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Dot/Cross product vector problem.

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data
    2zyd9qh.jpg

    2. Relevant equations

    3. The attempt at a solution
    Well, I have the whole thing drawn out, and calculated some information that will be needed.

    [itex]|A|=7.92[/itex]
    [itex]|B|=8.28[/itex]

    I am assuming that the question means that vector C is perpendicular with vector A, meaning, there is 90 degrees between them.

    a.) Find the x component of C

    It says that the scalar product of C and B is 13, so I am guessing that means I have to know the angle between those two vectors. Which would be the angle between B and A minus 90 degrees right?

    How do I find the angle between B and A ?

    Alright, so:
    [tex]A\bullet B = (4.8)(-3.9)+(-6.3)(7.3)[/tex]
    [tex]A\bullet B = -64.71[/tex]
    and since [itex]A\bullet B = |A||B|cos(\theta)[/itex]
    [tex]|A||B|cos(\theta) = -64.71[/tex]

    [tex]cos(\theta) = \frac{-64.71}{(7.92)(8.28)}[/tex]
    [tex]\theta = 170.67 degrees[/tex]

    Then the angle between B and C is 80.67 degrees, which looks about right.

    So then, the scalar/dot product of C and B is:
    [tex]C \bullet B = |8.28||C|cos(80.67) = 13[/tex]
    [tex]|C| = \frac{13}{cos(80.67)|8.28|}[/tex]
    [tex]|C| = 9.68[/tex]


    Now what?
    I don't see how the components of C are coming out of this.
     
    Last edited: Sep 7, 2011
  2. jcsd
  3. Sep 7, 2011 #2

    cepheid

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    The fact that the scalar product of C and B is 13 also means that CxBx + CyBy = 13 (by the definition of scalar product). Does that help at all?
     
  4. Sep 7, 2011 #3
    I was thinking of that but then I would still be stuck with two unknowns.
    [tex]C_{x}(-3.9)+C_{y}(7.3)=13[/tex]

    but you probably said that for a reason (thanks) so I will look for a second system perhaps.
     
  5. Sep 7, 2011 #4

    cepheid

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    Right, but you also know that C is perpendicular to A, which should give you a second equation involving Cx and Cy. Two equations and two unknowns ==> an exact solution can be found.
     
  6. Sep 7, 2011 #5
    Hmm,

    I found the angle from the positive x-axis to A to be -52.67 degrees (used arctan on vector A's components).

    If the angle between C and A is 90, then the angle between C and the positive x-axis must be 90-|-52.67| = 37.33

    Since I know |C| = 9.68 I could just get the x/y components of vector C right there?

    I get:
    [tex]C_{x} = 7.69[/tex]
    [tex]C_{y} = 5.87[/tex]
    Does that seem right? (Edit: No, it's not right at all, ugh. Looking in to cepheid's solution)
     
    Last edited: Sep 7, 2011
  7. Sep 7, 2011 #6

    cepheid

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    If two vectors are perpendicular, then their dot product is zero. You can see this just by noting that cos(90°) = 0, but it's also helpful to understand the geometric argument. Recall that when you're taking the dot product of A and B, you sort of "project" A onto B by drawing a line starting from the tip of A that is perpendicular to A and extend it until it lands on B. This marks out the component of A that is in the same direction as B. So the dot product can be interpreted as the magnitude of B multiplied by the component of A that is in the direction of B.

    However, if the two vectors are perpendicular, then there is NO component of A that is in the direction of B. When you draw a line from the tip of A perpendicular to A, it never intersects B.

    In any case, the dot product being zero gives you your second equation involving the x and y components.
     
  8. Sep 7, 2011 #7
    Ah that's right. Sorry, I'm kind of teaching myself ahead of the course some).

    My two systems of equations are:

    -3.9x + 7.3y = 13
    4.8x - 6.3y = 0

    x=-1.3739
    y= 1.047

    Which both to two significant digits, is still incorrect?
    x=-1.4
    y= 1.1
     
  9. Sep 7, 2011 #8

    cepheid

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    Yeah, I get different answers. My strategy was to use the lower equation to solve for x in terms of y, and plug that expression for x into the upper equation in order to solve for y. Once you have y, you have x.
     
  10. Sep 7, 2011 #9
    Yes, definitely made a mistake on the system of equations.

    The proper answer was x=7.8 and y=6.0 (to 2 s.f.)

    I sure made that one more complicated than it needed to be. Thanks for the help, I appreciate it.
     
  11. Sep 7, 2011 #10

    cepheid

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    You're welcome! :smile:
     
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