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Vectors and surfaces

  1. Nov 8, 2006 #1
    The question reads:
    Identify the following surfaces given that k, l, m, n are fixed values and [tex]\hat{u}[/tex] is a fixed unit vector.

    a) [tex]|\overrightarrow{r}|=k[/tex]
    b) [tex] \hat{r}\cdot \hat u=l[/tex]
    c) [tex] \overrightarrow{r} \cdot \hat{u} = m|\overrightarrow{r}|[/tex] for [tex] -1 \leq 1[/tex]
    d)[tex]|\overrightarrow{r} - (\overrightarrow{r}\cdot\hat{u})\hat{u}|=n[/tex]

    I am to consider the both the variability in magnitude and direction of [tex]\overrightarrow{r}[/tex]

    i was just wondering if the following are correct.
    for a) it seems pretty obvious that this is a sphere of radius k, and for b) i see that the cosine of the angle of r vector and u hat is a constant so i think this will lead to a cone.
    for c) i am not sure since i get the cosine of the angle between r and uhat is ranging between -1 and 1 so the angle is between 0 and pi, but considering the variability in magnitude of r i am not sure what this defines, and for d) i am not sure what to do

    any help would be appreciated
     
  2. jcsd
  3. Nov 11, 2006 #2
    no one can help with this?
     
  4. Nov 11, 2006 #3

    radou

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    In d), think about what the vector [tex](\vec{r}\cdot\hat{u})\hat{u}[/tex] represents.
     
  5. Nov 11, 2006 #4
    for d) this is just r minus the projection of r in the uhat direction? how should i interpret this? as a line?

    and also i am not sure how to interpret what i have found in part c) if i could get some explanation it would be appreciated...thanks
     
  6. Nov 11, 2006 #5

    radou

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    Regarding c) - you get that the cosine of the angle between r and u equals m. It's obvious that's some kind of cone. I only didn't understand the -1 to 1 part, but nevermind, you should be on the right track by now.
     
  7. Nov 11, 2006 #6
    i got that b) was a cone and it should be m goes from -1 to 1, so that means that the angle can vary from 0 to pi, so i am not sure how to look at this. Like r vector can be any direction from 0 to pi with the u unit vector, but given that r can vary in length it seems to be that it will be "half of all of space" from 0 to pi...
     
    Last edited: Nov 11, 2006
  8. Nov 11, 2006 #7
    if it is just the unit vector going from 0 to pi then this is a half sphere i guess, could this be it?
     
  9. Nov 11, 2006 #8

    radou

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    Decide what is going from 0 to Pi, i.e. what does -1 <= 1 in c) mean?
     
  10. Nov 11, 2006 #9
    i am not sure what is going from 0 to pi, i thought initially it was the r vector, but i dont think its correct. if it is the unit vector this would make more sense
     
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