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Vectors and the geometry of space problems

  1. May 3, 2005 #1
    First of all, I'd like to thank whoever made this website. I sometimes get so stressed with physics & math problems and had no one to ask for help since I'm taking the highest level classes out of all of my friends. I'm so glad I found this website.

    Here are the problems I need help on:

    1) Let r0 = <x0, y0, z0> and r = <x,y,z>. Describe the set of all points (x,y,z) for which

    a) r dot r0 = 0.
    b) (r - r0) dot r0 = 0

    The 0 next to r, x, y, and z should be subscript.

    2) Show that if u and v are vectors in 3-space, then |u x v|^2 = |u|^2 + |v|^2 - (u dot v)^2
    (hint: use the pythagorean identity)
    cos^2 theta + sin^2 theta = 1

    3) Show that if u and v are unit vectors and theta is the angle between them, then |u - v| = 2 sin (1/2 theta)

    Thanks in advance!!! :smile:
  2. jcsd
  3. May 3, 2005 #2
    well, let's look at the first one first. What does it mean (geometrically) to say that [itex]x \cdot y = 0[/itex] (where [itex]x, y[/itex] are vectors in [itex]\mathbb{R}^n[/itex] and [itex]\cdot[/itex] is the dot product)?

    Also, I think you need to check the second one over - it's not true the way you wrote it (assuming you mean that u x v is the cross product).
    Last edited: May 3, 2005
  4. May 3, 2005 #3
    wouldn't that mean they are orthogonal or perpendicular to each other?
  5. May 3, 2005 #4
    i'm not sure how i'm supposed to "describe" the set of all points (x,y,z)
  6. May 3, 2005 #5
    right, it means that they're orthogonal. Now, how can I get vectors orthogonal to some vector in 3-space? HINT: Think cross product!
  7. May 3, 2005 #6
    hmmm my book says "the vector a x b is orthogonal to both a and b"
    i'm kinda lost.. can u teach me how to do it?
  8. May 3, 2005 #7
    okay... indeed, a x b is orthogonal to both a and b. So if you look at u = (1, 0, 0) and v = (0, 1, 0) then u x v is some scalar multiple of (0, 0, 1). If you look at the set

    [tex]\{ u \times v | v \in \mathbb{R}^3 \},[/tex]

    then this just be the set of all vectors perpendicular to u, which is precisely what you're looking for for 1a (the set of all vectors perpendicular to [itex]r_0[/itex]).

    Now look at 1b. Try to work on it on your own (it's very similar). I'm afraid that I'm going to have to go to sleep now, but I'll be back tomorrow to help more. And as I said, please check that you transcribed question 2 correctly, because it doesn't work in general the way it is (for example for u = (1, 0, 0), v=(0, 0, 1) it doesn't hold) :smile:
    Last edited: May 3, 2005
  9. May 3, 2005 #8
    so the answer for 1a is "the set of all vectors perpendicular to r0"? or do i need to specify each vector? thanks for the help data.. i really appreciate it.. as for question 2, i've transcribed it correctly. My buddy and I spent hours trying to solve that question and gave up on it. I'm afraid that the professor has transcribed it wrong for the assignment.
    Last edited: May 3, 2005
  10. May 3, 2005 #9
    there are infinitely many such vectors (there's a whole plane of vectors perpendicular to u for any vector u in [itex]\mathbb{R}^3[/itex]), so you had better not be supposed to specify each one! There are certainly alternative methods of describing the appropriate set though. Using an expression analogous to the one in my last post should be sufficient.
  11. May 3, 2005 #10
    and are you sure for 2 that it's not

    [tex]|u \times v|^2 = |u|^2|v|^2 - (u\cdot v)^2[/tex]

    ? (I know, I'm too persistent!)
  12. May 3, 2005 #11
    i was thinking the same.. i tried to solve that problem over and over and that + sign between |u| and |v| caused me not to solve the problem.. i think my professor might have made a mistake.. or he knows a method that nobody knows.. lol how would u solve 1b and 3?

    for 1b, i'm guessing (r - r0) is perpendicular to r0?
    and for 3, do i have to use the half angle identity somehow?
    Last edited: May 3, 2005
  13. May 3, 2005 #12
    For 1b, indeed, [itex]r-r_0[/itex] must be orthogonal to r. So how can you specify the appropriate set (it will be similar to the one for part a)?

    Number 2 is simply false the way it is, so I'd ignore it until you can ask your professor about it.

    For 3, indeed, you will need the identity

    [tex]2\sin^2 (\theta / 2) = 1 - \cos \theta[/tex]

    and the definition of magnitude,

    [tex]\|x\| = \sqrt{ x \cdot x }[/tex]
  14. May 3, 2005 #13
    For 1b, is the answer "the vector <x-x0, y-y0, z-z0> is perpendicular to r0"?

    as for 3, i get stuck at |u - v|^2 = 2 - 2 cos theta after playing with the right side of the equation using the half angle identity.. i don't even know how to use the definition of magnitude to solve this problem.. do you think you can post how to get the answer? i sometimes find it easier to work backwards.. (from the answer to the first step) i appreciate the help so far..
  15. May 3, 2005 #14
    For 1b, remember that the set [itex]\{u \times v | v \in \mathbb{R}^3 \}[/itex] is the set of all vectors perpendicular to [itex]u[/itex]. In this question, [itex]r - r_0[/itex] can be any vector perpendicular to [itex]r_0[/itex]. So let's say I take a vector [itex]u[/itex] from the set [itex]S= \{ r_0 \times v | v \in \mathbb{R}^3 \}[/itex] and add [itex]r_0[/itex] to it, ie. take [itex]r = u + r_0[/itex]. Will [itex](r - r_0) \cdot r_0 = 0[/itex] be true? Say I consider all vectors of this form, ie. the set [itex]\{ u + r_0 | u \in S \}[/itex]. Can you prove that this is the set you're looking for to answer the question?

    Steps for question 3:

    i) Use the definition of magnitude that I posted above to get

    [tex]\|u-v\|^2 = \sqrt{(u-v) \cdot (u-v)} \Longrightarrow \|u-v\|^2 = (u-v)\cdot (u-v) = u \cdot u - 2 u \cdot v + v \cdot v= \|u\|^2 -2 u \cdot v + \|v\|^2[/tex]

    ii) Note that u, v are unit vectors so [itex]\|u\| = \|v\| = 1[/itex] and you get

    [tex]\|u-v\|^2 = 2 - 2 u \cdot v[/tex]

    iii) Recall that [itex] u \cdot v = \|u\|\|v\| \cos \theta[/itex] where theta is the angle between the vectors to get

    [tex]\|u-v\|^2 = 2 - 2 \|u\|\|v\| \cos \theta = 2 - 2\cos \theta[/tex]

    iv) Use the identity in my last post, [itex]2 \sin^2 (\theta / 2) = 1 - \cos \theta[/itex] to get

    [tex]\| u - v \|^2 = 2 - 2\cos \theta = 2(1 - \cos \theta) = 4 \sin^2 (\theta / 2)[/tex]

    v) Take the square root of both sides and reject the negative root since magnitudes are always positive, to get the result needed.
    Last edited: May 3, 2005
  16. May 3, 2005 #15
    hey data, i was able to understand the questions with ur explanation.. thanks a lot.. as for question 2, my professor announced that there was a typo (the + sign).. i'm looking forward to learn more from u..
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