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Find the vector algebraically

  1. Mar 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Find a unit vector in the direction of the given vector w=-i-2j

    2. Relevant equations

    3. The attempt at a solution


    vertical -2j_________________

    Could someone please show me how I would do it grafically and algebraically?

    Do I just need to find the magnitude? Wouldn't any vector that has a negative x component and a negative y component work?

    square root (-1)^2+(-2)^2=square root 5

    Thank you very much
  2. jcsd
  3. Mar 2, 2008 #2

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    Any vector A divided by magnitude of A is a unit vector.
  4. Mar 2, 2008 #3
    I haven't studied this in quite a while, but have you tried representing it with an Argand diagram?

    http://scholar.hw.ac.uk/site/maths/topic11.asp?outline= [Broken]
    Last edited by a moderator: May 3, 2017
  5. Mar 2, 2008 #4
    Thank you very much

    Isn't the magnitude, in this case, the square root of 5?

    So, would the unit vector be 1/square root of 5i? Aren't i and j the unit vectors? I'm already given the unit vectors, right? Don't I just need to find another one in the same direction?:confused:

    Thank you
  6. Mar 2, 2008 #5
    Well I had a quick check of old material, but the modulus of an Argand diagram or distance from 0 is the magnitude.

    if z=x+yi then [itex]|z|=\sqrt{x^2+y^2}.[/itex]


    And then.

    Like I say it's been a while but hope that helps.
    Last edited: Mar 2, 2008
  7. Mar 2, 2008 #6


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    Some straight answers …

    No-one gives you a straight answer, do they? :smile:

    Yes, the magnitude is √5. (btw, type alt-v and it gives you √)

    No, you mean the unit vector is (-i - 2j)/√5.

    Yes, i and j are always unit vectors in this sort of question.

    Yes, you're right. Why so puzzled?

    No - I don't see why you'd think that. :confused:

    Algebraically, you've understood it fine!

    Graphically: -i -2j is on the circle of radius √5; join it to the origin by a line; then you want the point where that line cuts the circle of radius 1. :smile:
  8. Mar 2, 2008 #7
    Thank you very much everyone

  9. Mar 2, 2008 #8

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    i and j are unit vectors along the positive x-axis and y-axis respectively, but you can find a unit vector in any direction. In this case, the given direction was along the vector w=-i-2j. So, you have to divide w by mod(w) to get the unit vector in the direction of w.

    You got that right. :biggrin: Everybody wants to make sure the OP does learn a bit at least by doing something himself/herself.
  10. Mar 3, 2008 #9
    Including me, I just learnt how to calculate the magnitude of a vector using an Argand diagram. Again. :smile:
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