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Vector Problem

  1. Dec 27, 2013 #1
    I am revising vectors at the moment. So, this is not a homework question.

    In 2-D cartesian space, I have two vectors, A = <a,2> and B = <1,3>. I want to find the limit of the angle between the vectors as a -> -∞.

    Geometrically, I know that direction of vector B approaches 180° as a→-∞. Thus, the angle between the vectors apporaches 180° - atan(3) = 108.4°.

    I now want to solve this using more algebraic means. The cosine of the angle θ, between the two vecors is given by,
    cosθ = [itex]\frac{\textbf{A . B}}{|\textbf{A}| |\textbf{B}|}[/itex] .
    This gives,
    cosθ = [itex]\frac{a + 6}{\sqrt{a^{2}+4} \sqrt{10}}[/itex]

    In this form, as a→-∞, cosθ should be negative, which makes sense.
    To algebraically get the limit of cosθ, I divide numerator and denominator by a, to get,
    cosθ = [itex]\frac{1+\frac{6}{a}}{\sqrt{1+\frac{4}{a^{2}}} \sqrt{10}}[/itex]
    When you take limits as a→-∞, you get,
    [itex]\frac{1+0^{-}}{\sqrt{1+0^{+}} \sqrt{10}}[/itex]
    Since the denominator terms were magnitudes of vectors, they have to be positive. This gives cosθ as positive. But this isn't true. In fact it is the negative of that limit.

    Can someone help me out with this. I have a feeling that I'm missing something blatantly obvious and would appreciate it you could point it out, if that's the case!
     
  2. jcsd
  3. Dec 27, 2013 #2

    Student100

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    Pull out your negative when you pull out the constants. That's your mistake I reckon. a/|a| is a negative one.
     
  4. Dec 27, 2013 #3

    Mark44

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    Mod note: Moved from the technical math sections.
     
  5. Jan 1, 2014 #4
    Hi Student100. Thanks for replying. If I try pulling out the constant a, the denominator technically contains a [itex]\sqrt{a^{2}}[/itex] term, which boils down to +/-a. This is ambiguous in the sign at this step itself. However, if I try the approach of dividing the numerator and denominator by 'a', I end up with a positive value for cosθ. I'm still confused.
     
  6. Jan 1, 2014 #5
    But ##\sqrt{a^2} = |a|##. So if you pull out this, then you get

    [tex]\cos(\theta) = \frac{a}{|a|}\frac{1 + \frac{6}{a}}{\sqrt{1 + \frac{4}{a^2}}\sqrt{10}}[/tex]

    So you need to decide what ##\frac{a}{|a|}##. This will be ##1## or ##-1## depending on whether ##a## is positive or negative. But since you want the limit ##a\rightarrow -\infty##, then you can assume ##a<0##, thus ##\frac{a}{|a|} = -1##, thus

    [tex]\cos(\theta) = -1 \frac{1 + \frac{6}{a}}{\sqrt{1 + \frac{4}{a^2}}\sqrt{10}}[/tex]

    Does that clear things up?
     
  7. Jan 1, 2014 #6

    vela

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    To be a bit more explicit, there is no ambiguity in sign here. ##\sqrt{x}## is defined to be the principal square root of ##x##, which is the positive square root.

    When ##a<0##, you have ##\sqrt{a^2} = -a##. When you divided the denominator by ##a##, it's initially outside of the radical. You have to put in the minus sign when you pull the ##a## into the radical because ##a## is negative, i.e. ##\frac1a = -(-\frac1a) = -\sqrt{\frac1{a^2}}##.
     
  8. Jan 1, 2014 #7

    PeroK

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    When a is -ve, you have:

    [itex]a = - √a^2[/itex]

    Whereas, you assumed that [itex]a = √a^2[/itex]

    That's where the minus sign went awol.
     
  9. Jan 1, 2014 #8

    Student100

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    I'm confused why you're annotating 0 with negative and poistive signs to be perfectly honest in your original question. When you take the limit the terms are zero, there is no sign.

    When ever you see |a| you should recognize immediately that you need to break it up into a piecewise function. |a|is a if a >0, and |a| is -a if a < 0. So when you pull out a/|a| you should notice this actually means a/-a = -1.

    When you actually take the limit there is no sign ambiguity, the terms that go to zero do just that, they go to zero.
     
    Last edited: Jan 1, 2014
  10. Jan 1, 2014 #9

    Mark44

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    No, a is a regardless of its sign, but |a| = a if a > 0, and |a| = -a if a < 0. Possibly that's what you meant but forgot to include the absolute value signs.
     
  11. Jan 1, 2014 #10

    Student100

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    Yep, thanks Mark.
     
  12. Jan 2, 2014 #11
    Thanks, Student100, Mark44, R136a1, vela and PeroK for the replies. From what I collectively gather here, the √ sign usually referes to the principal square root which is positive. Now, I wonder if this is a more general rule or perhaps because the denominator terms were actually lengths of vectors which are always positive, meaning that there is always a 'hidden' absolute value operator surrounding these terms and hence the |a| when you attempt to take 'a' outside the radical [itex]\sqrt{a^{2}+4}[/itex].

    I ask the above question because there are times when you'd like to consider the negative square root along with the positive root. For example, when finding the intersection of the parabola y = [itex]x^{2}[/itex] with the horizontal line y = 4. To solve it, you need to substitute for 'y' to get, 4 = [itex]x^{2}[/itex]. To get the value for 'x', you would use both +2 and -2, because indeed there are two intersection points here. The points would be (2,4) and (-2,4).

    Any input is much appreciated.

    The + and - superscripts are just a technicality in the theory of limits because [itex]\frac{6}{a}[/itex] never really attains the value 0 for any a[itex]\in[/itex][itex]\Re[/itex]. This limit of 0 is only approached from the left side (negative side) when 'a' is negative and a→-∞. You are right in the fact that once you put zero in the equation, you take it as zero itself and disregard the presence of the supersrcipts, in order to compute the overall limit.
     
  13. Jan 2, 2014 #12

    PeroK

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    You just have to be careful with the sign of square roots. Sticking to Real numbers:

    √a is defined for a ≥ 0 and is the positive square root.

    So, you have to be careful how you move things in and out of the square root.

    If a is negative, then -a = √a^2, so if a is negative, then:

    a√b = -√ba^2

    And, the same taking a out. If a is negative:

    √ba^2 = -a√b

    You just have to be careful with the sign.
     
  14. Jan 2, 2014 #13

    Student100

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    Anytime you pull out a |a| you need to be wary of the sign.

    Your ideas about the limit are also wrong. The limit isn't really -0.0000000000000000000000001 or 0.00000000000000000000000000000000000000000000000000001, it's simply 0. You only care about what the value approaches, it doesn't matter that it never actually attains that value. So the superscripts seem unneeded to me.
     
  15. Jan 2, 2014 #14

    Mark44

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    Yes, the expression √b ordinarily means the principal square root.
    No, there's no "hidden" absolute value signs. You just have to consider that a could be negative or positive (or zero). So ##\sqrt{a^2 + 4} = |a|\sqrt{1 + 4/a^2}##.
    Sure. When you solve the equation x2 = 4, there are two solutions.
     
  16. Jan 2, 2014 #15

    vela

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    When you take the (principal) square root of both sides, you get ##\sqrt{x^2} = \sqrt{4}## or equivalently ##\lvert x \rvert = 2##. This equation has two solutions, x=2 and x=-2.

    When you want to refer to both square roots, you typically use the ##\pm## symbol to denote both the positive and negative square roots and take the radical symbol to refer to only the positive square root. The quadratic equation is a common example of where this is done.
     
  17. Jan 3, 2014 #16
    Thanks all. I have gained a better understanding of the problem here and how its components fit into my overall picture of physics and mathematics.
     
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