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Homework Help: I feel so stupid

  1. Aug 25, 2008 #1
    I feel so stupid!!!

    OK, I know this is probably really annoying but I have a ton of questions. (I would go to the math tutor but they aren't here yet because the semester just started.)

    1. Find the angle POQ by vector methods if P=<1,1,0>, O=<0,0,0>, and Q=<1,2,-2>.

    What I have tried to use is cos(theta)=a.b/ab, but since there are three vectors I'm having trouble figuring out how to apply the theorem. The thing that confuses me most is O=<0,0,0>, it's not really a vector but somehow I'm supposed to use it to calculate an angle. The professor told us the answer is 45degrees, I got 20degrees.

    2. Describe all points (x,y) such that v=xi+yj satisfies:
    (a) |v|= 2
    (b) |v-i|= 2
    (c) v.i = 2
    (d) v.i=|v|

    This one really confuses me. I've tried just plugging the given v's into the equation, but the answers the professor gave us seem so far away from that I'm pretty certain that that's not the way to go about doing it. What'd be helpful for me would be the worked out solution to one and then I'll apply it to the other three.

    3. Find the angle between the diagonal of cube and (a) an edge (b) the diagonal of the face and (c) another diagonal of the cube.

    I think I somewhat understand how to get (a) and (b), but (c) makes absolutely no sense to me. I've drawn out a cube, and I've used unit vectors to find the angles of the first two, but I can't figure out what I'm supposed to be solving for for (c).

    4. Suppose I=(i+j)/sqrt(2) and J=(i-j)/sqrt(2). Check I.J=0 and write A=2i+3j as a combination aI+bJ. (a=A.I and b=A.J, solved in another problem.)

    I don't even know what this one is asking for. I have an answer key for this one, but it doesn't explain what it's doing and I can't see what it is. Here is the answer, I'd love it if someone would show me what it means.

    A = 2 i + 3 j = sqrt(2)(I+J)+(3sqrt(2)/2)(I-J)=aI+bJ with a=sqrt(2)+(3sqrt(2))/2 and b=sqrt(2)-(3sqrt(2))/2

    5. If |A+B|sq = |A|sq + |B|sq, prove that A is perpendicular to B.

    I assume this means I have to find a way to use the dot product to show that A.B=0, but I don't know where to begin. Would I use cos(theta)=(A.B)/(|A||B|)?

    I know this is a lot, it's one night's homework and I'm not terribly clever at math. The text I'm using is Strang which has about zero examples, and I will appreciate any and all advice (or answers!!).
    Thanks so much!!!!!!
    Last edited: Aug 25, 2008
  2. jcsd
  3. Aug 25, 2008 #2
    Re: I feel so stupid!!!

    Can someone move this back to calculus? I don't think it's going to get many replies here.
  4. Aug 25, 2008 #3


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    Homework Helper

    Re: I feel so stupid!!!

    1. Note that P and Q here really means OP and OQ, the position vector of P and Q from the origin. So you have two vectors here and not three. Just apply the formula you quoted to get the answer.

    2. You're asked to find the locus of points corresponding to the given conditions. Your approach is ok, plugging in xi+yj into the conditions will given you an equation of a graph. I suppose "describe" means you have to recognise what kind of graph you obtain.

    3. Well, you do know what is a cube diagonal is, right? The only question is from which vertex the other cube diagonal should originate from.

    4. You're given I and J. The first part is straightforward. The second one, you need to find a,b such that aI+bJ = A. Write this out explicitly:
    [tex]a\left( \begin{array}{cc}1\\1 \end{array} \right) + b\left( \begin{array}{cc}1\\-1 \end{array} \right) = \left( \begin{array}{cc}2\\3 \end{array} \right)[/tex]. Solve this for values of a,b.

    5. Note that [tex]|A|^2 = \mathbf{A} \cdot \mathbf{A}[/tex]. Make use of this to prove it. Note what happens to the dot product of two perpendicular vectors.
  5. Aug 25, 2008 #4
    Re: I feel so stupid!!!

    Thank you! A couple of things: I finished 1 and 2c correctly, but I don't know what a cube diagonal is really, and number four still makes no sense.

    2. (a) |v|=2=xi+yj. that means 2=<x,0>+<y,0>. (correct?) I don't understand how to get from there to 4=x^2+y^2 like my answer key tells me.
  6. Aug 25, 2008 #5
    Re: I feel so stupid!!!

    If I remember correctly, the 2 is actually the magnitude which means 2= square root(x^2 + y ^2) and I'm guessing they just squared the answer to make it look nicer
  7. Aug 25, 2008 #6
    Re: I feel so stupid!!!

    ohhh, that makes perfect sense!
  8. Aug 26, 2008 #7
    Re: I feel so stupid!!!

    I have gotten all but the last one!!! The thing is, I REALLY don't get the last one. I mean, I have that general idea but I have no clue as to where to start. Help!!
  9. Aug 26, 2008 #8


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    Homework Helper

    Re: I feel so stupid!!!

    Well you should start on the LHS and use this property:
    [tex]|A|^2 = \mathbf{A} \cdot \mathbf{A}[/tex]

    You also need to use the distributive law for dot product of vectors.
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