a point not with respect to the origin and I couldn't remember how to do it. I looked online and found only w.r.t (0,0,0) but who would have thought PennyArcade would have the answer! Thanks!

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